Asymptotic profile of solutions for semilinear wave equations with structural damping

Abstract

This paper is concerned with the initial value problem for semilinear wave equation with structural damping $u_{tt}+(-\Delta )^{\sigma }u_t -\Delta u =f(u)$, where $\sigma \in (0,\frac{1}{2})$ and $f(u) \sim |u|^p$ or $u |u|^{p-1}$ with $p> 1 + {2}/(n - 2 \sigma )$. We first show the global existence for initial data small in some weighted Sobolev spaces on ${{\mathbb {R}}}^n$ ($n \ge 2$). Next, we show that the asymptotic profile of the solution above is given by a constant multiple of the fundamental solution of the corresponding parabolic equation, provided the initial data belong to weighted $L^1$ spaces.

1 Introduction

In this paper, we consider the unique global existence of solutions and diffusion phenomina for the Cauchy problem of the semilinear wave equation with structural damping (damping term depends on the frequency) for $\sigma \in (0,\frac{1}{2})$:

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{tt} -\Delta u +(-\Delta )^{\sigma }u_t = f(u), &{} t \ge 0,\; x\in {\mathbb {R}}^n, \\ u(0,x)=u_0(x), \quad u_t(0,x)=u_1(x), &{} x\in {\mathbb {R}}^n, \end{array}\right. } \end{aligned}$$

(1.1)

where $f \in C^{[\bar{s}],1 }({{\mathbb {R}}})$ ($1 \le \bar{s},\; [\bar{s}] < p$) satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}|\dfrac{d^j}{du^j} f(u) | \le C |u|^{p - j} \quad (0 \le j \le [\bar{s}]), \\ &{}|\dfrac{d^j}{du^j} (f(u) - f(v)) | \le C |u - v|(|u| + |v|)^{p - [\bar{s}] } \quad ( j = [\bar{s}] ), \end{array}\right. } \end{aligned}$$

(1.2)

for a positive constant C. Here, $[\bar{s}]$ denotes the integer part of $\bar{s}$.

For linear wave equations with structural damping:

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{tt}+(-\Delta )^{\sigma }u_t -\Delta u = 0, &{} t \ge 0,\; x\in {\mathbb {R}}^n, \\ u(0,x)=u_0(x), \quad u_t(0,x)=u_1(x), &{} x\in {\mathbb {R}}^n, \end{array}\right. } \end{aligned}$$

(1.3)

with $\sigma \in (0,\frac{1}{2})$, Narazaki and Reissig [20] gave some $L^p-L^q$ ($1 \le p \le q \le \infty$) estimates of the solutions. D’Abbicco and Ebert [2] showed the diffusion phenomena, by giving the $L^p-L^q$ decay estimates of the difference between the low frequency part of the solution of (1.3) and that of the corresponding parabolic equation

$$\begin{aligned} v_t + (-\Delta )^{1 -\sigma } v = 0, \quad t \ge 0,\; x\in {\mathbb {R}}^n, \end{aligned}$$

(1.4)

with initial data $(-\Delta )^{\sigma }u_0 + u_1$. Ikehata and Takeda [11] showed that a constant multiple of the fundamental solution of the parabolic equation (1.4) gives the asymptotic profile of the solutions of (1.3) with $(u_0, u_1) \in ( L^1 \cap H^1) \times (L^1 \cap L^2)$ (see Remark 3).

For semilinear structural damped wave equation (1.1) with $\sigma \in (0, \frac{1}{2})$, D’Abbicco and Reissig [5] first showed global existence and decay estimates of the solution of (1.1) with small initial data for space dimension $1 \le n \le 4$ and $p \in [2,n/[n-2]_+]$ such that

$$\begin{aligned} p > p_{\sigma }:= 1 + \frac{2}{n - 2\sigma }. \end{aligned}$$

(1.5)

They showed the results by using $(L^1 \cap L^2)-L^2$ estimates of solutions of the linear wave equation with structural damping (1.3). In [5], they considered also for $\sigma \in [\frac{1}{2},1]$ and showed that $p_\sigma$ is critical in a particular case $u_{tt}+ 2(-\Delta )^{\sigma }u_t -\Delta u = 0$. Using the $L^p-L^q$ decay estimate ($1 \le p \le q \le \infty$) of solutions of the linear wave equations with structural damping (1.3) by [2] for low frequency part, D’Abbicco and Ebert [4] (see also [3]) showed the unique existence of solutions of (1.1) for small initial data in some Sobolev spaces and gave the decay estimates of the solutions, in the following two cases:

$$\begin{aligned} p_\sigma< p, \quad n< 1 + 2 \max \left\{ m \in {{\mathbb {N}}}; m < \frac{1+ 2 \sigma }{1 - 2\sigma } \right\} , \quad \end{aligned}$$

(1.6)

or

$$\begin{aligned} p_\sigma< p< 1 + \frac{2(1 + 2\sigma )}{[n - 2(1 +2\sigma )]_+}, \quad \left[ \frac{n}{2} \right] \left[ \frac{2}{p} - 1 \right] _+(1 - 2 \sigma ) < 1 + 2\sigma .\nonumber \\ \end{aligned}$$

(1.7)

In [4], they also treated the case where $-\Delta u$ is replaced by $(- \Delta )^\delta u$ with $\delta > 0$.

The assumption (1.6) and (1.7) for $p < 2$ restrict the space dimension from above. The first purpose of this paper is to remove restriction of the space dimension n from above for every $\sigma \in (0, \frac{1}{2})$.

The second purpose is to give the asymptotic profile of the solutions of (1.1) as $t \rightarrow \infty$, if small initial data belongs to some weighted $L^1$ spaces. We show that a constant multiple of the fundamental solution of the parabolic equation (1.4) gives the asymptotic profile of (1.1) (Theorem 3). As as far as the author knows, there seems to be no results on the asymptotic profile for semilinear wave equation with structural damping (1.1) for $\sigma \in (0,\frac{1}{2})$.

In the case $\sigma = 0$, the asymptotic profile for semilinear damped wave equation is investigated. Since we treat nonlinear term not necessarily absorbing, we only refer to the results for non-absorbing type nonlinear term. Then if $1 < p \le p_0$ where

$$\begin{aligned} p_0 := 1 + \frac{2}{n} \ :\text {Fujita Exponent}, \end{aligned}$$

then the solution of the semilinear damped wave equation blows up when $f(u) = |u|^p$ and the integrals of initial data on ${{\mathbb {R}}}^n$ are positive (see [10, 15, 23, 24] ). On the other hand, in the case $p > p_0$, small data global existence is widely studied, (see [7,8,9, 12, 16,17,18,19, 21, 23], for example, and the references therein). The asymptotic profiles of the solutions are obtained as follows. Galley and Raugel [6] ($n = 1$), Hosono and Ogawa [8] ($n = 2$), showed that the asymptotic profile of the solutions is given by a constant multiple of the heat kernel G(t, x), provided the initial data belong to some Sobolev spaces. (See also Kawakami and Takeda [14] for higher order asymptotic expansion in the case $n \le 3$.) For general space dimensions, Hayashi, Kaikina and Naumkin [7] proved the unique existence of global solution $u \in C([0,\infty );H^{\bar{s}} \cap H^{0,\delta })$ for small initial data belonging to some weighted $L^1$ spaces, and showed that a constant multiple of the heat kernel gives the asymptotic profile of the solutions (see Remark 9).

We consider the equation in weighted Sobolev spaces as in [7]. The high frequency part of the structural damped wave equation has a good regularizing property. However, unlike the damped wave equation ($\sigma = 0$), the Fourier transform of the kernel of the linear structural damped wave equation is singular at the origin. This fact causes the difficulty when we treat the equation in weighted Sobolev spaces. To get around this difficulty, we estimate the low frequency part in a new way employing Lorentz spaces (Lemma 1). For the estimate of nonlinear term, we use the method of [7, 9].

This paper is organized as follows.

• In Sect. 2, we list some notations and state main results.

• In Sect. 3, we list known preliminary lemmas.

• In Sect. 4, we estimate kernels.

• In Sect. 5, we prove Theorem 1.

• In Sect. 6, we estimate a nonlinear term.

• In Sect. 7, we estimate a convolution term.

• In Sect. 8, we prove Proposition 1 and Theorems 2 and 3. That is, we prove the global existence of the solution of semilinear wave equation with structural damping, and give the asymptotic profile of the solutions.

2 Main results

Before stating our results, we list some notations.

Notation 1

We write $\varphi (x) \lesssim \psi (x)$ on I if there exists a positive constant C such that

$$\begin{aligned}&\ \varphi (x) \le C\psi (x) \; \text {for every} \; x \in I. \end{aligned}$$

We write $\varphi (x) \sim \psi (x)$ on I, if $\varphi (x) \lesssim \psi (x)$ and $\psi (x) \lesssim \varphi (x)$ on I.

Notation 2

For $a \in {{\mathbb {R}}}$, $[a]_+: = \max \{a, 0\}$.

Notation 3

For every $q \in [1,\infty ]$, we abbreviate ${{\mathbb {R}}}^n$ in $L^q({{\mathbb {R}}}^n)$, and $L^q$ norm is denoted by $\Vert \cdot \Vert _q$.

Notation 4

Let $H^{s,\delta } = H^{s,\delta }({{\mathbb {R}}}^n)$ denote the weighted Sobolev space equipped with the norm

$$\begin{aligned} \Vert u \Vert _{{H^{s,\delta }}} = \Vert \langle x \rangle ^\delta ( 1 - \Delta )^{s/2} u\Vert _{{L^2}}. \end{aligned}$$

$H^{s,0}$ equals $H^s$. Let $\dot{H}^{s} = \dot{H}^{s}({{\mathbb {R}}}^n)$ denote the homogeneous Sobolev space equipped with the norm

$$\begin{aligned} \Vert u \Vert _{{\dot{H}^{s}}} = \Vert ( - \Delta )^{s/2} u\Vert _{{L^2}}. \end{aligned}$$

Notation 5

(see [1, section 1.3], for example) Let $q \in (1,\infty )$ and $r \in [1,\infty ]$. Let $\mu$ be the Lebesgue measure on ${{\mathbb {R}}}^n$. The distribution function $m(\tau , \varphi )$ is defined by

$$\begin{aligned} m(\tau , \varphi ) := \mu (\{x; |\varphi (x)| > \tau \}). \end{aligned}$$

The Lorentz space $L_{q,r}$ consists of all locally integrable function $\varphi$ on ${{\mathbb {R}}}^n$ such that

$$\begin{aligned}&\Vert \varphi \Vert _{{q,r}} := \left( \int \nolimits _0^\infty (t^{1/q} \varphi ^*(t))^r \frac{dt}{t} \right) ^{1/r}< \infty \quad \text {when} \quad r< \infty ,\\&\Vert \varphi \Vert _{{q,\infty }} := \sup _t t^{1/q} \varphi ^*(t) = \sup _\tau \tau m(\tau ,\varphi )^{1/q}< \infty , \end{aligned}$$

where $\varphi ^*(t) = \inf \{\tau ; m(\tau ,\varphi ) \le t \}$ (the rearrangement of $\varphi$).

Notation 6

For $\kappa \in (0,n)$, Riesz potential is the operator

$$\begin{aligned} I_{\kappa }f(x) := \frac{1}{|x|^{n-\kappa }}* f = \int \nolimits _{{\mathbb {R}}^n} \frac{f(y)}{|x-y|^{n-\kappa }} dy = C_{n,\kappa } {\mathcal {F}}^{-1}(|\xi |^{- \kappa } {\hat{f}}(\rho )) . \end{aligned}$$

First we give the asymptotic profile of the solutions to linear wave equation with structural damping.

Theorem 1

Let $n \ge 1$. Let $(u_0, u_1) \in ( L^1 \cap L^2) \times (L^1 \cap H^{-2 \sigma })$ such that $|\cdot |^{\theta _j} u_j \in L^1$ with $\theta _j \in [0,1]$ $(j = 0,1)$. Let $u \in C([0,\infty ); H^1) \cap C^1((0,\infty );L^2)$ be a unique global solution of (1.3). Then the following holds.

$$\begin{aligned}&\Vert u(t,\cdot ) - \vartheta _0 H_{\sigma }(t, \cdot ) - \vartheta _1 G_{\sigma }(t, \cdot ) \Vert _2 \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma }\left( -\frac{n}{4} + 2\sigma - 1 \right) }} \Vert u_0\Vert _1 + \langle t \rangle ^{{\max \{ \frac{1}{1 - \sigma }\left( -\frac{n}{4} + 3\sigma - 1 \right) , \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) \} }} \Vert u_1 \Vert _{{{1}}} \nonumber \\&\qquad + e^{-\varepsilon _{\sigma } t} (\Vert u_0 \Vert _2+ \Vert u_1 \Vert _{{H^{-2 \sigma } }}) \nonumber \\&\qquad + t ^{ \frac{1}{1-\sigma }(-\frac{n}{4} -\frac{\theta _0}{2} )} \Vert | \cdot |^{\theta _0} u_0 \Vert _1 + t ^{ \frac{1}{1-\sigma }(-\frac{n}{4} + \sigma -\frac{\theta _1}{2} )}\Vert | \cdot |^ {\theta _1} u_1 \Vert _1, \end{aligned}$$

(2.1)

where

$$\begin{aligned} H_{\sigma }(t, x)&:={{\mathcal {F}}}^{-1}[e^{- |\xi |^{2(1 - \sigma )} t}](x), \nonumber \\ G_{\sigma }(t, x)&:={{\mathcal {F}}}^{-1}[|\xi |^{-2 \sigma } e^{- |\xi |^{2(1 - \sigma )} t}](x) = C_{n,2 \sigma }^{-1} I_{2\sigma } H_{\sigma }(t, x), \end{aligned}$$

(2.2)

$$\begin{aligned} \vartheta _0&:= \int \nolimits _{{{\mathbb {R}}}^n}u_0 (y) dy, \quad \vartheta _1 := \int \nolimits _{{{\mathbb {R}}}^n} u_1(y) dy. \end{aligned}$$

(2.3)

Remark 1

$$\begin{aligned} \max \Big \{ \frac{1}{1 - \sigma }\Big (-\frac{n}{4} + 3\sigma - 1 \Big ), \frac{1}{\sigma } \Big (-\frac{n}{4} + \sigma \Big ) \Big \} = {\left\{ \begin{array}{ll} \frac{1}{1 - \sigma }\left( -\frac{n}{4} + 3\sigma - 1 \right) &{} \text {if} \; 8 \sigma \le n, \\ \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) &{} \text {if} \; 8 \sigma \ge n. \end{array}\right. } \end{aligned}$$

Remark 2

The function $H_\sigma (t,x)$ is the fundamental solution of the parabolic equation (1.4). We easily see that

$$\begin{aligned} \Vert G_{\sigma }(t,\cdot ) \Vert _{{2}} = \Vert \widehat{G_{\sigma }}(t,\cdot ) \Vert _{{2}} \sim t^{ \frac{1}{1-\sigma }(-\frac{n}{4} + \sigma )}, \quad \Vert H_{\sigma }(t,\cdot )\Vert = \Vert \widehat{H_{\sigma }}(t,\cdot ) \Vert _{{2}} \sim {t}^{ -\frac{n}{4(1-\sigma )}},\!\!\nonumber \\ \end{aligned}$$

(2.4)

(see (5.8) and (5.16)). Putting $\theta _0 = [\theta - 2 \sigma ]_+$ and $\theta _1 = \theta$ ($\theta \in (0,1]$) in (2.1), and taking (2.4) and the assumption $\sigma \in (0,\frac{1}{2})$ into consideration, we obtain

$$\begin{aligned} \begin{aligned}&\Vert u(t,\cdot ) - \vartheta _1 G_{\sigma }(t, \cdot ) \Vert _2 \\&\quad \lesssim t ^{{\max \left\{ \frac{1}{1 - \sigma }\left( -\frac{n}{4} + \sigma - \min \{1 - 2\sigma , \frac{\theta }{2} \} \right) , \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) \right\} }}\\&\qquad \times ( \Vert u_0 \Vert _2+ \Vert u_1 \Vert _{{H^{-2 \sigma } }} + \Vert \langle \cdot \rangle ^{{[\theta - 2 \sigma ]_+} } u_0 \Vert _1 + \Vert \langle \cdot \rangle ^{\theta } u_1 \Vert _1). \end{aligned} \end{aligned}$$

Thus, the decay order of $\Vert u(t,\cdot ) - \vartheta _1 G_{\sigma }(t, \cdot )\Vert _2$ is larger than that of $\Vert G_{\sigma }(t,\cdot )\Vert _2$ itself, and therefore, $\vartheta _1 G_{\sigma }(t,x)$ gives the asymptotic profile of the solution if $\vartheta _1 \ne 0$.

If $u_1 = 0$, then (2.1) implies

$$\begin{aligned} \begin{aligned}&\Vert u(t,\cdot ) - \vartheta _0 H_{\sigma }(t, \cdot ) \Vert _2 \lesssim t^{{ \frac{1}{1 - \sigma }\left( -\frac{n}{4} - \min \{ 1 - 2\sigma , \frac{\theta _0}{2}\} \right) }} ( \Vert u_0 \Vert _2 + \Vert \langle \cdot \rangle ^{\theta _0} u_0 \Vert _1). \end{aligned}\nonumber \\ \end{aligned}$$

(2.5)

Thus, the decay order of $\Vert u(t,\cdot ) - \vartheta _0 H_{\sigma }(t, \cdot )\Vert _2$ is larger than that of $\Vert H_{\sigma }(t,\cdot )\Vert _2$ itself if $\theta _0 > 0$, and therefore, $\vartheta _0 H_{\sigma }(t,x)$ gives the asymptotic profile if $\vartheta _0 \ne 0$.

Remark 3

Ikehata and Takeda [11, Theorem 1.2] showed

$$\begin{aligned} \Vert u(t,\cdot ) - \vartheta _1 G_{\sigma }(t,\cdot ) \Vert _2 = o( t^{ \frac{1}{1-\sigma }(-\frac{n}{4} + \sigma )}) \end{aligned}$$

as $t \rightarrow \infty$ for initial data in $(u_0, u_1) \in ( L^1 \cap H^1) \times (L^1 \cap L^2)$.

If $u_1 = 0$, Karch [13, Corollary 4.1] showed

$$\begin{aligned} \Vert u(t,\cdot ) - \vartheta _0 H_{\sigma }(t,\cdot ) \Vert _2 = o( t^{ -\frac{n}{4(1-\sigma )}}) \end{aligned}$$

as $t \rightarrow \infty$ for $u_0 \in L^1$.

Theorem 2

(Global existence of the solution) Let $n \ge 2$, and

$$\begin{aligned} p > p_{\sigma }:= 1 + \frac{2}{n - 2 \sigma }. \end{aligned}$$

(2.6)

Assume that $\bar{s} \ge 1$ and $[\bar{s}] < p$. If $2 \bar{s} < n$, assume moreover that

$$\begin{aligned}&p \le 1 + \frac{2}{n - 2\bar{s}}. \end{aligned}$$

(2.7)

Assume that $f \in C^{[\bar{s}],1 }({{\mathbb {R}}})$ satisfies (1.2). Let $q_j\; (j=0,1)$ be numbers such that

$$\begin{aligned} \left\{ \begin{aligned}&q_0 = \frac{2n}{n + 2 - 4\sigma }, \;&q_1 = \frac{2n}{n +2} \qquad \quad \text {if} \quad p_\sigma< p< 1 + \frac{4}{n + 2 - 4\sigma },&\\&1< q_0< \frac{n(p-1)}{2},\;&1< q_1< \frac{n(p-1)}{2(1 + (p-1)\sigma )}&\\&\;&\qquad \qquad \qquad \qquad \text {if} \quad 1 + \frac{4}{n + 2 - 4\sigma }< p \le 1 + \frac{4}{n},&\\&1< q_0< 2,\;&1< q_1< \frac{2n}{n + 4 \sigma } \qquad \quad \text {if} \quad 1 + \frac{4}{n} < p.&\end{aligned} \right. \!\!\!\!\!\nonumber \\ \end{aligned}$$

(2.8)

(Case 1). :

In the case $p_\sigma < p \le 1 + \frac{4}{n + 2 - 4\sigma }$, let $\delta$ be a number satisfying

$$\begin{aligned} 2 \left( \frac{1}{p-1} + \sigma \right) - \frac{n}{2} - 1 < \delta \le \frac{2}{p-1} - \frac{n}{2}. \end{aligned}$$

(2.9)

Then there exists a positive number $\varepsilon$ such that if initial data

$$\begin{aligned} u_0 \in H^{\bar{s}} \cap H^{0,\delta }, \langle \cdot \rangle ^\delta u_0 \in L^{q_0,2 }, \quad u_1 \in \dot{H}^{\bar{s}-1}, \langle \cdot \rangle ^\delta u_1 \in L^{{ q_1,2 }} \end{aligned}$$

(2.10)

satisfy

$$\begin{aligned} \begin{aligned}&\Vert \langle \cdot \rangle ^\delta u_0 \Vert _{q_0,2 } + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _2 + \Vert u_0 \Vert _{{H^{\bar{s}}}} + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{ q_1,2 }} + \Vert u_1 \Vert _{{\dot{H}^{s-1} }} \le \varepsilon , \end{aligned} \end{aligned}$$

(2.11)

then initial value problem (1.1) has a unique global solution $u \in C([0,\infty ); H^{\bar{s}}\cap H^{0,\delta }) \cap C^1((0,\infty );H^{\bar{s} - 1})$.

(Case 2). :

In the case $p >1 + \frac{4}{n + 2 - 4\sigma }$, there exists a positive number $\varepsilon$ such that if initial data

$$\begin{aligned} u_0 \in H^{\bar{s}} \cap L^{q_0,2}, \quad u_1 \in H^{\bar{s}-1} \cap L^{q_1,2}, \end{aligned}$$

satisfy

$$\begin{aligned}&\Vert u_0 \Vert _{q_0,2} + \Vert u_0 \Vert _{H^{\bar{s}}} + \Vert u_1 \Vert _{q_1,2} + \Vert u_1\Vert _{{ H^{\bar{s}-1}}} \le \varepsilon , \end{aligned}$$

(2.12)

then initial value problem (1.1) has a unique global solution $u \in C([0,\infty ); H^{\bar{s}}) \cap C^1((0,\infty );H^{\bar{s} - 1})$.

Remark 4

We note that $L^{q_j} = L^{q_j, q_j} \subset L^{q_j, 2}$ by Lemma A given later, since $q_j \le 2$.

Remark 5

If the space dimension $n = 2$, then $1 + \frac{4}{n + 2 - 4\sigma } \le p_{\sigma }$, and therefore, (Case 1) does not occur.

We prove Theorem 2 by using the following proposition.

Proposition 1

(Global existence of the solution) Let $n \ge 2$ and $r \in [1,\frac{2n}{n + 4 \sigma })$. Let

$$\begin{aligned} p > p_{\sigma ,r}:= 1 + \frac{2r}{n - 2r\sigma }. \end{aligned}$$

(2.13)

Assume that $\bar{s} \ge 1$ and $[\bar{s}] < p$. If $2 \bar{s} < n$, assume moreover (2.7). Assume that $f \in C^{[\bar{s}],1 }({{\mathbb {R}}})$ satisfies (1.2). Let $\delta$ be a non-negative constant satisfying

$$\begin{aligned} n\left( \frac{1}{r} - \frac{1}{2}\right) - 1 \le \delta < n \left( \frac{1}{r} - \frac{1}{2}\right) - 2\sigma . \end{aligned}$$

(2.14)

and

$$\begin{aligned} {\left\{ \begin{array}{ll} &{} n\left( \frac{1}{p} - \frac{1}{2}\right) < \delta , \quad \text {if}\quad r = 1,\\ &{} n\left( \frac{1}{pr} - \frac{1}{2}\right) \le \delta \quad \; \text {if} \quad r \in \left( 1,\frac{2n}{n + 4 \sigma }\right) . \end{array}\right. } \end{aligned}$$

(2.15)

Let

$$\begin{aligned} {\hat{q}}_0 = \frac{nr}{n - r(\delta + 2\sigma )}, \quad {\hat{q}}_1 = \frac{nr}{n - r \delta }. \end{aligned}$$

(2.16)

Then there exists a positive number $\varepsilon$ such that if initial data

$$\begin{aligned} u_0 \in H^{\bar{s}} \cap H^{0,\delta }, \langle \cdot \rangle ^\delta u_0 \in L^{{{\hat{q}}_0,2}}, \quad u_1 \in \dot{H}^{\bar{s}-1}, \langle \cdot \rangle ^\delta u_1 \in L^{{{\hat{q}}_1,2 }} \end{aligned}$$

(2.17)

satisfy

$$\begin{aligned} \begin{aligned}&\Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{\hat{q}}_0,2}} + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{2}} }} + \Vert u_0 \Vert _{{H^{\bar{s}}}} \\&\quad + \Vert u_1\Vert _{{{1}}} + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{\hat{q}}_1,2}} + \Vert (-\Delta )^{\frac{\bar{s}}{2}} (1 - \Delta )^{-\frac{1}{2}} u_1 \Vert _{{2}} \le \varepsilon , \end{aligned} \end{aligned}$$

(2.18)

in the case $r = 1$, and

$$\begin{aligned} \begin{aligned}&\Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\hat{q}}_0},2 }} + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{2}} + \Vert u_0\Vert _{{H^{\bar{s}}}} \\&\quad + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{{ {\hat{q}}_1 ,2}} }} + \Vert (-\Delta )^{\frac{\bar{s}}{2}} (1 - \Delta )^{-\frac{1}{2}} u_1 \Vert _{{2}} \le \varepsilon , \end{aligned} \end{aligned}$$

(2.19)

in the case $r \in (1,2]$, then initial value problem (1.1) has a unique global solution $u \in C([0,\infty ); H^{\bar{s}} \cap H^{0,\delta })) \cap C^1((0,\infty );H^{\bar{s}-1})$.

Furthermore, the solution satisfies estimate:

$$\begin{aligned}&\sup _{t > 0} \Big ( \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( \frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \sigma \right) }} {\left\| u(t,\cdot )\right\| _{2}} + \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( \frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{\delta }{2} - \sigma \right) }} {\left\| |\cdot |^\delta u(t,\cdot )\right\| _{2}} \nonumber \\&\quad + \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( \frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \sigma + \frac{\bar{s}}{2} \right) }} \Vert (-\Delta )^{\frac{\bar{s}}{2}} u(t,\cdot )\Vert _{{2}} \Big ) < \infty . \end{aligned}$$

(2.20)

Remark 6

The assumption $r < \frac{2n}{n + 4\sigma }$ implies that $n(\frac{1}{r} - \frac{1}{2}) - 2\sigma > 0$. The inequality

$$\begin{aligned} n\left( \frac{1}{pr} - \frac{1}{2}\right) < n\left( \frac{1}{r} - \frac{1}{2}\right) - 2\sigma \end{aligned}$$

is equivalent to

$$\begin{aligned} p > 1 + \frac{2 r \sigma }{n - 2 r \sigma }, \end{aligned}$$

which holds by (2.13) since $\sigma < 1$. Hence, we can take a non-negative number $\delta$ satisfying assumptions (2.14) and (2.15).

If initial data belong to weighted $L^1$ space, the asymptotic profile of the solution is given by a constant multiple of the fundamental solution of the parabolic equation (1.4).

Theorem 3

(Asymptotic profile) Assume the assumption of Proposition 1 with $r = 1$. Let $\varepsilon$ be a positive constant given by Proposition 1 for $r = 1$, and let $\theta \in [0,1]$. Assume that initial data satisfy (2.17) and (2.18) and that $\langle \cdot \rangle ^{[\theta - 2\sigma ]_+} u_0, \langle \cdot \rangle ^{\theta } u_1 \in L^1$. Let $\nu$ be an arbitrary number satisfying

$$\begin{aligned} 0< \nu < \min \Bigg \{ \frac{n}{4}(p-2) + \frac{1}{2}p \delta , \delta \Bigg \}. \end{aligned}$$

(2.21)

Assume moreover that

$$\begin{aligned} \nu< \frac{\delta }{2 \bar{s}} \left( n - \frac{p}{2}(n - 2 \bar{s})\right) \qquad \text {if} \quad \bar{s} < \frac{n}{2}. \end{aligned}$$

(2.22)

Then there is a constant C depending on

$$\begin{aligned}\begin{aligned}&\Vert \langle \cdot \rangle ^{[\theta - 2\sigma ]_+} u_0 \Vert _1+ \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\frac{n}{n - (\delta + 2\sigma )}}} }} + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{2}} }} + \Vert u_0 \Vert _{{H^{\bar{s} }}} \\&\quad + \Vert \langle \cdot \rangle ^\theta u_1 \Vert _1 + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{{ \frac{n}{n - \delta }}} }} + \Vert (-\Delta )^{\frac{\bar{s}}{2}} (1 - \Delta )^{-\frac{1}{2}} u_1 \Vert _{{2}} \end{aligned} \end{aligned}$$

such that the solution $u \in C([0,\infty ); H^{\bar{s}} \cap H^{0,\delta }) \cap C^1((0,\infty );H^{\bar{s} - 1})$ of (1.1), which is given by Proposition 1, satisfies the following:

$$\begin{aligned} \begin{aligned}&\Big \Vert u(t,\cdot ) - \varTheta G_{\sigma }(t, \cdot ) \Big \Vert _{{2}} \\&\quad \le C t^{{ \max \left\{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - \min \{( p - 1)\left( \frac{n}{2} - \sigma \right) - 1, 1 - 2 \sigma , \nu , \frac{\theta }{2} \} \right) , \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) \right\} }}, \end{aligned} \end{aligned}$$

(2.23)

where $G_{\sigma }$ is defined by (2.2) and

$$\begin{aligned} \varTheta&:= \int \nolimits _{{{\mathbb {R}}}^n} u_1(y)dy + \int \nolimits _0^\infty \int \nolimits _{{{\mathbb {R}}}^n}f(u(\tau ,y))dy d\tau . \end{aligned}$$

(2.24)

Remark 7

The right-hand sides of (2.21) and (2.22) are positive. In fact, assumtion (2.15) implies $\frac{n}{2}(p-2) + p \delta > 0$, and (2.7) implies $n - \frac{p}{2}(n - 2 \bar{s}) > 0$. Hence we can take $\nu$ satisfying (2.21) and (2.22).

Remark 8

Since (2.4) holds, (2.23) implies that $G_\sigma$ gives the asymptotic profile of the solution if $\varTheta \ne 0$.

Remark 9

In the case $\sigma = 0$, Hayashi, Kaikina and Naumkin [7] showed the existence of global solution $u \in C([0,\infty );H^{\bar{s}} \cap H^{0,\delta })$ of the semilinear damped wave (1.1) with $\sigma = 0$ for small initial data $u_0 \in H^{\bar{s}} \cap H^{0,\delta }, u_1 \in H^{\bar{s}-1} \cap H^{0,\delta }$ with $\delta > \frac{n}{2}$, and showed

$$\begin{aligned} \Big \Vert&u(t,\cdot ) - {\tilde{\varTheta }} G_0(t, \cdot ) \Big \Vert _{{q}} \le C t^{{ -\frac{n}{2}(1 - \frac{1}{q}) - \min \{\frac{n}{2}(p - 1) - 1, \frac{\delta }{2}- \frac{n}{4}, \nu \} }}, \end{aligned}$$

for $2 \le q \le \frac{2n}{n - 2 \bar{s}}$, where ${\tilde{\varTheta }} = \int \nolimits _{{{\mathbb {R}}}^n} (u_0(y) +u_1(y))dy + \int \nolimits _0^\infty \int \nolimits _{{{\mathbb {R}}}^n}f(u(\tau ,y))dy d\tau$, $G_0$ is the heat kernel ((2.2) with $\sigma = 0$) and $0< \nu < 1$.

3 Preliminary lemmas

We list some properties for weak $L^p$ and Lorentz spaces which are used in this paper (see [1, section 1.3], [22], for example).

Lemma A

Let $q \in (0,\infty )$. Then

$$\begin{aligned}&L^{q,q} = L^q, \quad L^{q,\infty } = L_q^*, \\&L^{q,\rho _1} \subset L^{q,\rho _2} \quad \text {if} \quad 1 \le \rho _1 \le \rho _2 \le \infty . \end{aligned}$$

Lemma B

Assume that $\mu , \rho , \nu \in (1,\infty )$ and ${\tilde{\mu }}, \tilde{\rho }, {\tilde{\nu }} \in [1,\infty ]$ satisfy

$$\begin{aligned} \frac{1}{\mu } = \frac{1}{\rho } + \frac{1}{\nu }, \quad \frac{1}{{\tilde{\mu }} } = \frac{1}{{\tilde{\rho }}} + \frac{1}{{\tilde{\nu }}}. \end{aligned}$$

Then

$$\begin{aligned} {\left\| f g\right\| _{{\mu ,{\tilde{\mu }}}}} \lesssim {\left\| f\right\| _{{\rho , \tilde{\rho }}}} {\left\| g\right\| _{{\nu , {\tilde{\nu }}}}}, \end{aligned}$$

provided the right-hand side is finite.

The next corollary immediately follows from Lemma B.

Corollary A

Let $\omega >0$, $\mu , \nu \in (1,\infty )$ and ${\tilde{\mu }} \in [1,\infty ]$. If

$$\begin{aligned} \frac{1}{\mu } = \frac{\omega }{n} + \frac{1}{\nu }, \end{aligned}$$

then the following hold.

$$\begin{aligned} {\left\| | x |^{-\omega } f\right\| _{{\mu , {\tilde{\mu }}}}}&\lesssim \Vert | x |^{-\omega }\Vert _{{{\frac{n}{\omega }, \infty } }} {\left\| f\right\| _{{\nu ,{\tilde{\mu }}}}} \lesssim {\left\| f\right\| _{{\nu ,{\tilde{\mu }}}}} \end{aligned}$$

Lemma C

Let $q \in (2,\infty )$, and let $q^\prime$ be the dual exponent of q, that is, $\frac{1}{q} + \frac{1}{q^\prime } = 1$. Let $\nu \in [1,\infty ]$. Then

$$\begin{aligned} \Vert {{\mathcal {F}}}[\varphi ]\Vert _{{{q,\nu } }} \lesssim \Vert \varphi \Vert _{{{q^{\prime },\nu }}}. \end{aligned}$$

Lemma D

(Young’s inequality) Let $q, \rho \in (1,2]$ such that $\frac{1}{q} + \frac{1}{\rho } = \frac{3}{2}$. Let $s,t \in [2,\infty )$ such that $\frac{1}{s} + \frac{1}{t} = \frac{1}{2}$. Then

$$\begin{aligned}&{\left\| \varphi *\psi \right\| _{2}} \lesssim {\left\| \varphi \right\| _{{\rho ,s}}} {\left\| \psi \right\| _{{q,t} }}. \end{aligned}$$

Lemma E

(sharp Sobolev embedding theorem) Let $q \in [2, \infty )$ and $s \ge 0$. If

$$\begin{aligned} \frac{n}{2} -s \le \frac{n}{q}, \end{aligned}$$

then

$$\begin{aligned} H^s({{\mathbb {R}}}^n) \subset L^{q,2}({{\mathbb {R}}}^n). \end{aligned}$$

4 Decay estimate for the kernels

In this section, we estimate the kernel of the following linear wave equation with structural damping (1.3).

By Fourier transform, the equation (1.3) is transformed to

$$\begin{aligned} {\hat{u}}_{tt}+|\xi |^{2\sigma } {\hat{u}}_t + |\xi |^2 {\hat{u}}=0 \quad (t > 0), \qquad {\hat{u}}(0) ={\hat{u}}_0, \quad {\hat{u}}_t(0) ={\hat{u}}_1. \end{aligned}$$

Hence the solution u of (1.3) is expressed as

$$\begin{aligned} u(t,x)&= (K_0(t,\cdot )*u_0)(x) + (K_1(t,\cdot )*u_1)(x), \end{aligned}$$

(4.1)

where

$$\begin{aligned} \widehat{K_0}(t,\xi )&=\frac{1}{\lambda _+(|\xi |) - \lambda _-(|\xi |)}(\lambda _+(|\xi |) e^{\lambda _-(|\xi |) t}-\lambda _-(|\xi |) e^{\lambda _+(|\xi |) t}), \end{aligned}$$

(4.2)

$$\begin{aligned} \widehat{K_1}(t,\xi )&=\frac{1}{\lambda _+(|\xi |) - \lambda _-(|\xi |)}(e^{\lambda _+(|\xi |) t}- e^{\lambda _-(|\xi |) t}), \end{aligned}$$

(4.3)

$$\begin{aligned} \lambda _{\pm }(|\xi |)&= \frac{1}{2} \left( -|\xi |^{2\sigma } \pm \sqrt{|\xi |^{4\sigma } - 4|\xi |^2 }\right) \end{aligned}$$

(4.4)

$$\begin{aligned}&= {\left\{ \begin{array}{ll} \frac{1}{2} |\xi |^{2 \sigma }\left( -1\pm \sqrt{1-4|\xi |^{2(1 - 2\sigma )} } \right) \quad \text {if} \quad |\xi |^{1 - 2\sigma } < \frac{1}{2}, \\ \frac{1}{2}|\xi |^{2\sigma } \left( -1 \pm i \sqrt{4|\xi |^{2(1 - 2\sigma )} -1} \right) \quad \text {if} \quad |\xi |^{1 - 2\sigma } > \frac{1}{2}. \end{array}\right. } \end{aligned}$$

(4.5)

We divide $K_0$ and $K_1$ into

$$\begin{aligned} \widehat{K_1^{\pm }}(t,\xi )&:= \pm \frac{e^{\lambda _{\pm }(|\xi |)t}}{\lambda _+(|\xi |) - \lambda _-(|\xi |)}, \end{aligned}$$

(4.6)

$$\begin{aligned} \widehat{K_0^{\pm }}(t,\xi )&:= \mp \frac{ \lambda _{\mp }(|\xi |) e^{\lambda _{\pm }(|\xi |)t}}{\lambda _+(|\xi |) - \lambda _-(|\xi |)} = -\lambda _{\mp }(\xi )\widehat{K_1^{\pm }}(t,\xi ). \end{aligned}$$

(4.7)

Let $\chi _{low}(\xi ) \in C^\infty ({{\mathbb {R}}}^n)$ be a function such that $\chi _{low}(\xi ) = 1$ for $|\xi | \le 2^{{ - \frac{3}{1 - 2 \sigma }}}$ and $\chi _{low}(\xi ) = 0$ for $|\xi | \ge 2^{{ - \frac{2}{1 - 2 \sigma }}}$. Let $\chi _{high}(\xi ) \in C^\infty ({{\mathbb {R}}}^n)$ be a function such that $\chi _{high}(\xi ) = 1$ for $|\xi | \ge 2$ and $\chi _{high}(\xi ) = 0$ for $|\xi | \le 1$.

We put

$$\begin{aligned} \chi _{mid}(\xi ) :&= 1 - \chi _{low}(\xi ) - \chi _{high}(\xi ), \\ \chi _{mh}(\xi ):=1-\chi _{low}(\xi )&= \chi _{mid}(\xi ) + \chi _{high}(\xi ). \end{aligned}$$

Here we note that

$$\begin{aligned} {{\,\mathrm{supp}\,}}\chi _{mid} \subset \{\xi ; |\xi | \in [2^{{-\frac{3}{1 - 2\sigma } }}, 2] \}, \quad {{\,\mathrm{supp}\,}}\chi _{hm} \subset \{\xi ; |\xi | \in [2^{{-\frac{3}{1 - 2\sigma } }}, \infty ) \}.\nonumber \\ \end{aligned}$$

(4.8)

We put

$$\begin{aligned} \begin{aligned} K_{j,low}(t,x)&:= {{\mathcal {F}}}^{-1}[ \hat{K_j}(t,\cdot )\chi _{low}(\cdot ) ], \\ K_{j,mid}(t,x)&:= {{\mathcal {F}}}^{-1}[ \hat{K_j}(t,\cdot )\chi _{mid}(\cdot ) ], \\ K_{j,high}(t,x)&:= {{\mathcal {F}}}^{-1}[ \hat{K_j}(t,\cdot )\chi _{high}(\cdot ) ], \\ K_{j,mh}(t,x)&:= K_{j,mid}(t,x) +K_{j,high}(t,x), \end{aligned} \end{aligned}$$

for $j = 0,1$. Dividing the kernel into

$$\begin{aligned} \begin{aligned} K_j =K_{j,low} + K_{j,mid } + K_{j,high} = K_{j,low} + K_{j,mh} \end{aligned} \end{aligned}$$

for $j = 0,1$, we estimate each part.

4.1 Estimate of the kernels for low frequency part

In this subsection, we consider low frequency region: $|\xi | \le 2^{{ - \frac{2}{1 - 2 \sigma }}}$.

Lemma 1

Let $\alpha > -\frac{n}{2}$ and $\beta > 0$. Let $a > 2^{{ - \frac{2}{1 - 2 \sigma }}}$. Let $g(t,\rho )$ be a smooth function on $[0,\infty ) \times (0,a)$ satisfying

$$\begin{aligned}&\left| \dfrac{\partial ^k}{\partial \rho ^k} g(t,\rho ) \right| \lesssim \rho ^{\alpha - k} e^{-\frac{1}{2}\rho ^\beta t} \end{aligned}$$

(4.9)

on $[0,\infty ) \times (0,a)$ for every $k = 0,1,\cdots$. Put

$$\begin{aligned} K(t,x):= {{\mathcal {F}}}^{-1}[g(t,|\xi |) \chi _{low}](x). \end{aligned}$$

Then for every $q_j \in [1,2)$ $(j = 0,1)$ and $\vartheta \in [ 0,\frac{n}{2} + \alpha )$ satisfying

$$\begin{aligned} \frac{1}{q_1} \ge \frac{1}{2} + \frac{\vartheta - \alpha }{n}, \; \frac{1}{q_2} \ge \frac{1}{2} - \frac{\alpha }{n}, \; \end{aligned}$$

(4.10)

the following holds.

$$\begin{aligned} {\left\| | x |^\vartheta \left( K(t,\cdot ) * \varphi (\cdot ) \right) \right\| _{2}} \lesssim \,&\langle \, t \rangle ^{{ \frac{1}{\beta } \left( -n(\frac{1}{q_1} - \frac{1}{2}) + \vartheta - \alpha \right) }} \Vert \varphi \Vert _{q_1}^\prime \nonumber \\&+ \langle t \rangle ^{{ \frac{1}{\beta } \left( -n(\frac{1}{q_2} - \frac{1}{2}) - \alpha \right) }} \Vert | x |^\vartheta \varphi \Vert _{q_2}^\prime , \end{aligned}$$

(4.11)

where $\Vert \cdot \Vert ^\prime _q$ denote

$$\begin{aligned} \Vert \cdot \Vert ^\prime _q = {\left\{ \begin{array}{ll} \Vert \cdot \Vert _1 \quad &{}\text {if} \quad q = 1 \\ \Vert \cdot \Vert _{q,2} \quad &{}\text {if} \quad q \in (1,2]. \end{array}\right. } \end{aligned}$$

(4.12)

Before proving Lemma 1, we state two corollaries:

Corollary 1

Let $\alpha > -\frac{n}{2}$ and $\beta > 0$. Let $a > 2^{{ - \frac{2}{1 - 2 \sigma }}}$. Let $\upsilon$ and $\lambda$ be smooth functions on some interval (0, a) such that

$$\begin{aligned} |\upsilon ^{(j)}(\rho ) |&\lesssim \rho ^{\alpha -j}, \end{aligned}$$

(4.13)

$$\begin{aligned} |\lambda ^{(j)}(\rho ) |&\lesssim \rho ^{\beta -j}, \quad -\lambda (\rho ) \sim \rho ^{\beta } \end{aligned}$$

(4.14)

on (0, a) for every $j = 0,1,\cdots$. Put

$$\begin{aligned} K(t,x):= {{\mathcal {F}}}^{-1}[\upsilon (|\xi |) e^{\lambda (|\xi |)t}(x) \chi _{low}](x). \end{aligned}$$

(4.15)

Then the conclusion of Lemma 1 holds.

In fact, we easily see that

$$\begin{aligned}&\left| \dfrac{\partial ^k}{\partial \rho ^k} \left( \upsilon (\rho )e^{\lambda (\rho )t}\right) \right| \lesssim \rho ^{\alpha - k} \left( \sum _{j=0}^k (\rho ^{\beta }t)^j \right) e^{-\rho ^\beta t} \lesssim \rho ^{\alpha - k} e^{-\frac{1}{2}\rho ^\beta t} \end{aligned}$$

(4.16)

on (0, a) for every $k = 0,1,\ldots$. Hence, $g(t,\rho ) = \upsilon (\rho )e^{\lambda (\rho )t}$ satisfies the assumption (4.9) of Lemma 1, and thus the conclusion holds.

Corollary 2

Let $\alpha , \beta , \gamma$ be numbers such that $\alpha - \beta + \gamma > -\frac{n}{2}$, $\beta > 0$ and $\gamma > 0$. Let $a > 2^{{ - \frac{2}{1 - 2 \sigma }}}$. Let $\upsilon$ and $\lambda$ be smooth functions on (0, a) such that

$$\begin{aligned} |\upsilon ^{(j)}(\rho ) |&\lesssim \rho ^{\alpha -j}, \end{aligned}$$

(4.17)

$$\begin{aligned} |\lambda ^{(j)}(\rho ) |&\lesssim \rho ^{\beta -j}, \quad -\lambda (\rho ) \sim \rho ^{\beta } \end{aligned}$$

(4.18)

$$\begin{aligned} |\mu ^{(j)}(\rho ) |&\lesssim \rho ^{\gamma -j}, \quad -\mu (\rho ) \sim \rho ^{\gamma } \end{aligned}$$

(4.19)

on (0, a) for every $j = 1,2,\cdots$. Put

$$\begin{aligned} K(t,x):= {{\mathcal {F}}}^{-1}[\upsilon (|\xi |) e^{\lambda (|\xi |)t} (1 - e^{\mu (|\xi |)t})\chi _{low}](x). \end{aligned}$$

Then for every $q_j \in [1,2)\,(j = 1,2)$ and $\vartheta \in [ 0,\frac{n}{2} + \alpha - \beta + \gamma )$ satisfying

$$\begin{aligned} \frac{1}{q_1} \ge \frac{1}{2} + \frac{\vartheta - \alpha + \beta - \gamma }{n}, \; \frac{1}{q_2} \ge \frac{1}{2} + \frac{-\alpha + \beta - \gamma }{n}, \; \end{aligned}$$

(4.20)

the following holds.

$$\begin{aligned} {\left\| | x |^\vartheta \left( K(t,\cdot ) * \varphi (\cdot ) \right) \right\| _{2}} \lesssim \,&\langle t \rangle ^{{ \frac{1}{\beta } \left( -n(\frac{1}{q_1} - \frac{1}{2}) + \vartheta - \alpha + \beta - \gamma \right) }} \Vert \varphi \Vert _{q_1}^\prime \nonumber \\&+ \langle t \rangle ^{{ \frac{1}{\beta } \left( -n(\frac{1}{q_2} - \frac{1}{2}) - \alpha + \beta - \gamma \right) }} \Vert | x |^\vartheta \varphi \Vert _{q_2}^\prime . \end{aligned}$$

(4.21)

Remark 10

D’Abbicco and Ebert [2] considered the kernels:

$$\begin{aligned} K(t,x) = {{\mathcal {F}}}^{-1}[\upsilon (|\xi |) e^{\lambda (|\xi |)t} \chi _{low}](x), \end{aligned}$$

where $\upsilon$ and $\lambda$ satisfy the assumptions (4.13) and (4.14) for $\alpha > -1$ (see [2, Lemma 3.1]), and

$$\begin{aligned} K(t,x) = {{\mathcal {F}}}^{-1}\left[ \upsilon (|\xi |) e^{\lambda (|\xi |)t} \frac{1 - e^{\mu (|\xi |)t}}{\mu (|\xi |)t}\chi _{low}\right] (x), \end{aligned}$$

where $\upsilon$, $\lambda$ and $\mu$ satisfy (4.17), (4.18) and (4.19) for $\alpha> -1, \beta> 0, \gamma > 0$ (see [2, Lemma 3.2]), and showed $L^p-L^q$ estimates of $\varphi \mapsto K(t,\cdot )*\varphi$ for $1 \le p \le q \le \infty$ such that

• (i) $p \ne q$ if $\alpha = 0$ and $\upsilon$ is not a constant,

• (ii) $\frac{1}{p} - \frac{1}{q} \ge - \frac{\alpha }{n}$ if $\alpha \in (-1,0)$,

by using the description of kernels by Bessel functions.

In this paper, we show weighted $L^2$ estimates of $K(t,\cdot )*\varphi$ in a way different from [2] by employing Lorentz spaces.

Proof of Corollary 2

By the Leibniz rule, we have

$$\begin{aligned}&\dfrac{\partial ^k}{\partial \rho ^k} \left( \upsilon (\rho )e^{\lambda (\rho )t}(1 - e^{\mu (\rho )t}) \right) = \sum _{j=0}^k C_{k,j} \dfrac{\partial ^j}{\partial \rho ^j} (\upsilon (\rho )e^{\lambda (\rho )t}) \dfrac{\partial ^{k-j}}{\partial \rho ^{k-j}}(1 - e^{\mu (\rho )t}). \end{aligned}$$

(4.22)

By assumtion (4.19), we have

$$\begin{aligned} \left| \dfrac{\partial ^{k-j}}{\partial \rho ^{k-j}}(1 - e^{\mu (\rho )t}) \right|&= \left| \dfrac{\partial ^{k-j}}{\partial \rho ^{k-j}} e^{\mu (\rho )t}\right| \nonumber \\&\lesssim \rho ^{-(k-j)} \left( \sum _{i=1}^{k-j} (\rho ^{\gamma }t)^i \right) e^{-\rho ^\gamma t} = \rho ^{-k+j} \rho ^\gamma t \sum _{i=0}^{k-j-1}(\rho ^{\gamma }t)^i e^{-\rho ^\gamma t} \nonumber \\&\lesssim \rho ^{-k+j+\gamma } t e^{-\frac{\rho ^\gamma t}{2}} \le \rho ^{-k+j+\gamma } t, \end{aligned}$$

(4.23)

if $j \le k - 1$, and

$$\begin{aligned} \left| \dfrac{\partial ^{k-j}}{\partial \rho ^{k-j}}(1 - e^{\mu (\rho )t}) \right| = | 1 - e^{\mu (\rho )t}| = |\mu (\rho )t e^{\theta \mu (\rho )t} | \lesssim \rho ^\gamma t, \end{aligned}$$

(4.24)

with $\theta \in (0,1)$ if $j = k$. From (4.22), (4.16) with k replaced by j, (4.23) and (4.24), it follows that

$$\begin{aligned}&\left| \dfrac{\partial ^k}{\partial \rho ^k} \left( \upsilon (\rho )e^{\lambda (\rho )t}(1 - e^{\mu (\rho )t}) \right) \right| \lesssim \rho ^{\alpha + \gamma - k}t e^{-\frac{1}{2}\rho ^\beta t} \lesssim \rho ^{\alpha + \gamma - \beta - k} e^{-\frac{1}{4}\rho ^\beta t} \end{aligned}$$

on (0, a). Hence, $g(t,\rho ) = \upsilon (\rho )e^{\lambda (\rho )t}(1 - e^{\mu (\rho )t})$ satisfies the assumption (4.9) with $\alpha$ replaced by $\alpha - \beta + \gamma$, and therefore, Lemma 1 implies the assertion.

$\square$

Now we prove Lemma 1.

Proof of Lemma 1

(Step 1) Let k be a non-negative integer and $\nu \in (0,\infty )$. We show that

$$\begin{aligned} \begin{aligned} \left\| (-\Delta )^{\frac{k}{2}} {\hat{K}}(t,\cdot ) \right\| _{{{\nu } }}&\lesssim \langle t \rangle ^{\frac{1}{\beta }(-\alpha + k - \frac{n}{\nu })} \end{aligned} \end{aligned}$$

(4.25)

for every $t \ge 0$ if $(-\alpha + k)\nu < n$, and

$$\begin{aligned} \begin{aligned} \left\| (-\Delta )^{\frac{k}{2}} {\hat{K}}(t,\cdot ) \right\| _{{{\nu ,\infty } }}&\lesssim \langle t \rangle ^{\frac{1}{\beta }(-\alpha + k - \frac{n}{\nu })} = 1, \end{aligned} \end{aligned}$$

(4.26)

for every $t \ge 0$ if $(-\alpha + k)\nu =n$.

First, we assume that $(-\alpha + k)\nu < n$. Using assumtion (4.9) and changing variables by $t^{1/\beta }\rho = r$, we have

$$\begin{aligned} \left\| (-\Delta )^{\frac{k}{2}} {\hat{K}}(t,\cdot ) \right\| _{{{\nu } }}^{\nu }&\lesssim \left\| \sum _{|\gamma | = k} \partial _{\xi }^\gamma {\hat{K}}(t,\xi ) \right\| _{{\nu }}^{\nu } \nonumber \\&\lesssim \int \nolimits _{0}^a \rho ^{(\alpha - k)\nu } e^{{-\frac{1}{2}\nu \rho ^{\beta } t }} \rho ^{n-1} d\rho \end{aligned}$$

(4.27)

$$\begin{aligned}&= t^{\frac{1}{\beta }((-\alpha + k) \nu - n)} \int \nolimits _{0}^{t a} r^{-(-\alpha + k)\nu } e^{{-\frac{1}{2}\nu r^{\beta } }} r^{n-1} dr \nonumber \\&\lesssim t^{\frac{1}{\beta }((-\alpha + k) \nu - n)}. \end{aligned}$$

(4.28)

By (4.27), we have

$$\begin{aligned} \left\| (-\Delta )^{\frac{k}{2}} {\hat{K}}(t,\cdot ) \right\| _{{{\nu } }}^{\nu }&\lesssim \int \nolimits _{0}^a \rho ^{(\alpha - k)\nu } \rho ^{n-1} d\rho < \infty , \end{aligned}$$

for $0 < t \le 1$, which together with (4.28) yields (4.25).

Next we assume that $(-\alpha + k)\nu = n$. By (4.9), we have

$$\begin{aligned} \left| \partial _{\xi }^\gamma {\hat{K}}(t,\xi ) \right|&\lesssim |\xi |^{\alpha - |\gamma |} \end{aligned}$$

for every $t \ge 0$. Hence,

$$\begin{aligned} s \mu ( \{ \xi ; | \partial _{\xi }^\gamma {\hat{K}}(t,\xi )| > s \} )^{{\frac{1}{\nu }}} \lesssim s^{{1 -\frac{n}{(-\alpha +k)\nu } }} = 1, \end{aligned}$$

if $|\gamma | = k$, and therefore,

$$\begin{aligned} \begin{aligned} \left\| (-\Delta )^{\frac{k}{2}} {\hat{K}}(t,\cdot ) \right\| _{{\nu ,\infty }} \lesssim \sum _{|\gamma | = k} \left\| \partial _{\xi }^\gamma {\hat{K}}(t,\xi ) \right\| _{\nu ,\infty }&\lesssim 1, \end{aligned} \end{aligned}$$

for every $t \ge 0$, that is, (4.26) holds in the case $(-\alpha + k)\nu = n$.

(Step 2) Let $\vartheta \in [0,\delta ]$ and $\kappa \in (1,2]$. We prove that

$$\begin{aligned} \begin{aligned} {\left\| |\cdot |^{\vartheta } K(t,\cdot ) \right\| _{{\kappa , 2}}}&\lesssim \langle t \rangle ^{\frac{1}{^\beta }(\vartheta - \alpha - n( 1 - \frac{1}{\kappa }))}, \end{aligned} \end{aligned}$$

(4.29)

for every $t > 0$ if $-\alpha +\vartheta < n(1 - \frac{1}{\kappa })$, and

$$\begin{aligned} \begin{aligned} {\left\| |\cdot |^{\vartheta } K(t,\cdot ) \right\| _{{\kappa , \infty }}}&\lesssim \langle t \rangle ^{\frac{1}{^\beta }(\vartheta - \alpha - n( 1 - \frac{1}{\kappa }))} = 1, \end{aligned} \end{aligned}$$

(4.30)

for every $t > 0$ if $-\alpha +\vartheta = n(1 - \frac{1}{\kappa })$.

Let $\omega$ be a non-negative number such that $\vartheta + \omega$ becomes an integer and that

$$\begin{aligned} n\left( \frac{1}{\kappa } -\frac{1}{2} \right) \le \omega < \frac{n}{\kappa }. \end{aligned}$$

(4.31)

Since $n \ge 2$, we can take $\omega$ satisfying above conditions. Let $\nu$ and its dual exponent $\nu ^\prime$ be the numbers defined by

$$\begin{aligned} \frac{1}{\kappa } = \frac{\omega }{n} + 1 - \frac{1}{\nu } = \frac{\omega }{n} + \frac{1}{\nu ^\prime }. \end{aligned}$$

(4.32)

Since $0 < 1/\nu ^\prime = 1/\kappa - \omega /n \le 1/2$ by assumtion (4.31), we have

$$\begin{aligned} 1< \nu \le 2 \le \nu ^\prime < \infty . \end{aligned}$$

(4.33)

Now we prove (4.29) under the assumption $-\alpha +\vartheta < n(1 - \frac{1}{\kappa })$. By Corollary A and Lemmas A and C together with the relation (4.33), we have

$$\begin{aligned} {\left\| |\cdot |^{\vartheta } K(t,\cdot ) \right\| _{{\kappa , 2}}}&={\left\| |\cdot |^{-\omega }|\cdot |^{\omega + \vartheta } K(t,\cdot ) \right\| _{{\kappa , 2}}} \lesssim {\left\| |\cdot |^{-\omega }\right\| _{{{\frac{n}{\omega },\infty }} }} {\left\| |\cdot |^{\vartheta + \omega } K(t,\cdot ) \right\| _{{\nu ^\prime ,2}}} \nonumber \\&\lesssim {\left\| |\cdot |^{\vartheta + \omega } K(t,\cdot ) \right\| _{{\nu ^\prime ,2}}} = {\left\| {{\mathcal {F}}}^{-1}[(-\Delta )^{\frac{\vartheta +\omega }{2}} {\hat{K}}(t,\cdot )] \right\| _{{\nu ^\prime ,2}}} \nonumber \\&\lesssim {\left\| (-\Delta )^{\frac{\vartheta +\omega }{2}} {\hat{K}}(t,\cdot ) \right\| _{{\nu ,2}}} \nonumber \\&\lesssim {\left\| (-\Delta )^{\frac{\vartheta +\omega }{2}} {\hat{K}}(t,\cdot ) \right\| _{{\nu ,\nu }}} ={\left\| (-\Delta )^{\frac{\vartheta +\omega }{2}} {\hat{K}}(t,\cdot ) \right\| _{{\nu }}}. \end{aligned}$$

(4.34)

Since the assumption $-\alpha +\vartheta < n(1 - \frac{1}{\kappa })$ and (4.32) imply that $- \alpha + \vartheta + \omega < \frac{n}{\nu }$, we can take $k = \vartheta + \omega$ in (4.25). Substituting the inequality into (4.34), we obtain (4.29).

Next we prove (4.30) under the assumption $-\alpha +\vartheta = n(1 - \frac{1}{\kappa })$. By Corollary A and Lemma C together with (4.33), we have

$$\begin{aligned} {\left\| |\cdot |^{\vartheta } K(t,\cdot ) \right\| _{{\kappa ,\infty }}}&={\left\| |\cdot |^{-\omega } |\cdot |^{\omega + \vartheta } K(t,\cdot ) \right\| _{{\kappa ,\infty }}} \nonumber \\&\lesssim {\left\| |\cdot |^{\vartheta + \omega } K(t,\cdot ) \right\| _{{\nu ^\prime ,\infty }}} = {\left\| {{\mathcal {F}}}^{-1}[(-\Delta )^{\frac{\vartheta +\omega }{2}} {\hat{K}}(t,\cdot )] \right\| _{{\nu ^\prime ,\infty }}} \nonumber \\&\lesssim {\left\| (-\Delta )^{\frac{\vartheta +\omega }{2}} {\hat{K}}(t,\cdot ) \right\| _{{\nu ,\infty }}}. \end{aligned}$$

(4.35)

Since the assumption $-\alpha +\vartheta = n(1 - \frac{1}{\kappa })$ implies $- \alpha + \vartheta + \omega = \frac{n}{\nu }$, we can take $k = \vartheta + \omega$ in (4.26). Substituting the inequality into (4.35), we obtain (4.30).

(Step 3) We define $r_j \in (1,2]$ by $\frac{1}{q_j} + \frac{1}{r_j} = \frac{3}{2}$ $(j = 1,2)$. We estimate each term of the right-hand side of

$$\begin{aligned} {\left\| |\cdot |^{\vartheta }(K(t,\cdot ) * \varphi ) \right\| _{2}}&\lesssim {\left\| (|\cdot |^{\vartheta } K(t,\cdot ) )* \varphi \right\| _{2}} +{\left\| K(t,\cdot ) *(|\cdot |^\vartheta \varphi )\right\| _{2}}. \end{aligned}$$

(4.36)

If $q_1 \in (1,2)$, Lemma D yields

$$\begin{aligned} {\left\| (|\cdot |^{\vartheta } K(t,\cdot ))* \varphi \right\| _{2}}&\lesssim {\left\| |\cdot |^{\vartheta } K(t,\cdot ) \right\| _{{r_1,\infty }}} {\left\| \varphi \right\| _{{q_1,2} }}. \end{aligned}$$

(4.37)

The assumption (4.10) implies $-\alpha +\vartheta \le n (\frac{1}{q_1} - \frac{1}{2}) = n(1 - \frac{1}{r_1})$. Hence, noting that ${\left\| |\cdot |^{\vartheta } K(t,\cdot ) \right\| _{{r_1, \infty }}} \le {\left\| |\cdot |^{\vartheta } K(t,\cdot ) \right\| _{{r_1, 2}}}$ and substituting (4.29) or (4.30) with $(\vartheta , \kappa ) =(\vartheta , r_1)$ into (4.37), we obtain

$$\begin{aligned} {\left\| (|\cdot |^{\vartheta } K(t,\cdot ) )* \varphi \right\| _{2}}&\lesssim \langle t \rangle ^{{ \frac{1}{\beta } \left( -n(\frac{1}{q_1} - \frac{1}{2}) + \vartheta - \alpha \right) }} \Vert \varphi \Vert _{q_1}^\prime . \end{aligned}$$

(4.38)

In the case $q_1 = 1$, Young’s inequality yields

$$\begin{aligned} {\left\| (|\cdot |^{\vartheta } K(t,\cdot ))* \varphi \right\| _{2}}&\lesssim {\left\| |\cdot |^{\vartheta } K(t,\cdot ) \right\| _2} {\left\| \varphi \right\| _{{1} }} = {\left\| |\cdot |^{\vartheta } K(t,\cdot ) \right\| _{2,2}} {\left\| \varphi \right\| _{{1} }}. \end{aligned}$$

(4.39)

The assumption $\vartheta < \frac{n}{2} + \alpha$ implies $- \alpha + \vartheta < \frac{n}{2} = n(1 - \frac{1}{2})$. Hence, substituting (4.29) with $(\vartheta , \kappa ) =(\vartheta , 2)$ into (4.39), we see that (4.38) holds also for $q_1 = 1$.

We can estimate the second term of (4.36) in the same way: If $q_2 \in (1,2)$, Lemma D yields

$$\begin{aligned} {\left\| K(t,\cdot ) *(|\cdot |^\vartheta \varphi )\right\| _{2}}&\lesssim {\left\| K(t,\cdot )\right\| _{{r_2,\infty } }} {\left\| |\cdot |^\vartheta \varphi \right\| _{{q_2,2}}}. \end{aligned}$$

(4.40)

The assumption (4.10) implies $-\alpha \le n (\frac{1}{q_2} - \frac{1}{2}) = n(1 - \frac{1}{r_2})$. Hence, substituting (4.29) or (4.30) with $(\vartheta , \kappa ) =(0, r_2)$ into (4.40), we obtain

$$\begin{aligned} {\left\| K *(|\cdot |^\vartheta \varphi )\right\| _{2}}&\lesssim \langle t \rangle ^{{ \frac{1}{\beta } \left( -n(\frac{1}{q_2} - \frac{1}{2}) - \alpha \right) }} \Vert | x |^\vartheta \varphi \Vert _{q_2}^\prime . \end{aligned}$$

(4.41)

In the case $q_2 = 1$, Young’s inequality yields

$$\begin{aligned} {\left\| K(t,\cdot ) *(|\cdot |^\vartheta \varphi )\right\| _{2}}&\lesssim {\left\| K(t,\cdot )\right\| _{2}} {\left\| |\cdot |^\vartheta \varphi \right\| _1} = {\left\| K(t,\cdot )\right\| _{2,2}} {\left\| |\cdot |^\vartheta \varphi \right\| _1}. \end{aligned}$$

(4.42)

Since $-\alpha < \frac{n}{2} = n (1 - \frac{1}{2})$, we have (4.29) with $(\vartheta , \kappa ) =(0, 2)$, which together with (4.42) yields (4.41) with $q_2 = 1$.

Hence, (4.38) and (4.41) hold for every case. Substituting (4.38) and (4.41) into (4.36), we obtain (4.11). $\square$

Lemma 2

Assume that $0 \le s_2 \le s_1$ and $\vartheta \ge 0$ satisfy $\vartheta - s_1 + s_2 < \frac{n}{2} - 2 \sigma$. If $q_j \in [1,2)$ $(j = 1,2,3,4)$ satisfy

$$\begin{aligned} \frac{1}{q_1} \ge \frac{1}{2} + \frac{2\sigma + \vartheta - s_1 + s_2 }{n}, \; \frac{1}{q_2} \ge \frac{1}{2} + \frac{2 \sigma - s_1 + s_2}{n}, \; \frac{1}{q_3} \ge \frac{1}{2} + \frac{ \vartheta - s_1 + s_2 }{n}, \;\nonumber \\ \end{aligned}$$

(4.43)

then the following hold provided the right-hand sides are finite:

$$\begin{aligned}&{\left\| |\cdot |^{\vartheta }(-\Delta )^{\frac{s_1}{2}} \left( {{\mathcal {F}}}^{-1}[\widehat{K_1^+}(t,\cdot ) \chi _{low}] * \varphi \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_1} - \frac{1}{2}) + \frac{\vartheta - s_1 + s_2 }{2} + \sigma \right) }} \Vert (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{q_1}}}^\prime \nonumber \\&\qquad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_2}} - \frac{1}{2}) - \frac{s}{2} + \sigma \right) }} \Vert |\cdot |^{\vartheta }(-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{{q_2}} }}^\prime , \end{aligned}$$

(4.44)

$$\begin{aligned}&{\left\| |\cdot |^{\vartheta } (-\Delta )^{\frac{s_1}{2}} \left( {{\mathcal {F}}}^{-1}[\widehat{K_1^-}(t,\cdot ) \chi _{low}] * \varphi \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{ \sigma } \left( -\frac{n}{2}(\frac{1}{q_1} - \frac{1}{2}) + \frac{{\vartheta - s_1 + s_2} }{2} + \sigma \right) }} \Vert (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{q_1}}}^\prime \nonumber \\&\qquad + \langle t \rangle ^{{ \frac{1}{\sigma } \left( -\frac{n}{2}(\frac{1}{{q_2}} - \frac{1}{2}) - \frac{s}{2} + \sigma \right) }} \Vert |\cdot |^{\vartheta } (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{{q_2}} }}^\prime , \end{aligned}$$

(4.45)

$$\begin{aligned}&{\left\| |\cdot |^{\vartheta } (-\Delta )^{\frac{s_1}{2}} \left( K_{1,low}(t,\cdot ) * \varphi \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_1} - \frac{1}{2}) + \frac{{\vartheta - s_1 + s_2} }{2} + \sigma \right) }} \Vert (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{q_1}}}^\prime \nonumber \\&\qquad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_2}} - \frac{1}{2}) - \frac{s}{2} + \sigma \right) }} \Vert |\cdot |^{\vartheta } (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{{q_2}} }}^\prime , \end{aligned}$$

(4.46)

$$\begin{aligned}&{\left\| |\cdot |^{\vartheta } (-\Delta )^{\frac{s_1}{2}} \left( {{\mathcal {F}}}^{-1}[\widehat{K_0^+}(t,\cdot ) \chi _{low}] * \varphi \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_3} - \frac{1}{2}) + \frac{{\vartheta - s_1 + s_2} }{2} \right) }} \Vert (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{q_3}}}^\prime \nonumber \\&\qquad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_4}} - \frac{1}{2}) - \frac{s}{2} \right) }} \Vert |\cdot |^{\vartheta } (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{q_4}}^\prime , \end{aligned}$$

(4.47)

$$\begin{aligned}&{\left\| |\cdot |^{\vartheta } (-\Delta )^{\frac{s_1}{2}} \left( {{\mathcal {F}}}^{-1}[\widehat{K_0^-}(t,\cdot ) \chi _{low}] * \varphi \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{\sigma } \left( -\frac{n}{2}(\frac{1}{q_3} - \frac{1}{2}) + \frac{{\vartheta - s_1 + s_2} }{2} + 2 \sigma - 1 \right) }} \Vert (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{q_3}}}^\prime \nonumber \\&\qquad + \langle t \rangle ^{{ \frac{1}{\sigma } \left( -\frac{n}{2}(\frac{1}{{q_4}} - \frac{1}{2} ) - \frac{s}{2}+ 2 \sigma - 1) \right) }} \Vert |\cdot |^{\vartheta } (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{{q_4}} }}^\prime , \end{aligned}$$

(4.48)

$$\begin{aligned}&{\left\| |\cdot |^{\vartheta } (-\Delta )^{\frac{s_1}{2}} \left( K_{0,low}(t,\cdot ) * \varphi \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_3} - \frac{1}{2}) + \frac{{\vartheta - s_1 + s_2} }{2} \right) }} \Vert (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{q_3}}}^\prime \nonumber \\&\qquad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_4}} - \frac{1}{2}) - \frac{s}{2} \right) }} \Vert |\cdot |^{\vartheta } (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{{q_4}} }}^\prime , \end{aligned}$$

(4.49)

where $\Vert \cdot \Vert ^\prime _q$ is defined by (4.12).

Proof

Since

$$\begin{aligned} \begin{aligned}&{\left\| |\cdot |^{\vartheta }(-\Delta )^{\frac{s_1}{2}} \left( {{\mathcal {F}}}^{-1}[{\hat{K}}(t,\cdot ) \chi _{low}] * \varphi \right) \right\| _{2}} \\&\quad ={\left\| |\cdot |^\vartheta {{\mathcal {F}}}^{-1} [|\xi |^{s_1-s_2} {\hat{K}}(t,\xi ) \chi _{low} |\xi |^{s_2} {\hat{\varphi }}(\xi )] \right\| _{2}} \\&\quad = {\left\| |\cdot |^{\vartheta }(-\Delta )^{\frac{s_1-s_2}{2}} {{\mathcal {F}}}^{-1} [{\hat{K}}(t,\cdot )\chi _{low}]*(-\Delta ) ^{{\frac{s_2}{2} }} \varphi \right\| _2}, \end{aligned} \end{aligned}$$

the conclusion reduces to the case $s_2 = 0$ by taking $s_1 - s_2$ and $(-\Delta ) ^{s_2} \varphi$ as $s_1$ and $\varphi$ respectively.

Let $\lambda _\pm$ be the functions defined by (4.4). From (4.5), it follows that

$$\begin{aligned} -2 \rho ^{2 - 2 \sigma } \le \lambda _+(\rho )&= -\frac{2 \rho ^{2 - 2 \sigma }}{1 +\sqrt{1-4\rho ^{2(1 - 2\sigma )} } } \le - \rho ^{2 - 2 \sigma }, \end{aligned}$$

(4.50)

$$\begin{aligned} - \rho ^{2 \sigma } \le \lambda _- (\rho )&\le -\frac{1}{2} \rho ^{2 \sigma }, \end{aligned}$$

(4.51)

$$\begin{aligned} \lambda _+(\rho ) - \lambda _- (\rho )&= \rho ^{2 \sigma } \sqrt{1-4\rho ^{2(1 - 2\sigma )} } \sim \rho ^{2 \sigma }, \end{aligned}$$

(4.52)

on the support of $\chi _{low}$.

We first prove (4.44). It is written that

$$\begin{aligned} (-\Delta )^{\frac{s_1}{2}} \left( {{\mathcal {F}}}^{-1}[\widehat{K_1^+}(t,\cdot ) \chi _{low}] * \varphi \right)&= {{\mathcal {F}}}^{-1}[|\xi |^{s_1} \widehat{K_1^+}(t,\cdot ) \chi _{low}]* \varphi , \end{aligned}$$

where $K_1^+$ is defined by (4.6). By definition, $K(t,x) = {{\mathcal {F}}}^{-1}[|\xi |^{s_1} \widehat{K_1^+}(t,\cdot ) \chi _{low}]$ has the form (4.15) with

$$\begin{aligned} \upsilon (\rho ) = \frac{\rho ^{s_1}}{\lambda _+(\rho ) - \lambda _-(\rho )} \chi _{low}, \quad \lambda (\rho ) = \lambda _{+}(\rho ). \end{aligned}$$

By using (4.50) and (4.52), we easily see that $\upsilon$ and $\lambda$ above satisfy the assumption of Corollary 1 with $\alpha = s_1 - 2 \sigma$, $\beta = 2(1 -\sigma )$. The assumption $n \ge 2$ and $0< 2 \sigma < 1$ implies $\alpha > -\frac{n}{2}$ and $\beta > 0$, that is, $\alpha$ and $\beta$ satisfy the assumption of Corollary 1. Definition of $\alpha$ and (4.43) imply (4.10) (here we note that we assume $s_2 = 0$). Hence, applying Corollary 1, we obtain (4.44).

$K(t,x) = {{\mathcal {F}}}^{-1}[|\xi |^{s_1} \widehat{K_1^-}(t,\cdot ) \chi _{low}] \,(K_1^-$ is defined by (4.6)) has the form (4.15) with

$$\begin{aligned} \upsilon (\rho ) = \frac{-\rho ^{s_1}}{\lambda _+(\rho ) - \lambda _-(\rho )} \chi _{low}, \quad \lambda (\rho ) = \lambda _{-}(\rho ). \end{aligned}$$

By using (4.51) and (4.52), we easily see that $\upsilon$ and $\lambda$ above satisfy the assumption of Corollary 1 for $\alpha =s_1 - 2 \sigma (> -\frac{n}{2}), \beta = 2 \sigma (> 0 )$ and therefore, (4.45) holds in the same way as in the proof of (4.44).

Since $\sigma < 1- \sigma$ by the assumption that $\sigma \in (0,1/2)$, the estimate (4.46) follows from (4.44) and (4.45).

$K(t,x) = {{\mathcal {F}}}^{-1}[|\xi |^{s_1} \widehat{K_0^+}(t,\cdot ) \chi _{low}] \,(K_0^+$ is defined by (4.7)) has the form (4.15) with

$$\begin{aligned} \upsilon (\rho ) = \frac{-\rho ^{s_1} \lambda _-(\rho )}{\lambda _+(\rho ) - \lambda _-(\rho )} \chi _{low}, \quad \lambda (\rho ) = \lambda _{+}(\rho ), \end{aligned}$$

By using (4.50)–(4.52), we easily see that $\upsilon$ and $\lambda$ satisfy the assumption of Corollary 1 with $\alpha = s_1, \beta = 2(1 -\sigma )$. Definition of $\alpha$ and (4.43) imply assumtion (4.10) of $q_1$ for $q_1 = q_3$. Since $q_4 < 2$, the assumption on $q_2$ of (4.10) holds for $q_2 = q_4$. Hence, we can apply Corollary 1 to obtain (4.47).

Kernel $K = {{\mathcal {F}}}^{-1}[|\xi |^{s_1} \widehat{K_0^-}(t,\cdot ) \chi _{low}]$ $(K_0^-$ is defined by (4.7)) has the form (4.15) with

$$\begin{aligned} \upsilon (\rho ) = \frac{\rho ^{s_1} \lambda _+(\rho )}{\lambda _+(\rho ) - \lambda _-(\rho )} \chi _{low}, \quad \lambda (\rho ) = \lambda _{-}(\rho ), \end{aligned}$$

By using (4.50)–(4.52), we easily see that $\upsilon$ and $\lambda$ above satisfy the assumption of Corollary 1 with $\alpha =s_1 + 2(1- 2 \sigma ) (> 0), \beta = 2 \sigma (> 0)$. Definition of $\alpha$ and (4.43) imply the assumption on $q_1$ of (4.10) for $q_1 = q_3$. The assumption on $q_2$ of (4.10) holds for $q_2 = q_4$ in the same reason as above. Hence, (4.48) holds by (4.11).

Since $\sigma \in (0,1/2)$, inequality (4.49) follows from (4.47) and (4.48). $\square$

Lemma 3

Let $\vartheta \ge 0$ and $0 \le s_2 \le s_1$ such that $\vartheta - s_1 + s_2 < \frac{n}{2} - 2 \sigma$. Assume that $q_j \in [1,2)$ $(j = 1,2,3,4)$ satisfy

$$\begin{aligned} \frac{1}{q_1} \ge \frac{1}{2} + \frac{2\sigma + \vartheta - s_1 + s_2 }{n}, \; \frac{1}{q_2} \ge \frac{1}{2} + \frac{2 \sigma - s_1 + s_2}{n}, \; \frac{1}{q_3} \ge \frac{1}{2} + \frac{ \vartheta - s_1 + s_2 }{n}. \;\nonumber \\ \end{aligned}$$

(4.53)

Then the following hold provided the right-hand sides are finite:

$$\begin{aligned}&{\left\| |\cdot |^{\vartheta } (-\Delta )^{\frac{s_1}{2}} \left( {{\mathcal {F}}}^{-1}[(\widehat{K_1^+}(t,\cdot ) - |\xi |^{-2 \sigma } e^{- |\xi |^{2(1 - \sigma )} t} )\chi _{low}] * \varphi \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_1} - \frac{1}{2}) + \frac{{\vartheta - s_1 + s_2} }{2} + 3\sigma - 1 \right) }} \Vert (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{{{q_1}}}^\prime \nonumber \\&\qquad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_2}} - \frac{1}{2}) + 3\sigma - 1\right) }} \Vert |\cdot |^{\vartheta } (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{q_2}^\prime , \end{aligned}$$

(4.54)

$$\begin{aligned}&{\left\| |\cdot |^{\vartheta } (-\Delta )^{\frac{s_1}{2}} \left( {{\mathcal {F}}}^{-1}[(\widehat{K_0^+}(t,\xi ) - e^{- |\xi |^{2(1 - \sigma )} t} )\chi _{low}] * \varphi \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_1} - \frac{1}{2}) + \frac{{\vartheta - s_1 + s_2} }{2} + 2\sigma - 1 \right) }} \Vert (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{q_1}^\prime \nonumber \\&\qquad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_2}} - \frac{1}{2}) + 2\sigma - 1 \right) }} \Vert |\cdot |^{\vartheta } (-\Delta )^{\frac{s_2}{2}}\varphi \Vert _{q_2}^\prime , \end{aligned}$$

(4.55)

where $\Vert \cdot \Vert ^\prime _q$ is defined by (4.12).

Proof

We first prove (4.54). Let $\lambda _\pm$ be the functions defined by (4.5). By the same reason as in the proof of Lemma 2, we may assume that $s_2 = 0$. It follows from the definition that

$$\begin{aligned}&|\xi |^{s_1} \widehat{K_1^+}(t,\xi ) - |\xi |^{s_1 -2 \sigma } e^{- |\xi |^{2(1 - \sigma )} t} \nonumber \\&\quad = |\xi |^{s_1 -2 \sigma } \Bigg ( \frac{1}{\sqrt{1-4|\xi |^{2(1 - 2\sigma )}}} \exp \Bigg (\frac{- 2 |\xi |^{2(1 - \sigma )}t}{1 + \sqrt{1-4|\xi |^{2(1 - 2\sigma )}} } \Bigg ) - e^{- |\xi |^{2(1 - \sigma )} t} \Bigg ) \nonumber \\&\quad = \frac{|\xi |^{s_1 -2 \sigma } e^{- |\xi |^{2(1 - \sigma )} t} }{\sqrt{1-4|\xi |^{2(1 - 2\sigma )}}} \left( \exp \Big (\frac{ - 4|\xi |^{2(2 - 3 \sigma ) } t}{(1+\sqrt{1-4|\xi |^{2(1 - 2\sigma )}})^2 } \Big ) -1 \right) \nonumber \\&\qquad - \frac{4 |\xi |^{s_1 + 2(1 - 3 \sigma ) } e^{- |\xi |^{2(1 - \sigma )} t} }{(1 + \sqrt{1-4|\xi |^{2(1 - 2\sigma )}})\sqrt{1-4|\xi |^{2(1 - 2\sigma )}}}. \nonumber \\&\quad := M_{1,1} + M_{1,2} \quad \text {(we put)}. \end{aligned}$$

(4.56)

We easily see that

$$\begin{aligned} \upsilon (\rho ) = \frac{- \rho ^{s_1 -2 \sigma }}{\sqrt{1-4\rho ^{2(1 - 2\sigma )}}}, \quad \lambda (\rho ) = -\rho ^{2(1 - \sigma )}, \quad \mu (\rho ) = \frac{4\rho ^{2(2 - 3 \sigma ) }}{(1+\sqrt{1-4\rho ^{2(1 - 2\sigma )}})^2 } \end{aligned}$$

satisfy the assumption (4.17)–(4.19) of Corollary 2 with $\alpha = s_1 - 2 \sigma , \beta = 2(1 -\sigma ), \gamma = 2(2 - 3 \sigma )$. Then the assumption $n \ge 2$ and $2 \sigma < 1$ implies $\alpha - \beta + \gamma = s_1 + 2(1 - 3\sigma ) > -\frac{n}{2}$ and $\beta > 0$, that is, $\alpha$, $\beta$ and $\gamma$ satisfy the assumption of Corollary 2. The assumption (4.53) and the definition of $\alpha , \beta , \gamma$ above imply

$$\begin{aligned} \frac{1}{q_1}&\ge \frac{1}{2} + \frac{2\sigma + \vartheta - s_1 }{n} \ge \frac{1}{2} + \frac{ \vartheta - \alpha + \beta - \gamma }{n}, \; \\ \frac{1}{q_2}&\ge \frac{1}{2} + \frac{2 \sigma - s_1 }{n} \ge \frac{1}{2} + \frac{ - \alpha + \beta - \gamma }{n}, \; \end{aligned}$$

that is, (4.20) holds. Hence, we can apply Corollary 2 for the above choice to obtain

$$\begin{aligned} \begin{aligned}&|M_{1,1}| \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_1} - \frac{1}{2}) + \frac{\vartheta - s_1 }{2} + 3\sigma - 1 \right) }} \Vert \varphi \Vert _{{q_1}}^\prime \\&\qquad \quad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_2}} - \frac{1}{2}) - \frac{s}{2} + 3\sigma - 1 \right) }} \Vert |\cdot |^{\vartheta } \varphi \Vert _{q_2}^\prime . \end{aligned} \end{aligned}$$

(4.57)

We also see that

$$\begin{aligned} \upsilon (\rho ) = \frac{4 \rho ^{s_1 + 2(1 - 3 \sigma ) }}{(1+\sqrt{1-4\rho ^{2(1 - 2\sigma )}})\sqrt{1-4\rho ^{2(1 - 2\sigma )}}}, \quad \lambda (\rho ) = -\rho ^{2(1 - \sigma )} \end{aligned}$$

satisfy the assumption of Corollary 1 with $\alpha = s_1 + 2(1 - 3 \sigma ) (>- \frac{n}{2})$ and $\beta = 2(1 -\sigma ) (> 0)$. The assumption (4.10) is satisfied by (4.53) together with the definition of $\alpha$. Hence, we can apply Corollary 1 to obtain

$$\begin{aligned} \begin{aligned}&|M_{1,2}| \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_1} - \frac{1}{2}) + \frac{\vartheta - s_1 }{2} + 3\sigma - 1 \right) }} \Vert \varphi \Vert _{{{q_1}}}^\prime \\&\qquad \quad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_2}} - \frac{1}{2}) - \frac{s_1}{2} + 3\sigma - 1 \right) }} \Vert |\cdot |^{\vartheta } \varphi \Vert _{q_2}^\prime . \end{aligned} \end{aligned}$$

(4.58)

Inequality (4.54) follows from (4.56), (4.57) and (4.58).

Next we prove (4.55). It follows from (4.5), (4.50) and (4.52) that

$$\begin{aligned}&|\xi |^{s_1} \widehat{K_0^+}(t,\xi ) - |\xi |^{s_1} e^{- |\xi |^{2(1 - \sigma )} t}\nonumber \\&\quad = |\xi |^{s_1} \left( \frac{- \lambda _-(|\xi |)}{\lambda _+(|\xi |) - \lambda _-(|\xi |) } - 1 \right) e^{\lambda _+ (|\xi |) t} + |\xi |^{s_1} \left( e^{\lambda _+ (|\xi |)t }- e^{-|\xi |^{2 (1- \sigma )}t} \right) \nonumber \\&\quad = - \frac{2 |\xi |^{s_1 + 2(1 - 2 \sigma ) }}{(1 + \sqrt{1-4|\xi |^{2(1 - 2\sigma )}})\sqrt{1-4|\xi |^{2(1 - 2\sigma )}}} \exp (\lambda _+(|\xi |) t) \nonumber \\&\qquad + |\xi |^{s_1 } \exp (-|\xi |^{2(1 - \sigma )}t) \left( \exp \Bigg (\frac{- 4|\xi |^{2(2 - 3 \sigma ) }t}{(1+\sqrt{1-4|\xi |^{2(1 - 2\sigma )}})^2 } \Bigg ) -1 \right) \end{aligned}$$

(4.59)

$$\begin{aligned}&\quad =: M_{2,1} + M_{2,2} \quad (\text {we put}). \end{aligned}$$

(4.60)

We easily see that

$$\begin{aligned} \upsilon (\rho )&= \frac{2 \rho ^{s_1 + 2(1 -2 \sigma ) }}{(1+\sqrt{1-4\rho ^{2(1 - 2\sigma )}})\sqrt{1-4\rho ^{2(1 - 2\sigma )}}}, \\ \lambda (\rho )&=\lambda _+(\rho ) = \frac{- 2 \rho ^{2 - 2 \sigma }t}{1 +\sqrt{1-4\rho ^{2(1 - 2\sigma )} } } \end{aligned}$$

satisfy the assumption of Corollary 1 with $\alpha = s_1 + 2(1 - 2 \sigma ) , \beta = 2(1 -\sigma )$. The assumption (4.53) and the definition of $\alpha$ above yield

$$\begin{aligned}&\frac{1}{q_3} \ge \frac{1}{2} + \frac{ \vartheta - s_1 }{n} \ge \frac{1}{2} + \frac{ \vartheta - \alpha }{n}, \\&\quad \frac{1}{q_4} > \frac{1}{2} \ge \frac{1}{2} + \frac{- \alpha }{n}, \end{aligned}$$

that is, (4.10) holds for $q_1 = q_3$ and $q_2 = q_4$. Hence, we can apply Corollary 1 to obtain

$$\begin{aligned} |M_{2,1}|&\lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_3} - \frac{1}{2}) + \frac{{\vartheta - s_1 } }{2} + 2\sigma - 1 \right) }} \Vert \varphi \Vert _{q_3}^\prime \nonumber \\&\quad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_4}} - \frac{1}{2}) - \frac{s_1}{2}+ 2\sigma - 1 \right) }} \Vert |\cdot |^{\vartheta } \varphi \Vert _{q_4}^\prime . \end{aligned}$$

(4.61)

We also see that

$$\begin{aligned} \upsilon (\rho ) = \rho ^{s_1}, \quad \lambda (\rho ) = -\rho ^{2(1 - \sigma )}, \quad \mu (\rho ) = \frac{4\rho ^{2(2 - 3 \sigma ) }}{(1+\sqrt{1-4\rho ^{2(1 - 2\sigma )}})^2 } \end{aligned}$$

satisfy the assumption of Corollary 2 with $\alpha = s_1, \beta = 2(1 -\sigma ), \gamma = 2(2 - 3 \sigma )$. The assumption (4.53) and the definition of $\alpha , \beta , \gamma$ above yield

$$\begin{aligned}&\frac{1}{q_3} \ge \frac{1}{2} + \frac{\vartheta - s_1 }{n}> \frac{1}{2} + \frac{\vartheta - s_1-2(1 - 2\sigma ) }{n} = \frac{1}{2} + \frac{ \vartheta - \alpha + \beta - \gamma }{n}, \; \\&\frac{1}{q_4} > \frac{1}{2} \ge \frac{1}{2} + \frac{- s_1-2(1 - 2\sigma ) }{n} = \frac{1}{2} + \frac{ - \alpha + \beta - \gamma }{n}, \; \end{aligned}$$

that is, (4.20) holds for $q_1 = q_3, q_2 = q_4$. Hence, we can apply Corollary 2 to obtain

$$\begin{aligned} |M_{2,2}|&\lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{q_3} - \frac{1}{2}) + \frac{{\vartheta - s_1 } }{2} + 2\sigma - 1 \right) }} \Vert \varphi \Vert _{q_3}^\prime \nonumber \\&\quad + \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{{q_4}} - \frac{1}{2}) - \frac{s_1}{2}+ 2\sigma - 1 \right) }} \Vert |\cdot |^{\vartheta } \varphi \Vert _{q_4}^\prime . \end{aligned}$$

(4.62)

Inequality (4.55) follows from (4.60), (4.61) and (4.62). $\square$

4.2 Estimate of the kernels for high frequency part$(|\xi | \ge 1)$

In this subsection, we consider high frequency region: $|\xi | \ge 1$.

Lemma 4

For every $s \ge 0, \delta \ge 0$, the following hold.

$$\begin{aligned}&{\left\| (-\Delta )^{\frac{s }{2}} \left( K_{1,high}(t, \cdot ) * \varphi \right) \right\| _{2}} \lesssim e^{-\frac{t}{2}} \Vert (-\Delta )^{\frac{s }{2}} (1 - \Delta )^{-\frac{1}{2}} \varphi \Vert _{{2}}, \end{aligned}$$

(4.63)

$$\begin{aligned}&{\left\| \langle \cdot \rangle ^\delta \left( K_{1,high}(t, \cdot ) * \varphi \right) \right\| _{2}} \lesssim e^{-\frac{t}{2}} \Vert (1 - \Delta )^{-\frac{1}{2}} \langle \cdot \rangle ^\delta \varphi \Vert _{{2}}, \end{aligned}$$

(4.64)

$$\begin{aligned}&{\left\| (-\Delta )^{\frac{s }{2}} \left( K_{0,high}(t, \cdot ) * \varphi \right) \right\| _{2}} \lesssim e^{-\frac{t}{2}} \Vert (-\Delta )^{\frac{s }{2}} \varphi \Vert _{{2}}, \end{aligned}$$

(4.65)

$$\begin{aligned}&{\left\| \langle \cdot \rangle ^\delta \left( K_{0,high}(t, \cdot ) * \varphi \right) \right\| _{2}} \lesssim e^{-\frac{t}{2}} \Vert \langle \cdot \rangle ^\delta \varphi \Vert _{{2}}, \end{aligned}$$

(4.66)

provided the right-hand sides are finite.

Proof

We easily see that

$$\begin{aligned} \langle \xi \rangle |\widehat{K_1^{\pm }}(t,\xi ) |&\lesssim e^{-\frac{|\xi |^{2 \sigma }}{2}t} \le e^{-\frac{t}{2}}, \end{aligned}$$

(4.67)

on the support of $\chi _{high}$. Hence,

$$\begin{aligned} \begin{aligned}&{\left\| (-\Delta )^{\frac{s }{2}} \left( K_{1,high}(t, \cdot ) * \varphi \right) \right\| _{2}} ={\left\| |\xi |^{s } \widehat{K_1}(t,\xi ) \chi _{high}(\xi ) {\hat{\varphi }}(\xi ) \right\| _{2}} \\&\quad \le {\left\| \widehat{K_1}(t,\xi ) \chi _{high}(\xi )\langle \xi \rangle \right\| _{\infty }} {\left\| |\xi |^{s } \langle \xi \rangle ^{ - 1 } \hat{\varphi }(\xi ) \right\| _{2}} \lesssim e^{-\frac{t}{2}} \Vert (-\Delta )^{\frac{s }{2}} (1 - \Delta )^{-\frac{1}{2}} \varphi \Vert _{{2}}, \end{aligned} \end{aligned}$$

that is, (4.63) holds.

In the proof of [9, p. 10] (see also [7, p. 643]), the following Leibniz rule is shown:

$$\begin{aligned} \Vert (1 - \Delta )^{\frac{\vartheta }{2} } (\varphi \psi ) \Vert _{{2}} \le \sum _{j = 1}^n \sum _{k = 1}^{[\vartheta ]+1}\Vert \partial _j^k \varphi \Vert _{{\infty }} \Vert (1 - \Delta )^{\frac{\vartheta }{2} } \psi \Vert _{{2}}. \end{aligned}$$

(4.68)

Since $|\partial _j^k (\widehat{K_1}(t,\xi ) \chi _{high}(\xi ) \langle \xi \rangle )| \le C_{k} e^{-\frac{t}{2}}$ for every nonnegative integer k, we have

$$\begin{aligned} \begin{aligned}&{\left\| |\cdot |^\vartheta \left( K_{1,high} * \varphi \right) \right\| _{2}} ={\left\| |(-\Delta )^{\frac{\vartheta }{2}} \left( (\widehat{K_1}(t,\xi ) \chi _{high}(\xi )\langle \xi \rangle ) ( \langle \xi \rangle ^{-1} {\hat{\varphi }}(\xi ) ) \right) \right\| _{2}} \\&\quad \lesssim e^{-\frac{t}{2}} {\left\| \langle \xi \rangle ^{-1} (1 - \Delta )^{\vartheta /2} {\hat{\varphi }}(\xi ) \right\| _{2}} \sim e^{-\frac{t}{2}} \Vert (1 - \Delta )^{-\frac{1}{2}} \langle \cdot \rangle ^\vartheta \varphi \Vert _{{2}}. \end{aligned} \end{aligned}$$

Taking $\vartheta = 0$ and $\delta$ in this inequality, we obtain (4.64).

We can prove (4.65) and (4.66) in the same way. $\square$

4.3 Estimate of the kernels for middle frequency part

In this subsection, we consider the region: $|\xi | \in [2^{{-\frac{3}{1 - 2\sigma } }}, 2]$.

Lemma 5

There is a constant $\varepsilon _{\sigma } \in (0, \frac{1}{2})$ such that the following hold for every $s \ge 0$, $\delta \ge 0$:

$$\begin{aligned}&{\left\| (-\Delta )^{\frac{s}{2}} \left( K_{1,mid}(t, \cdot ) * \varphi \right) \right\| _{2}} \lesssim e^{-\varepsilon _{\sigma } t} \Vert (1 - \Delta )^{-\frac{1}{2}} \varphi \Vert _{{2}}, \end{aligned}$$

(4.69)

$$\begin{aligned}&{\left\| \langle \cdot \rangle ^\delta \left( K_{1,mid}(t, \cdot ) * \varphi \right) \right\| _{2}} \lesssim e^{-\varepsilon _{\sigma } t}\Vert (1 - \Delta )^{-\frac{1}{2}} \langle \cdot \rangle ^\delta \varphi \Vert _{{2}}, \end{aligned}$$

(4.70)

$$\begin{aligned}&{\left\| (-\Delta )^{\frac{s}{2}} \left( K_{0,mid}(t, \cdot ) * \varphi \right) \right\| _{2}} \lesssim e^{-\varepsilon _{\sigma } t} \Vert \varphi \Vert _{{2}}, \end{aligned}$$

(4.71)

$$\begin{aligned}&{\left\| \langle \cdot \rangle ^\delta \left( K_{0,mid}(t, \cdot ) * \varphi \right) \right\| _{2}} \lesssim e^{-\varepsilon _{\sigma } t} \Vert \langle \cdot \rangle ^\delta \varphi \Vert _{{2}}, \end{aligned}$$

(4.72)

provided the right-hand sides are finite.

Proof

By definitions (4.3), (4.4) and (4.5), we have

$$\begin{aligned} \begin{aligned} | \widehat{K_1}(t,\xi ) |&= \left| \frac{e^{\lambda _+(|\xi |) t}- e^{\lambda _-(|\xi |) t}}{\lambda _+(|\xi |) - \lambda _-(|\xi |)} \right| = t \left| e^{(\theta \lambda _+(|\xi |) + (1 - \theta )\lambda _-(|\xi |))t } \right| \\&= t \left| e^{\frac{t}{2} (- |\xi |^{2 \sigma } + (2 \theta - 1) |\xi |^{2 \sigma } \sqrt{1-4|\xi |^{2(1 - 2\sigma )}} )} \right| \end{aligned} \end{aligned}$$

(4.73)

for some $\theta \in (0,1)$. Hence,

$$\begin{aligned} | \widehat{K_1}(t,\xi ) | = t e^{-\frac{1}{2} |\xi |^{2 \sigma }t} \le t e^{-2^{- \frac{1}{1 - 2\sigma }}t} \lesssim e^{-2 \varepsilon _{\sigma }t} \end{aligned}$$

(4.74)

in the case $2|\xi |^{1 - 2 \sigma } \ge 1$, where $\varepsilon _{\sigma } = 2^{- \frac{6}{1 - 2 \sigma }- 1}$. Next, we consider the case $2|\xi |^{1 - 2 \sigma } \le 1$. Then

$$\begin{aligned}&\frac{1}{2} |\xi |^{2 \sigma } (-1 + (2 \theta - 1) \sqrt{1-4|\xi |^{2(1 - 2\sigma )}} ) \le \frac{1}{2} |\xi |^{2 \sigma }(-1 + \sqrt{1-4|\xi |^{2(1 - 2\sigma )}} ) \\&\quad = \frac{ - 2|\xi |^{2 - 2\sigma }}{ 1 + \sqrt{1-4|\xi |^{2(1 - 2\sigma )}} } \le - |\xi |^{2 - 2\sigma } \le -2^{- \frac{6(1-\sigma )}{1 - 2 \sigma }} \le - 2 \varepsilon _{\sigma }, \end{aligned}$$

and thus,

$$\begin{aligned} | \widehat{K_1}(t,\xi ) | \le t e^{-2 \varepsilon _{\sigma } t } \lesssim e^{- \varepsilon _{\sigma } t} \end{aligned}$$

on $[2^{-\frac{3}{1-2\sigma }}, 2^{-\frac{1}{1-2\sigma }}]$, which together with (4.74) yields

$$\begin{aligned} \langle \xi \rangle | \widehat{K_1}(t,\xi )\chi _{mid}(\xi ) | \lesssim e^{- \varepsilon _{\sigma } t}. \end{aligned}$$

(4.75)

Calculating in the same way as in the proof of (4.63) by using (4.75) instead of (4.67), and noting that $-\Delta$ is bounded operator on the support $\chi _{mid}$, we obtain (4.69).

In the same way as in the proof of (4.75), we see that

$$\begin{aligned} \Vert (1 - \Delta )^{\frac{k}{2}} \widehat{K_1}(t,\cdot ) \chi _{mid} \Vert _{{\infty }} \le C_k e^{- \varepsilon _{\sigma } t}, \end{aligned}$$

(4.76)

for every $k \in {{\mathbb {N}}}\cup \{0\}$. Then by the same calculation as in the proof of (4.64), we obtain (4.70).

We can estimate

$$\begin{aligned} \widehat{K_0}&=\frac{ \lambda _+ e^{\lambda _- t}- \lambda _- e^{\lambda _+ t}}{\lambda _+ - \lambda _-} = \frac{ e^{\lambda _+} - e^{\lambda _-} }{\lambda _+ - \lambda _-}\lambda _+ + \lambda _-, \end{aligned}$$

in the same way, and obtain the assertion for $K_{0,mid}$. $\square$

5 Asymptotic profile of the solutions of linear equation

In this section, we prove Theorem 1.

Since the solution u of (1.3) is written as

$$\begin{aligned} u(t,x)&= \left( K_0(t,\cdot )*u_0 \right) (x) + \left( K_1(t,\cdot )*u_1 \right) (x), \end{aligned}$$

the conclusion of Theorem 1 follows from the following lemma.

Lemma 6

Let $u_j \in L^1 \cap L^2$ for $j = 0,1$. Then the following hold.

$$\begin{aligned}&\left\| K_1(t,\cdot )*u_1 - G_{\sigma }(t,x) \int \nolimits _{{{\mathbb {R}}}^n}u_1 (y)dy \right\| _2 \nonumber \\&\quad \lesssim \langle t \rangle ^{{\max \{ \frac{1}{1 - \sigma }\left( -\frac{n}{4} + 3\sigma - 1 \right) , \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) \} }} \Vert u_1 \Vert _{{{1}}} + e^{{-\varepsilon _\sigma t}} \Vert u_1 \Vert _{{H^{-2 \sigma } }}\nonumber \\&\qquad + {t}^{ \frac{1}{1-\sigma }(-\frac{n}{4} + \sigma -\frac{\theta }{2} )} \left\| | \cdot |^\theta u_1 \right\| _1, \end{aligned}$$

(5.1)

$$\begin{aligned} \nonumber \\&\left\| K_0(t,\cdot )*u_0 - H_{\sigma }(t,x) \int \nolimits _{{{\mathbb {R}}}^n} u_0(y)dy \right\| _2 \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma }\left( -\frac{n}{4} + 2\sigma - 1 \right) }} \Vert u_0\Vert _1 + e^{-\varepsilon _{\sigma } t} \Vert u_0 \Vert _2 + {t}^{ \frac{1}{1-\sigma }(-\frac{n}{4} - \frac{\theta }{2} )} \left\| | \cdot |^\theta u_0 \right\| _1. \end{aligned}$$

(5.2)

Proof

First we prove (5.1). We have

$$\begin{aligned}&\left\| K_1(t,\cdot )*u_1 - G_{\sigma }(t,\cdot ) \int \nolimits _{{{\mathbb {R}}}^n}u_1 (y)dy \right\| \nonumber \\&\quad \le {\left\| {{\mathcal {F}}}^{-1}[\widehat{K_1^+}(t,\xi ) \chi _{low}] * u_1 - {{\mathcal {F}}}^{-1}[ |\xi |^{-2\sigma } e^{- |\xi |^{2(1 - \sigma )} t} \chi _{low}] * u_1\right\| _2} \nonumber \\&\qquad + {\left\| {{\mathcal {F}}}^{-1}[\widehat{K_1^-}(t,\xi ) \chi _{low}] * u_1\right\| _{2}} +{\left\| K_{1,mh}(t,\cdot ) * u_1 \right\| _{2}} \nonumber \\&\qquad + {\left\| {{\mathcal {F}}}^{-1}\left[ \widehat{G_{\sigma }}(t,\xi )\chi _{mh} \right] * u_1 \right\| _{2}} + \Vert G_{\sigma }(t,\cdot )*u_1 - G_{\sigma }(t,\cdot ) \int \nolimits _{{{\mathbb {R}}}^n}u_1 (y)dy \Vert _2 \nonumber \\&\quad =: I_{1,1} + I_{1,2}+ I_{1,3} + I_{1,4} + I_{1,5}. \end{aligned}$$

(5.3)

By (4.54) and (4.45) for $q_1 = q_2 = 1$ and $\vartheta = s_1 = s_2 = 0$, we have

$$\begin{aligned} I_{1,1} + I_{1,2} \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4}+ 3\sigma - 1 \right) }} \Vert u_1\Vert _{{1}} + \langle t \rangle ^{{ \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) }} \Vert u_1 \Vert _1. \end{aligned}$$

(5.4)

By (4.63) and (4.69), we have

$$\begin{aligned}&I_{1,3} \lesssim e^{-\varepsilon _{\sigma } t} \Vert u_1 \Vert _{{H^{-1}}}. \end{aligned}$$

(5.5)

Since the support of $\chi _{mh}$ is included in $[2^{-\frac{3}{1 - 2\sigma }}, \infty )$ and $2^{{ -\frac{6(1 - \sigma )}{1 - 2\sigma } }} > \varepsilon _\sigma$ , we have

$$\begin{aligned} I_{1,4}&={\left\| \widehat{G_{\sigma }}(t,\cdot )\chi _{mh} {\hat{u}}_1 \right\| _{2}} = {\left\| |\xi |^{- 2 \sigma }e^{-|\xi |^{2(1-\sigma )}t }\chi _{mh} {\hat{u}}_1 \right\| _{2}} \lesssim e^{-\varepsilon _{\sigma } t} \Vert u_1 \Vert _{{H^{-2 \sigma }}}. \end{aligned}$$

(5.6)

It is written that

$$\begin{aligned}&I_{1,5} = \Vert \widehat{G_{\sigma }}(t,\cdot )({\hat{u}}_1(\cdot )- {\hat{u}}_1(0)) \Vert _{{2}}. \end{aligned}$$

(5.7)

Since $\widehat{G_{\sigma }}(t,\xi ) = |\xi |^{-2\sigma }e^{-|\xi |^{2(1-\sigma )}t }$, we have by the transformation $t^{\frac{1}{2(1-\sigma )}} r = \rho$ that

$$\begin{aligned} \begin{aligned} \Vert \widehat{G_{\sigma }}(t,\cdot )|\cdot |^{\theta } \Vert _{{2}}^2&= \int \nolimits _0^\infty r^{2( \theta -2\sigma ) + n - 1} e^{-2 r^{2(1-\sigma )} t} dr \\&=t^{- \frac{n+2( \theta -2\sigma )}{2(1-\sigma )}} \int \nolimits _0^\infty \rho ^{2(\theta -2\sigma ) + n-1} e^{-2 \rho ^{2(1-\sigma )} } d\rho \sim t^{- \frac{n+2(\theta -2\sigma )}{2(1-\sigma )}}, \end{aligned} \end{aligned}$$

that is,

$$\begin{aligned} \Vert \widehat{G_{\sigma }}(t,\cdot )|\cdot |^{\theta } \Vert _{{2}} \sim {t}^{ \frac{1}{1-\sigma }(-\frac{n}{4}-\frac{\theta }{2} + \sigma )}. \end{aligned}$$

(5.8)

On the other hand, since $\theta \in [0,1]$, we have

$$\begin{aligned} |{\hat{u}}_1(\xi )- {\hat{u}}_1(0) |&\le \int \nolimits _{{{\mathbb {R}}}^n} \left| (e^{i x \cdot \xi } - 1)u_1(x) \right| dx = \int \nolimits _{{{\mathbb {R}}}^n}\left| (e^{\frac{i x \cdot \xi }{2}} - e^{\frac{-i x \cdot \xi }{2}} )u_1(x) \right| dx \nonumber \\&= 2 \int \nolimits _{{{\mathbb {R}}}^n}\left| \sin \left( \frac{ x \cdot \xi }{2} \right) u_1(x) \right| dx \nonumber \\&\le 2 \int \nolimits _{{{\mathbb {R}}}^n} \left( \frac{|x \cdot \xi |}{2}\right) ^\theta |u_1(x)| dx = 2^{1-\theta } |\xi |^\theta \left\| | \cdot |^\theta u_1 \right\| _1 \end{aligned}$$

(5.9)

for every $\xi \in {{\mathbb {R}}}^n$. From (5.7), (5.8) and (5.9), it follows that

$$\begin{aligned} I_{1,5}\le & {} 2 \left\| \widehat{G_{\sigma }}(t,\cdot )|\cdot |^{\theta } \; \left\| | \cdot |^\theta u_1 \right\| _1 \right\| _{{2}} = 2 \Vert \widehat{G_{\sigma }}(t,\cdot )|\cdot |^{\theta } \Vert _{{2}} \; \left\| | \cdot |^\theta u_1 \right\| _1\nonumber \\\lesssim & {} {t}^{ \frac{1}{1-\sigma }(-\frac{n}{4}-\frac{\theta }{2} + \sigma )} \left\| | \cdot |^{\theta } u_1 \right\| _1. \end{aligned}$$

(5.10)

Substituting (5.4)–(5.6) and (5.10) into (5.3), we obtain (5.1).

Next we prove (5.2). We have

$$\begin{aligned}&\left\| K_0(t,\cdot )*u_0 - H_{\sigma }(t,x) \int \nolimits _{{{\mathbb {R}}}^n}u_0(y)dy \right\| _2 \nonumber \\&\quad \le {\left\| {{\mathcal {F}}}^{-1}[\widehat{K_0^+}(t,\cdot ) \chi _{low}] * u_0 - {{\mathcal {F}}}^{-1}[ \widehat{H_{\sigma }}(t,\cdot )\chi _{low}] * u_0\right\| _2} \nonumber \\&\qquad + {\left\| {{\mathcal {F}}}^{-1}[\widehat{K_0^-}(t,\cdot ) \chi _{low}] * u_0\right\| _{2}} +{\left\| K_{0,mh}(t,\cdot ) * u_0 \right\| _{2}} \nonumber \\&\qquad + {\left\| {{\mathcal {F}}}^{-1}\left[ \widehat{H_{\sigma }}(t,\cdot )\chi _{mh} \right] * u_0 \right\| _{2}} + \Vert H_{\sigma }(t,\cdot )*u_0 - H_{\sigma }(t,\cdot ) \int \nolimits _{{{\mathbb {R}}}^n}u_0 (y)dy \Vert _2 \nonumber \\&\quad =: I_{0,1} + I_{0,2}+ I_{0,3} + I_{0,4} + I_{0,5}. \end{aligned}$$

(5.11)

By (4.55) with $q_j =1, s_j = 0 \;(j = 1,2)$ and $\vartheta = 0$, we have

$$\begin{aligned} I_{0,1}= & {} {\left\| {{\mathcal {F}}}^{-1}\left[ \left( \widehat{K_0^+}(t,\xi ) - e^{- |\xi |^{2(1 - \sigma )} t} \right) \chi _{low}\right] *u_0\right\| _2} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4}+ 2\sigma - 1 \right) }} \Vert u_0\Vert _{{1}}. \end{aligned}$$

(5.12)

Inequality (4.48) implies

$$\begin{aligned}&I_{0,2} \lesssim \langle t \rangle ^{{ \frac{1}{\sigma } \left( -\frac{n}{4} + 2 \sigma - 1 \right) }} \Vert u_0 \Vert _1, \end{aligned}$$

(5.13)

and inequalities (4.65) and (4.71) imply

$$\begin{aligned}&I_{0,3} \lesssim e^{-\varepsilon _\sigma t} \Vert u_0 \Vert _2. \end{aligned}$$

(5.14)

Since the support of $\chi _{mh}$ is included in $[2^{-\frac{3}{1 - 2\sigma }}, \infty )$ and $2^{{ -\frac{6(1 - \sigma )}{1 - 2\sigma } }} > \varepsilon _\sigma$ , we have

$$\begin{aligned} I_{0,4}&={\left\| e^{-|\xi |^{2(1-\sigma )}t }\chi _{mh} {\hat{u}}_0 \right\| _{2}} \lesssim e^{-\varepsilon _\sigma t} \Vert u_0 \Vert _2. \end{aligned}$$

(5.15)

By (5.8) with $\theta$ replaced by $2 \sigma + \theta$, we have

$$\begin{aligned} \begin{aligned} \Vert \widehat{H_{\sigma }}(t,\cdot )|\cdot |^{\theta } \Vert _{{2}} = \Vert \widehat{G_{\sigma }}(t,\cdot )|\cdot |^{2 \sigma + \theta } \Vert _{{2}}&\sim {t}^{ \frac{1}{1-\sigma }(-\frac{n}{4}-\frac{ \theta }{2} )}. \end{aligned} \end{aligned}$$

(5.16)

Then, in the same way as in the proof of (5.10), by using (5.16) instead of (5.8), we have

$$\begin{aligned} I_{0,5}&\lesssim {t}^{ \frac{1}{1-\sigma }(-\frac{n}{4} - \frac{\theta }{2} )} \left\| | \cdot |^\theta u_0 \right\| _1. \end{aligned}$$

(5.17)

Since $\sigma < 1 - \sigma$, (5.2) follows from (5.11)–(5.15) and (5.17). $\square$

6 Estimate of the nonlinear term

Throughout this section, we suppose assumtion (1.2), and estimate nonlinear terms by using the argument of [7] and [9]. For $r \in [1,2)$, $\delta \in [ 0,\frac{n}{2} - 2 \sigma )$ and $\bar{s} \ge 1$, we define

$$\begin{aligned} X_{r,\delta ,\bar{s}} :=\{ u \in C((0,\infty );H^{\bar{s}} \cap H^{0,\delta } ); {\left\| u\right\| _X}_{r,\delta ,\bar{s}} < \infty \}, \end{aligned}$$

(6.1)

where

$$\begin{aligned} \begin{aligned} {\left\| \varphi \right\| _{X_{r,\delta ,\bar{s}}}}&:= \sup _{t > 0} \Big ( \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( \frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \sigma + \frac{\bar{s}}{2}\right) }} {\left\| (-\Delta )^{{\bar{s}/2}} \varphi (t)\right\| _{2}} \\&\quad + \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( \frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{\delta }{2} - \sigma \right) }} {\left\| \langle \cdot \rangle ^\delta \varphi (t)\right\| _{2}} \Big ). \end{aligned} \end{aligned}$$

(6.2)

For $\vartheta \in [0, \frac{n}{2}- 2 \sigma )$, we put

$$\begin{aligned} \zeta _{r,\vartheta } :&= \frac{1}{1 - \sigma } \left( -\frac{n}{2}\Bigg (\frac{1}{r} - \frac{1}{2} \Bigg ) + \sigma + \frac{\vartheta }{2} - (p-1)\left( \frac{n}{2r}- \sigma \right) + \frac{1}{2} \right) \nonumber \\&= \frac{1}{1 - \sigma } \left( -\Bigg (\frac{n}{2r} - \sigma \Bigg )p + \frac{n}{4} + \frac{\vartheta }{2} + \frac{1}{2} \right) . \end{aligned}$$

(6.3)

For $s \ge 0$, we define

$$\begin{aligned} {\tilde{q}}_{s} :&= \frac{2n}{n + 2 +2 [s] - 2 s}, \quad \text {that is,}\quad \; \frac{1}{{\tilde{q}}_s} = \frac{1}{2} + \frac{1 + [s] - s }{n}. \end{aligned}$$

(6.4)

Lemma 7

Let $r \in [1,2)$, $\delta \in [ 0,\frac{n}{2} - 2 \sigma )$ and $\bar{s} > 2 \sigma$. Let $X = X_{r,\delta ,\bar{s}}$. Then the following holds for every $\vartheta \in [0,\delta ]$, $s \in [0,\bar{s}]$ and $u \in X$:

• (i) We have

  $$\begin{aligned}&{\left\| (- \Delta )^{\frac{s}{2}} u(t,\cdot )\right\| _{2}} \lesssim \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \sigma - \frac{s }{2} \right) }}{\left\| u\right\| _X}, \end{aligned}$$

  (6.5)
  $$\begin{aligned}&{\left\| |\cdot |^\vartheta u(t,\cdot )\right\| _{2}} \lesssim \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\vartheta }{2} + \sigma \right) }}{\left\| u\right\| _X}. \end{aligned}$$

  (6.6)

• (ii) We have

  $$\begin{aligned}&{\left\| u(t,\cdot )\right\| _{q,2}} \lesssim \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{q}) + \sigma \right) }}\Vert {u}\Vert _{X}, \; \quad \text {if} \quad q = \frac{2n}{n + 2\delta }, \end{aligned}$$

  (6.7)
  $$\begin{aligned}&{\left\| u(t,\cdot )\right\| _{q}} \lesssim \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{q}) + \sigma \right) }}{\left\| u\right\| _X} \nonumber \\&\qquad \qquad \qquad \qquad \qquad \quad \text {if} \quad q \in {\left\{ \begin{array}{ll} (\frac{2n}{n + 2\delta }, \frac{2n}{n - 2 \bar{s}}] \quad &{}(2\bar{s} < n), \\ (\frac{2n}{n + 2\delta }, \infty ) \qquad \quad &{}(2\bar{s} \ge n). \end{array}\right. } \end{aligned}$$

  (6.8)

• (iii) We have

  $$\begin{aligned}&\Vert (-\Delta )^{\frac{[s ]}{2}} f(u(t,\cdot ))\Vert _{{{{\tilde{q}}_s }}} \lesssim \langle t \rangle ^ {{ \frac{1}{1- \sigma } \left( (-\frac{n}{2r} + \sigma )p + \frac{n}{4} + \frac{1}{2} - \frac{s }{2} \right) }} \Vert u \Vert _X^p, \end{aligned}$$

  (6.9)
  $$\begin{aligned}&{\left\| (- \Delta )^{\frac{s}{2}} (1 - \Delta )^{-\frac{1}{2}} f(u(t,\cdot ))\right\| _{2}} \lesssim \langle t \rangle ^ {{ \frac{1}{1 - \sigma } \left( (-\frac{n}{2r}+\sigma )p + \frac{n}{4} + \frac{1}{2} - \frac{s }{2} \right) }} \Vert {u} \Vert _{X}^p \end{aligned}$$

  (6.10)
  $$\begin{aligned}&{\left\| \langle \cdot \rangle ^\vartheta f(u(t,\cdot )) \right\| _{ {2n/(n+2)} } } \lesssim \langle t \rangle ^{\zeta _{r,\vartheta }} \Vert u\Vert _{X}^p. \end{aligned}$$

  (6.11)

Proof

Except use of the weak $L^p$ estimate, we follow the argument of [9, Lemmas 2.3 and 2.5], which is originated in [7, Lemma 2.1, 2.3 and 2.5].

• (i) By Plancherel’s theroem and Hölder’s inequality, we have

  $$\begin{aligned}&{\left\| (- \Delta )^{\frac{s}{2}} u(t,\cdot )\right\| _{2}} = \Vert {|\cdot |^{s} {\hat{u}}(t,\cdot )}\Vert _{{2}} \\&\quad \le \Vert {|\cdot |^{\bar{s}} {\hat{u}}(t,\cdot )}\Vert _{{2}}^{\frac{s }{\bar{s}}} \Vert {\hat{u}}(t,\cdot )\Vert _{{2}}^{1 -\frac{s }{\bar{s}}} = \Vert {(-\Delta )^{\frac{\bar{s}}{2}} u(t,\cdot )}\Vert _{{2}}^{\frac{s }{\bar{s}}} \Vert u(t,\cdot ) \Vert _{{2}}^{1 -\frac{s }{\bar{s}}}, \end{aligned}$$

  which together with the definition of $\Vert \cdot \Vert _X$ implies (6.5). In the same way, we see that (6.6) holds by Hölder’s inequality.

• (ii) We first consider the case $\frac{2n}{n + 2\delta } = q$, that is, $n(\frac{1}{q} - \frac{1}{2}) = \delta$. Then

  $$\begin{aligned} \Vert u(t,\cdot ) \Vert _{{q,2}} \lesssim \Vert |\cdot |^{-\delta } \Vert _{{\frac{n}{\delta }, \infty }} \Vert |\cdot |^{\delta }u(t,\cdot )\Vert _{{2}} \lesssim \Vert |\cdot |^{\delta }u \Vert _{{2}}, \end{aligned}$$

  (6.12)

  which together with the definition of $\Vert \cdot \Vert _X$ implies (6.7). Next we consider the case $\frac{2n}{n + 2\delta }< q < 2$, that is, $0< n \left( \frac{1}{q} - \frac{1}{2}\right) < \delta$. In [9, (2.12)] (see also [7, (2.5)]), the following is shown

  $$\begin{aligned} \Vert u \Vert _{{q}} \lesssim \Vert u\Vert _{{2}}^{1 - \frac{n}{\delta }\frac{2 - q}{2q}} \Vert |\cdot |^\delta u\Vert _{{2}}^{\frac{n}{\delta }\frac{2 - q}{2q}}, \end{aligned}$$

  (6.13)

  when $0< n \left( \frac{1}{q} - \frac{1}{2}\right) < \delta$. This together with the definition of $\Vert \cdot \Vert _X$ implies (6.8). We consider the case $2 \le q \le \frac{2n}{n - 2 \bar{s}}$ and $\bar{s} < 2 n$. Let ${\tilde{s}} = \frac{n}{2} - \frac{n}{q} (\le \bar{s})$. Then Sobolev’s embedding theorem together with (6.5) implies

  $$\begin{aligned} \Vert u \Vert _{{q}} \le \Vert u \Vert _{{\dot{H}^{{\tilde{s}}}}} \le \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \sigma - \frac{s}{2} \right) }}\Vert {u}\Vert _{X} = \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{q}) + \sigma \right) }}\Vert {u}\Vert _{X}, \end{aligned}$$

  that is, (6.8) holds. In the same way, (6.8) holds also in the case $2 \le q < \infty$ and $\bar{s} \le 2 n$.

• (iii) We put

  $$\begin{aligned} \kappa := \frac{n}{2} - \frac{1}{p-1}. \end{aligned}$$

  (6.14)

  By the Leibniz rule together with assumtion (1.2), we have

  $$\begin{aligned} \begin{aligned} \Vert \nabla ^{[s]} f(u(t,\cdot )) \Vert _{{{{\tilde{q}}_s}}} \le \Big \Vert u(t,\cdot )^{p-[s]} \sum _{{\sum _{j=1}^{[s]}|\nu _j |= [s] }} \prod _{j=1}^{[s]} | D_x^{\nu _j} u(t,\cdot )| \Big \Vert _{{{\tilde{q}_s}}}, \end{aligned} \end{aligned}$$

  (6.15)

  where $\nu _j$ is a multi index. Put $k_j = |\nu _j|$. Then, as in the proof of [9, Lemma 2.5], we can choose $s_j \in [0,k_j - \frac{1}{p-1})$ such that $q_j$ $(j =1,\ldots ,[s])$ defined by

  $$\begin{aligned} \frac{1}{q_0}&= \left( \frac{1}{2} - \frac{\kappa }{n} \right) (p - [s]), \end{aligned}$$

  (6.16)
  $$\begin{aligned} \frac{1}{q_j}&= \frac{1}{2} - \frac{\kappa + s_j - k_j}{n} \;\; (j = 1,\ldots ,[s]) \end{aligned}$$

  (6.17)

  satisfies

  $$\begin{aligned}&\sum _{j = 0}^{[s]}\frac{1}{q_j} = \frac{1}{{\tilde{q}}_s}, \end{aligned}$$

  (6.18)
  $$\begin{aligned}&(p - [s])q_0 \in [2,\infty ) \; \text {and} \; q_j \in [2,\infty ) \; \text {for} \; j = 1,\ldots ,[s]. \end{aligned}$$

  (6.19)

  Since

  $$\begin{aligned}&\frac{1}{\tilde{q_{s}}} - \frac{1}{q_0} = \frac{1}{2} + \frac{1}{n} + \frac{[s] - s}{n} - \left( \frac{1}{2} - \frac{\kappa }{n} \right) (p - [s]), \\&\quad \sum _{j=1}^{[s]}\frac{1}{q_j} = \sum _{j=1}^{[s]} \left( \frac{1}{2} - \frac{\kappa + s_j - k_j}{n} \right) = \Bigg (\frac{1}{2} - \frac{\kappa }{n}\Bigg )[s] - \frac{1}{n} \sum _{j=1}^{[s]} s_j + \frac{[s]}{n}, \end{aligned}$$

  the condition (6.18) is equivalent to

  $$\begin{aligned} \sum _{j=1}^{[s]} s_j = s - \kappa , \end{aligned}$$

  (6.20)

  and thus, $\kappa + s_j \le s$. Taking (6.16)–(6.19) into account, we apply Hölder’s inequality and Sobolev’s embedding theorem to (6.15). Then we obtain

  $$\begin{aligned} \begin{aligned} \Vert \nabla ^{[s]} f(u(t,\cdot )) \Vert _{{{{\tilde{q}}_s}}}&\lesssim \Vert u^{p-[s]}\Vert _{{{q_0}}} \sum _{{k_j \ge 0, \sum _{j=1}^{[s]}k_j = [s] }} \prod _{j=1}^{[s]} \Vert |\nabla |^{k_j} u(t,\cdot ) \Vert _{{{q_j}}} \\&\lesssim \Vert \nabla ^{\kappa } u \Vert _{{2}}^{p - [s] } \sum _{{k_j \ge 0, \sum _{j=1}^{[s]}k_j = [s] }} \prod _{j=1}^{[s]} \Vert |\nabla |^{\kappa + s_j} u(t,\cdot ) \Vert _{{2}}, \end{aligned} \end{aligned}$$

  (6.21)

  where $|\nabla | := (-\Delta )^{1/2}$. Then estimating the right-hand side of (6.21) by the definition of $\Vert \cdot \Vert _X$, and using (6.20) and (6.14), we obtain

  $$\begin{aligned} \begin{aligned} \Vert \nabla ^{[s]} f(u(t,\cdot )) \Vert _{{{{\tilde{q}}_s}}}&\lesssim \langle t \rangle ^ {{ \frac{1}{1- \sigma } \left( (-\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \sigma )p - \frac{\kappa }{2}(p - [s]) - \frac{1}{2} \sum _{j=1}^{[s]} (\kappa + s_j) \right) }} \Vert u \Vert _X^p \\&= \langle t \rangle ^ {{ \frac{1}{1- \sigma } \left( (-\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \sigma - \frac{\kappa }{2})p - \frac{1}{2}\sum _{j=1}^{[s]} s_j \right) }} \Vert u \Vert _X^p \\&= \langle t \rangle ^ {{ \frac{1}{1- \sigma } \left( (-\frac{n}{2r} + \sigma )p + \frac{n}{4} + \frac{1}{2} - \frac{s}{2} \right) }} \Vert u \Vert _X^p, \end{aligned} \end{aligned}$$

  that is, (6.9) holds. Sobolev’s embedding theorem together with inequality (6.9) implies (6.10). By Hölder’s inequality and assumtion (1.2), we have

  $$\begin{aligned} \begin{aligned}&\Vert \langle \cdot \rangle ^\vartheta f(u(t,\cdot )) \Vert _{{2n/(n+2)} } \lesssim \Vert \langle \cdot \rangle ^\vartheta u(t,\cdot )\Vert _2 \Vert u(t,\cdot )\Vert ^{p-1}_{{{{n(p-1)}} }}. \end{aligned} \end{aligned}$$

  (6.22)

  The assumption (2.13) implies

  $$\begin{aligned} n(p-1) \ge \frac{2rn}{n - 2r\sigma } > \frac{2n}{n+ 2 \delta }, \end{aligned}$$

  and (2.7) implies

  $$\begin{aligned} n(p-1) \le \frac{2n}{n - 2 \bar{s}} \end{aligned}$$

  if $2 \bar{s} < n$. Thus, we can apply (6.6) and (6.8) with $q = n(p-1)$, which together with (6.22) yields (6.11). $\square$

7 Estimates of a convolution term

Throughout this section, we suppose assumtion (1.2).

7.1 Decay estimates

Throughout this subsection, we suppose the assumption of Proposition 1.

Lemma 8

Let $\vartheta \in [0,\delta ]$. For every $u \in X = X_{r,\delta ,\bar{s}}$, we have

$$\begin{aligned} \begin{aligned}&\int \nolimits _0^t \Vert |\cdot |^\vartheta {{\mathcal {F}}}^{-1} \left[ \widehat{K_1^{\pm }} \chi _{mh}\right] (t - \tau ,\cdot ) * f(u(\tau ,\cdot )) \Vert _2 d\tau \lesssim \langle t \rangle ^{\zeta _{r, \vartheta } } \Vert u\Vert _X^p, \end{aligned} \end{aligned}$$

(7.1)

where $\zeta _{r,\vartheta }$ is the number defined by (6.3).

Proof

By (4.64) and (4.70), Sobolev’s embedding theorem and (6.11), we have

$$\begin{aligned} \begin{aligned}&{\left\| |\cdot |^\vartheta {{\mathcal {F}}}^{-1} \left[ \widehat{K_1^{\pm }} \chi _{mh}\right] (t - \tau ,\cdot ) * f(u(\tau ,\cdot )) \right\| _{2}} \\&\quad \lesssim e^{-\varepsilon _\sigma (t - \tau )} \Vert (1 - \Delta )^{-\frac{1}{2}} \langle \cdot \rangle ^\vartheta f(u(\tau ,\cdot ))\Vert _{{2}} \\&\quad \lesssim e^{-\varepsilon _\sigma (t - \tau )} \Vert \langle \cdot \rangle ^\vartheta f(u(\tau ,\cdot ))\Vert _{{{{\frac{2n}{n+2}}} }} \lesssim e^{-\varepsilon _\sigma (t - \tau )} \langle \tau \rangle ^{\zeta _{r,\vartheta }} \Vert u \Vert _X^p, \end{aligned} \end{aligned}$$

which yields (7.1). $\square$

Lemma 9

For every $u,v \in X = X_{r,\delta ,\bar{s}}$, we have

$$\begin{aligned}&\int \nolimits _0^t {\left\| |\cdot |^\delta \left( K_{1}(t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.2)

$$\begin{aligned}&\int \nolimits _0^t {\left\| |\cdot |^\delta \left( K_{1}(t - \tau ,\cdot ) * (f(u(\tau ,\cdot )) - f(v(\tau ,\cdot ))) \right) \right\| _{2}}d\tau \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} (\Vert u \Vert _X +\Vert v\Vert _{X})^{p-1} \Vert u - v\Vert _X. \end{aligned}$$

(7.3)

Proof

First, we estimate the low frequency part. By (4.46) with $q_1 = r$, $q_2 = \frac{2n}{n+2}$ and $\vartheta = \delta$, $s_1 = s_2 = 0$, we have

$$\begin{aligned}&\int \nolimits _0^t {\left\| |\cdot |^\delta \left( K_{1,low }(t - \tau ,\cdot ) * f(u(\tau ,\cdot )) \right) \right\| _{2}}d\tau \le I_1 + I_2, \end{aligned}$$

(7.4)

where

$$\begin{aligned}&I_1: = {\left\{ \begin{array}{ll} &{} \int \nolimits _0^{t} \langle t -\tau \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} {\left\| f(u(\tau ,\cdot ))\right\| _{{r}}}d\tau \; \quad \text {if} \quad r \ge 1, \\ &{} \int \nolimits _0^{t} \langle t -\tau \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} {\left\| f(u(\tau ,\cdot ))\right\| _{{r,2}}}d\tau \quad \text {if} \quad r > 1,\end{array}\right. } \\&I_2 := \int \nolimits _0^{t} \langle t -\tau \rangle ^{{ \frac{1}{1 - \sigma } \left( - \frac{1}{2} + \sigma \right) }} {\left\| |\cdot |^\delta f(u(\tau ,\cdot )) \right\| _{{2n/n+2}}} d\tau . \end{aligned}$$

Since $\bar{s} \ge 1$, assumtion (2.7) implies that $pr \le 2 p \le \frac{2n}{n - 2 \bar{s}}$ if $2 s_o < n$. From (2.15), it follows that $\frac{2n}{n + 2 \delta }< pr$ in the case $r = 1$, and $\frac{2n}{n + 2 \delta } \le pr$ in the case $r > 1$. In the case $\frac{2n}{n + 2 \delta } < pr$, we can apply (6.8) with $q = pr$ to obtain

$$\begin{aligned} {\left\| f(u(\tau ,\cdot )) \right\| _{{r}}} \lesssim \Vert u(\tau ,\cdot )\Vert _{{{pr}}}^p&\lesssim \langle \tau \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{p-1}{r}) + p \sigma \right) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.5)

In the case $pr = \frac{2n}{n + 2 \delta }$, making use of the equality $(|u|^p)^* = (u^*)^p$, we have

$$\begin{aligned} {\left\| f(u(\tau ,\cdot )) \right\| _{{r,2}}} \lesssim&\Vert |u(\tau ,\cdot )|^p \Vert _{{{r,2}}} = \Vert u(\tau ,\cdot ) \Vert _{{{pr,2p}}}^p \lesssim \Vert u(\tau ,\cdot ) \Vert _{{{pr,2}}}^p. \end{aligned}$$

Hence, by (6.7) with $q = pr$, we obtain

$$\begin{aligned} {\left\| f(u(\tau ,\cdot )) \right\| _{{r,2}}} \lesssim \langle \tau \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{p-1}{r}) + p \sigma \right) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.6)

Substituting (7.5) or (7.6) into $I_1$, we obtain

$$\begin{aligned} I_1 \lesssim&\int \nolimits _0^{t} \langle t - \tau \rangle ^ {{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} \langle \tau \rangle ^{\frac{1}{1-\sigma }(-\frac{n}{2}(\frac{p}{r} - \frac{1}{r}) + p \sigma )} d\tau \Vert u\Vert _{X}^p. \end{aligned}$$

The following inequality is commonly used to estimate the nonlinear term.

$$\begin{aligned} \int \nolimits _0^t \langle t - \tau \rangle ^{\rho } s^{\eta }d\tau \lesssim {\left\{ \begin{array}{ll} \langle t \rangle ^{\max \{\rho ,\eta \} }&{} \text {if}\quad \min \{\rho , \eta \} < - 1,\\ \langle t \rangle ^{\max \{\rho ,\eta \} }\log (2+t)&{} \text {if} \quad \min \{\rho , \eta \} = -1, \\ \langle t \rangle ^{1+\rho +\eta }&{} \text {if} \quad \min \{\rho , \eta \} > -1. \end{array}\right. } \end{aligned}$$

(7.7)

The assumption that $\delta \ge n(\frac{1}{r} - \frac{1}{2}) - 1$ implies

$$\begin{aligned} \frac{1}{1 - \sigma } \left( -\frac{n}{2}\Big (\frac{1}{r} - \frac{1}{2}\Big ) + \frac{\delta }{2} + \sigma \right) \ge -1. \end{aligned}$$

(7.8)

The assumption (2.13) is equivalent to $p \left( \frac{n}{2r} - \sigma \right) > \frac{n}{2r} - \sigma + 1,$ which is equivalent to

$$\begin{aligned} \frac{1}{1-\sigma }\left( -\frac{n}{2}\Bigg (\frac{p - 1}{r}\Bigg ) + p \sigma \right) < -1. \end{aligned}$$

(7.9)

Hence, by using (7.7), we obtain

$$\begin{aligned} I_1 \lesssim&\langle t \rangle ^{{ \frac{1}{1 - \sigma } \big (-\frac{n}{2}\big (\frac{1}{r} - \frac{1}{2}\big ) + \frac{\delta }{2} + \sigma \big ) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.10)

Since $\frac{1}{1 - \sigma } (- \frac{1}{2}+ \sigma ) > -1$, it follows from (6.11) and (7.7) that

$$\begin{aligned} I_2&\lesssim \int \nolimits _0^{t} \langle t - \tau \rangle ^{{ \frac{1}{1 - \sigma } (- \frac{1}{2}+ \sigma )}} \langle \tau \rangle ^{\zeta _{r,\delta }}d\tau \Vert u\Vert _{X}^p \\&\lesssim {\left\{ \begin{array}{ll} \langle t \rangle ^{{ \frac{1}{1 - \sigma } (- \frac{1}{2}+ \sigma )}}\Vert u\Vert _{X}^p &{} \quad \text {if} \; \zeta _{r,\delta } < - 1 \\ \langle t \rangle ^{{ \frac{1}{1 - \sigma } (- \frac{1}{2}+ \sigma ) + \zeta _{r,\delta } + 1}} \log (t + 2) \Vert u\Vert _{X}^p &{} \quad \text {if} \; \zeta _{r,\delta } \ge -1. \end{array}\right. } \end{aligned}$$

The assumption that $\delta \ge n(\frac{1}{r} - \frac{1}{2}) - 1$ implies $- \frac{1}{2}+ \sigma \le -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma$. Hence, we have

$$\begin{aligned} I_2 \lesssim&\langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} \Vert u\Vert _{X}^p \end{aligned}$$

(7.11)

in the case $\zeta _{r,\delta } < - 1$. By definition (6.3) and assumption (2.13), we have

$$\begin{aligned}&\frac{1}{1 - \sigma } \Big (- \frac{1}{2}+ \sigma \Big ) + \zeta _{r,\delta } + 1 \\&\quad = \frac{1}{1 - \sigma } \left( -\frac{n}{2}\Big (\frac{1}{r} - \frac{1}{2}\Big ) + \frac{\delta }{2} + \sigma + 1 - (p-1)\left( \frac{n}{2r} -\sigma \right) \right) \\&\quad < \frac{1}{1 - \sigma } \left( -\frac{n}{2}\Big (\frac{1}{r} - \frac{1}{2}\Big ) + \frac{\delta }{2} + \sigma \right) . \end{aligned}$$

Hence, (7.11) holds also in the case $\zeta _{r,\delta } \ge -1$. Substituting (7.10) and (7.11) into (7.4), we obtain

$$\begin{aligned} \begin{aligned} \int \nolimits _0^t {\left\| |\cdot |^\delta \left( K_{1,low}(t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} \Vert u\Vert _{X}^p, \end{aligned} \end{aligned}$$

which together with (7.1) yields (7.2).

The assumption (1.2) implies

$$\begin{aligned} |f(u(\tau ,x)) - f(v(\tau ,x))| \lesssim ( |u(\tau ,x)| + | v(\tau ,x)|)^{p-1} |u(\tau ,x)- v(\tau ,x)|, \end{aligned}$$

and we can prove (7.3) in the same way. $\square$

Lemma 10

Let $s \in [0,\bar{s}]$. For every $u,v \in X = X_{r,\delta ,\bar{s}}$, we have

$$\begin{aligned}&\int \nolimits _0^t {\left\| (- \Delta )^{\frac{s}{2}} \left( K_{1}^+(t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{s }{2} + \sigma \right) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.12)

$$\begin{aligned}&\int \nolimits _0^t {\left\| (- \Delta )^{\frac{s}{2}} \left( K_{1}^-(t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \max \{ \frac{1}{ \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{s }{2} + \sigma \right) , \frac{1}{1- \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2})- \frac{s }{2} + \sigma - (p-1)(\frac{n}{2r} - \sigma ) + 1 \right) \} }} \Vert u\Vert _{X}^p \end{aligned}$$

(7.13)

$$\begin{aligned}&\int \nolimits _0^t {\left\| (- \Delta )^{\frac{s}{2}} \left( K_{1}(t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{s }{2} + \sigma \right) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.14)

$$\begin{aligned}&\int \nolimits _0^t {\left\| (- \Delta )^{\frac{s}{2}} \left( K_{1}(t - \tau ,\cdot ) * (f(u(\tau ,\cdot )) - f(v(\tau ,\cdot )) ) \right) \right\| _{2}}d\tau \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{s }{2} + \sigma \right) }} (\Vert u\Vert _{X} + \Vert v\Vert _{X})^{p-1} \Vert u - v\Vert _{X}. \end{aligned}$$

(7.15)

for every $u,v \in X$,

Proof

We first prove (7.12). We divide the left-hand side of (7.12) into three parts:

$$\begin{aligned}&\int \nolimits _0^t {\left\| (- \Delta )^{\frac{s}{2}} \left( K_{1}^+(t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \nonumber \\&\quad = \int \nolimits _0^{t/2} {\left\| (- \Delta )^{\frac{s}{2}} \left( {{\mathcal {F}}}^{-1}[{\hat{K}}_{1}^+(t - \tau ,\cdot ) \chi _{low }] (t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \nonumber \\&\qquad + \int \nolimits _{t/2}^t {\left\| (- \Delta )^{\frac{s}{2}} \left( {{\mathcal {F}}}^{-1}[{\hat{K}}_{1}^+(t - \tau ,\cdot ) \chi _{low }] (t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \nonumber \\&\qquad + \int \nolimits _0^t {\left\| (- \Delta )^{\frac{s}{2}} \left( {{\mathcal {F}}}^{-1}[{\hat{K}}_{1}^+(t - \tau ,\cdot ) \chi _{mh} ](t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \nonumber \\&\quad := J_1^+ + J^+_2+ J^+_3 \;(\text {we put}). \end{aligned}$$

(7.16)

Substituting (4.44) with $\vartheta = 0$, $q_1 = q_2 = r$, $s_1 = s$ and $s_2 = 0$ and (7.5) into $J_1^+$, and using (7.9), we obtain

$$\begin{aligned} J_1^+\lesssim & {} \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{s }{2} + \sigma \right) }} \int \nolimits _0^{t/2} \langle \tau \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{p-1}{r}) + p \sigma \right) }}d\tau \Vert u\Vert _{X}^p \nonumber \\\lesssim & {} \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{s }{2} + \sigma \right) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.17)

By (4.44) with $\vartheta = 0$, $s_1 = s, \; s_2 =[s ]$, $q_1 = q_2 = \tilde{q}_s$ (defined by (6.4)), and (6.9), we have

$$\begin{aligned} J_2^+&\lesssim \int \nolimits _{t/2}^{t} \langle t - \tau \rangle ^ {{ \frac{1}{2(1 - \sigma )} \left( - n (\frac{1}{{\tilde{q}}_s } - \frac{1}{2}) - s + [s ] + 2 \sigma \right) }} {\left\| (-\Delta )^{\frac{[s ]}{2}} f(u(\tau ,\cdot ) )\right\| _{{{\tilde{q}}_s } }}d\tau \nonumber \\&\lesssim \langle t \rangle ^ {{ \frac{1}{1- \sigma } \left( (-\frac{n}{2r} + \sigma )p + \frac{n}{4} + \frac{1}{2} - \frac{s }{2} \right) }} \Vert u\Vert _{X}^p \int \nolimits _0^{t/2} \langle \tau \rangle ^ {{ \frac{1}{2(1 - \sigma )} \left( - 1 + 2 \sigma \right) }} d\tau \nonumber \\&\sim \langle t \rangle ^ {{ \frac{1}{1- \sigma } \left( (-\frac{n}{2r} + \sigma )p + \frac{n}{4} - \frac{s }{2} + 1 \right) }} \Vert u\Vert _{X}^p \nonumber \\&= \langle t \rangle ^ {{ \frac{1}{1- \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2})- \frac{s }{2} + \sigma - (p-1)(\frac{n}{2r} - \sigma ) + 1 \right) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.18)

Last we estimate $J^+_3$. Combining (4.63), (4.69) and (6.10), we have

$$\begin{aligned} J^+_3&\lesssim \int \nolimits _0^t e^{-\varepsilon _\sigma (t-\tau )} \langle \tau \rangle ^ {{ \frac{1}{1 - \sigma } \left( (-\frac{n}{2r}+\sigma )p + \frac{n}{4} + \frac{1}{2} - \frac{s }{2} \right) }} d \tau \Vert {u} \Vert _{X}^p \nonumber \\&\lesssim \langle t \rangle ^ {{ \frac{1}{1 - \sigma } \left( (-\frac{n}{2r}+\sigma )p + \frac{n}{4} + \frac{1}{2} - \frac{s }{2} \right) }} \Vert u\Vert _{X}^p \nonumber \\&= \langle t \rangle ^ {{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2})+\sigma - \frac{s }{2} - (p-1)(\frac{n}{2r} - \sigma ) + \frac{1}{2} \right) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.19)

The assumption (2.13) implies $- (p-1)(\frac{n}{2r} - \sigma ) + 1 < 0$. Thus, (7.12) follows from (7.16)–(7.19).

We divide the left-hand side of (7.13) into three parts:

$$\begin{aligned}&\int \nolimits _0^t {\left\| (- \Delta )^{\frac{s}{2}} \left( K_{1}^-(t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \nonumber \\&\quad = \int \nolimits _0^{t/2} {\left\| (- \Delta )^{\frac{s}{2}} \left( {{\mathcal {F}}}^{-1}[{\hat{K}}_{1}^-(t - \tau ,\cdot ) \chi _{low }] (t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \nonumber \\&\qquad + \int \nolimits _{t/2}^t {\left\| (- \Delta )^{\frac{s}{2}} \left( {{\mathcal {F}}}^{-1}[{\hat{K}}_{1}^-(t - \tau ,\cdot ) \chi _{low }] (t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \nonumber \\&\qquad + \int \nolimits _0^t {\left\| (- \Delta )^{\frac{s}{2}} \left( {{\mathcal {F}}}^{-1}[{\hat{K}}_{1}^-(t - \tau ,\cdot ) \chi _{mh} ](t - \tau ,\cdot ) * f(u(\tau ,\cdot ))\right) \right\| _{2}}d\tau \nonumber \\&\quad =: J^-_1 + J^-_2+ J^-_3 \;(\text {we put}). \end{aligned}$$

(7.20)

Substituting (4.45) with $\vartheta = 0$, $q_1 = q_2 = r$, $s_1 = s$ and $s_2 = 0$ and (7.5) into $J^-_1$, and using (7.9), we obtain

$$\begin{aligned} \begin{aligned} J^-_1&\lesssim \langle t \rangle ^{{ \frac{1}{ \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{s }{2} + \sigma \right) }} \int \nolimits _0^{t/2} \langle \tau \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{p-1}{r}) + p \sigma \right) }}d\tau \Vert u\Vert _{X}^p\\&\lesssim \langle t \rangle ^{{ \frac{1}{\sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{s }{2} + \sigma \right) }} \Vert u\Vert _{X}^p. \end{aligned} \end{aligned}$$

(7.21)

Since $\sigma < 1-\sigma$, the right-hand side of (4.45) is dominated by that of (4.44). Hence, $J_2^-$ and $J_3^-$ are estimated by the right-hand sides of (7.18) and (7.19), respectively, and thus,

$$\begin{aligned} J^-_2&\lesssim \langle t \rangle ^ {{ \frac{1}{1- \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2})- \frac{s }{2} + \sigma - (p-1)(\frac{n}{2r} - \sigma ) + 1 \right) }} \Vert u\Vert _{X}^p, \end{aligned}$$

(7.22)

$$\begin{aligned} J^-_3&\lesssim \langle t \rangle ^ {{ \frac{1}{1- \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2})- \frac{s }{2} + \sigma - (p-1)(\frac{n}{2r} - \sigma ) + \frac{1}{2} \right) }} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.23)

Substituting (7.21), (7.22) and (7.23) into (7.20), we obtain (7.13).

Inequality (7.14) follows from (7.12) and (7.13), since $\sigma < 1 -\sigma$ and $- (p-1)(\frac{n}{2r} - \sigma ) + 1 < 0$.

By using assumtion (1.2), we can prove (7.15) in the same way. $\square$

7.2 Diffusion estimate

Lemma 11

Let $\delta$ and $\nu$ be an arbitrary number satisfying the assumption of Theorem 3. Let $X = X_{1,\delta ,\bar{s}}$, where $X_{1,\delta ,\bar{s}}$ is defined by (6.1). Then we have

$$\begin{aligned} \begin{aligned}&\Big \Vert \int \nolimits _0^t K_1(t-\tau ,\cdot ) * f(u(\tau ,\cdot )) d\tau - G_{\sigma }(t,\cdot ) \int \nolimits _0^\infty \int \nolimits _{{{\mathbb {R}}}^n}f(u(\tau ,y))dy d\tau \Big \Vert _{{2}} \\&\quad \lesssim t ^{{\max \{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - \min \{( p - 1)\left( \frac{n}{2} \right) - 1, 1 - 2 \sigma , \nu \} \right) , \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) \} }} \Vert u\Vert _{X}^p. \end{aligned}\nonumber \\ \end{aligned}$$

(7.24)

Proof

We have

$$\begin{aligned}&\Big \Vert \int \nolimits _0^t K_1(t -\tau ,\cdot ) * f(u(\tau ,\cdot )) d\tau - G_{\sigma }(t,\cdot )\int \nolimits _0^\infty \int \nolimits _{{{\mathbb {R}}}^n}f(u(\tau ,y))dy d\tau \Big \Vert _2 \nonumber \\&\quad \le L_1 + L_2 + L_3 + L_4, \end{aligned}$$

(7.25)

where

$$\begin{aligned} L_1&:= \int \nolimits _{\frac{t}{2}}^{t} \Vert K_1(t-\tau ,\cdot )*f(u(\tau ,\cdot )) \Vert _2 d\tau , \\ L_{2}&:= \int \nolimits _0^{t/2} \left\| K_1(t-\tau ,\cdot ) * f(u(\tau ,\cdot )) - G_{\sigma }(t - \tau ,\cdot ) \int \nolimits _{{{\mathbb {R}}}^n} f(u(\tau ,y))dy \right\| _2 d\tau , \\ L_3&:= \int \nolimits _0^{t/2} \Big \Vert G_{\sigma }(t-\tau ,\cdot ) - G_{\sigma }(t,\cdot ) \Big \Vert _2 \left| \int \nolimits _{{{\mathbb {R}}}^n} f(u(\tau ,y))dy \right| d\tau , \\ L_4&:= \Vert G_{\sigma }(t,\cdot ) \Vert _2 \left| \int \nolimits _{t/2}^\infty \int \nolimits _{{{\mathbb {R}}}^n}f(u(\tau ,y))dy d\tau \right| . \end{aligned}$$

First we estimate $L_1$ by dividing the integrand as

$$\begin{aligned} \Vert K_1(t-\tau ,\cdot )*f(u(\tau ,\cdot )) \Vert _2&= \Vert K_{1,low}(t-\tau ,\cdot )* f(u(\tau ,\cdot )) \Vert _2 \\&\quad + \Vert K_{1,mh}(t-\tau ,\cdot ) * f(u(\tau ,\cdot )) \Vert _2. \end{aligned}$$

Taking $q_1 = q_2 = \frac{2n}{n + 4 \sigma }$ and $s_1 = s_2 = \vartheta = 0$ in (4.46), we obtain

$$\begin{aligned} \Vert K_{1,low}(t-\tau , \cdot )* f(u(\tau ,\cdot )) \Vert _2 \lesssim \Vert f(u(\tau ,\cdot ))\Vert _{\frac{2n}{n + 4 \sigma }} \lesssim \Vert u(\tau ,\cdot )\Vert _{{\frac{2np}{n + 4 \sigma }}}^p.\nonumber \\ \end{aligned}$$

(7.26)

Since $\bar{s} \ge 1$, assumtion (2.13) implies

$$\begin{aligned} \frac{2np}{n + 4 \sigma } > \frac{2n}{n + 2\delta }, \end{aligned}$$

and (2.7) implies

$$\begin{aligned} \frac{2np}{n + 4 \sigma } \le \frac{2n}{n - 2 \bar{s}} \end{aligned}$$

if $2 \bar{s} < n$. Thus, we can apply (6.8) with $r = 1$ and $q = \frac{2np}{n + 4 \sigma }$ to obtain

$$\begin{aligned} \begin{aligned} \Vert u(\tau ,\cdot )\Vert _{\frac{2np}{n + 4 \sigma }}^p&\lesssim \langle \tau \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}( p - \frac{1}{2} - \frac{2 \sigma }{n}) + p \sigma \right) }} \Vert u\Vert _{X}^p \\&= \langle \tau \rangle ^{{ \frac{1}{1 - \sigma } \left( - ( p - 1)\left( \frac{n}{2} - \sigma \right) - \frac{n}{4} +2 \sigma \right) }}\Vert u\Vert _{X}^p. \end{aligned} \end{aligned}$$

From the inequality above and (7.26), it follows that

$$\begin{aligned} \begin{aligned}&\int \nolimits _{\frac{t}{2}}^{t} \Vert K_{1,low}(t - \tau ,\cdot )*f(u(\tau ,\cdot )) \Vert _2 d\tau \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - ( p - 1)\left( \frac{n}{2} - \sigma \right) + 1 \right) }}\Vert u\Vert _{X}^p. \end{aligned}\nonumber \\ \end{aligned}$$

(7.27)

By (7.19) and (7.23) with $r = 1$ and $s = 0$, we have

$$\begin{aligned} \int \nolimits _{\frac{t}{2}}^{t} \Vert K_{1,mh} (t-\tau , \cdot )*f(u(\tau ,\cdot )) \Vert _2 d\tau&\lesssim \langle t \rangle ^ {{ \frac{1}{1 - \sigma } \left( -\frac{n}{4}+\sigma + \frac{1}{2} - (p-1)(\frac{n}{2} - \sigma ) \right) }} \Vert u\Vert _{X}^p, \end{aligned}$$

which together with (7.27) yields

$$\begin{aligned} L_1 \lesssim \, \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - ( p - 1)\left( \frac{n}{2} - \sigma \right) + 1 \right) }}\Vert u\Vert _{X}^p. \end{aligned}$$

(7.28)

By (5.1) with $u_1 =f(u(\tau ,\cdot ))$ and $\theta = 2 \tilde{\nu }$, we have

$$\begin{aligned} \begin{aligned} L_2 \lesssim \,&\langle t \rangle ^{{ \max \{ \frac{1}{1 - \sigma }\left( -\frac{n}{4} + 3\sigma - 1 \right) , \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) \} }} \int \nolimits _0^{t/2} \Vert f(u(\tau ,\cdot )) \Vert _{1} d\tau \\&+ e^{{-\frac{\varepsilon _\sigma t}{2}}} \int \nolimits _0^{t/2} \Vert f(u(\tau ,\cdot )) \Vert _{{H^{-2 \sigma }}} d\tau \\&+ t^{ \frac{1}{1-\sigma }(-\frac{n}{4} + \sigma - \tilde{\nu })} \int \nolimits _0^{t/2} \Vert |\cdot |^{2 {\tilde{\nu }}} f(u(\tau ,\cdot )) \Vert _{{{1}}} d\tau \\ =:&L_{2,1} + L_{2,2} + L_{2,3} \, (\text {we put}). \end{aligned} \end{aligned}$$

(7.29)

Inequality (7.5) and (7.9) with $r = 1$ yield

$$\begin{aligned} \int \nolimits _0^{t/2} {\left\| f(u(\tau ,\cdot )) \right\| _1} d\tau&\lesssim \int \nolimits _0^{t/2}\langle \tau \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(p-1) + p \sigma \right) }}d \tau \Vert u\Vert _{X}^p \lesssim \Vert u\Vert _{X}^p. \end{aligned}$$

Thus

$$\begin{aligned} L_{2,1} \lesssim \langle t \rangle ^{{ \max \{ \frac{1}{1 - \sigma }\left( -\frac{n}{4} + 3\sigma - 1 \right) , \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) \} }}\Vert u\Vert _{X}^p. \end{aligned}$$

(7.30)

Since $\bar{s} \ge 1$, (2.7) and (2.15) with $r = 1$ imply $2p > \frac{2n}{n + 2\delta }$, and moreover $2p \le \frac{2n}{n - 2 \bar{s}}$ if $2 \bar{s} < n$. Hence, we can use (6.8) with $r = 1$ and $q = 2p$ to obtain

$$\begin{aligned} \begin{aligned} \Vert f(u(\tau ,\cdot )) \Vert _{{H^{-2 \sigma }}}&\lesssim \Vert u(t,\cdot )\Vert _{2p}^p \lesssim \langle t \rangle ^{{\frac{p}{1 - \sigma } \left( -\frac{n}{2}(1 - \frac{1}{2p}) + \sigma \right) }}\Vert u\Vert _{X}^p \lesssim \Vert u\Vert _{X}^p. \end{aligned} \end{aligned}$$

Thus

$$\begin{aligned} L_{2,2} \le e^{{-\varepsilon _\sigma t/2}} \Vert u\Vert _{X}^p. \end{aligned}$$

(7.31)

We estimate $L_{2,3}$. Let $\tilde{\nu }$ be an arbitrary number satisfying

$$\begin{aligned} 0< \nu< \tilde{\nu } < \min \left\{ \frac{n}{4}(p-2) + \frac{1}{2}p \delta , \delta \right\} . \end{aligned}$$

(7.32)

Assume moreover that

$$\begin{aligned} \tilde{\nu } \le \frac{\delta }{2 \bar{s}}\left( n - \frac{p}{2}(n - 2 \bar{s})\right) , \end{aligned}$$

(7.33)

if $\bar{s} < \frac{n}{2}$. By Hölder’s inequality, we have

$$\begin{aligned} \left\| | \cdot |^{\tilde{2 \nu }} f(u(\tau ,\cdot )) \right\| _1 \lesssim \Vert |\cdot |^{\frac{2 \tilde{\nu }}{p} }u(\tau ,\cdot ) \Vert _p^p\le & {} \Vert (|\cdot |^\delta |u(\tau ,\cdot )|)^{\frac{2 \tilde{\nu }}{p\delta }}\Vert _{\frac{p\delta }{\tilde{\nu }}}^p \; \Vert |u(\tau ,\cdot )|^{1 -\frac{2 {\tilde{\nu }}}{p\delta }} \Vert _q^p \nonumber \\= & {} \Vert |\cdot |^\delta u(\tau ,\cdot )\Vert _{2}^{\frac{2 \tilde{\nu }}{\delta }} \; \Vert u(\tau ,\cdot )\Vert _{{\tilde{q}}}^{p -\frac{2 \tilde{\nu }}{\delta }}, \end{aligned}$$

(7.34)

where $q = \frac{ p \delta }{ \delta - \tilde{\nu }}$ and ${\tilde{q}} = q(1 -\frac{2 {\tilde{\nu }}}{p\delta })$. The assumption (7.32) implies

$$\begin{aligned} {\tilde{q}} = q \Bigg (1 -\frac{2\tilde{\nu }}{p\delta }\Bigg )= \frac{p \delta -2 \tilde{\nu }}{ \delta - \tilde{\nu }} > \frac{2n}{n + 2\delta }. \end{aligned}$$

In fact, the condition $\tilde{\nu } < \frac{n}{4}(p-2) + \frac{p \delta }{2}$ is equivalent to ${\tilde{q}} = \frac{p \delta - 2 \tilde{\nu }}{ \delta - \tilde{\nu }} > \frac{2n}{n + 2\delta }$. The condition (7.33) is equivalent to ${\tilde{q}}= \frac{p \delta - 2 \tilde{\nu }}{ \delta - \tilde{\nu }} \le \frac{2n}{n - 2 \bar{s}}$ in the case $n > 2 \bar{s}$. Hence, using (6.8) with taking q as ${\tilde{q}}$, and definition of $\Vert \cdot \Vert _X$ with $r = 1$ (see (6.2)) in the right-hand side of (7.34), we obtain

$$\begin{aligned} \left\| | \cdot |^{\tilde{2 \nu }} f(u(\tau ,\cdot )) \right\| _1\lesssim & {} \langle \tau \rangle ^ {{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(1 - \frac{1}{2}) + \sigma + \frac{\delta }{2} \right) \frac{2 {\tilde{\nu }}}{\delta } }} \langle \tau \rangle ^{{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(1 - \frac{1}{{\tilde{q}}}) + \sigma \right) (p -\frac{2 {\tilde{\nu }}}{\delta }) }} \Vert u\Vert _X^p\nonumber \\= & {} \langle \tau \rangle ^{{\frac{1}{1 - \sigma } \left( (-\frac{n}{2} + \sigma )p + \frac{n}{2} + {\tilde{\nu }} \right) }} \Vert {u}\Vert _{X}^p. \; \end{aligned}$$

(7.35)

Thus,

$$\begin{aligned} L_{2,3} \lesssim \langle t \rangle ^{\frac{1}{1-\sigma }(-\frac{n}{4} + \sigma - {\tilde{\nu }})} \; \int \nolimits _0^{t/2}\langle \tau \rangle ^{{\frac{1}{1 - \sigma } \left( \left( -\frac{n}{2} + \sigma \right) p + \frac{n}{2} + {\tilde{\nu }}\right) }} d\tau \Vert {u}\Vert _{X}^p, \end{aligned}$$

which yields

$$\begin{aligned} L_{2,3}&\lesssim {\left\{ \begin{array}{ll} \langle t \rangle ^ {{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - ( p - 1)\left( \frac{n}{2} - \sigma \right) + 1 \right) }} &{}\text {if} \quad \frac{1}{1 - \sigma } \left( -\frac{n}{2} (p-1) + p \sigma + {\tilde{\nu }}\right) > -1, \nonumber \\ \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - {\tilde{\nu }}\right) }} \log (\langle t \rangle + 1) &{}\text {if} \quad \frac{1}{1 - \sigma } \left( -\frac{n}{2} (p-1) + p \sigma + {\tilde{\nu }}\right) = -1, \nonumber \\ \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - {\tilde{\nu }}\right) }} &{}\text {if} \quad \frac{1}{1 - \sigma } \left( -\frac{n}{2} (p-1) + p \sigma + {\tilde{\nu }}\right) < -1. \end{array}\right. } \nonumber \\&\lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - \min \{( p - 1)\left( \frac{n}{2} - \sigma \right) - 1, \nu \} \right) }}. \end{aligned}$$

(7.36)

Inequalities (7.30), (7.31) and (7.36) yield

$$\begin{aligned} L_{2} \lesssim \langle t \rangle ^{{ \max \{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - \min \{1 - 2\sigma , ( p - 1)\left( \frac{n}{2} - \sigma \right) - 1, \nu \} \right) ,\; \frac{1}{\sigma } \left( -\frac{n}{4} + \sigma \right) \} }} \Vert u\Vert _{X}^p.\nonumber \\ \end{aligned}$$

(7.37)

We estimate $L_3$. By the definition of $G_\sigma$,

$$\begin{aligned} {{\mathcal {F}}}( G_{\sigma }(t-\tau ,\cdot ) - G_{\sigma }(t,\cdot ))&= \tau \int \nolimits _0^{1}\frac{\partial \widehat{G_{\sigma }}}{\partial t}(t- \theta \tau , \xi ) d \theta \\&= \tau \int \nolimits _0^{1} |\xi |^{2(1-2\sigma ) } e^{- |\xi |^{2(1 - \sigma )} (t- \theta \tau )} d \theta . \end{aligned}$$

Inequality (5.8) with $\theta = 2 - 2 \sigma$ implies

$$\begin{aligned} \begin{aligned} \left\| |\xi |^{2(1-2\sigma ) } e^{- |\xi |^{2(1 - \sigma )} (t- \theta \tau )} \right\| _2&\sim t^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} - 1 + 2 \sigma \right) }} \end{aligned} \end{aligned}$$

(7.38)

uniformly to $\theta \in [0,1]$ and $\tau \in [0,t/2]$. Then by using (7.5) with $r = 1$ together, we have

$$\begin{aligned} \begin{aligned} L_3&\lesssim t^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} - 1 + 2 \sigma \right) }} \int \nolimits _0^{t/2} \tau \Vert f(u(\tau ,\cdot ))\Vert _1 d\tau \\&\lesssim t^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} - 1 + 2 \sigma \right) }} \int \nolimits _0^{t/2} \tau \langle \tau \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2} (p-1) + p \sigma \right) }} d\tau , \end{aligned} \end{aligned}$$

(7.39)

which yields

$$\begin{aligned} L_3&\lesssim {\left\{ \begin{array}{ll} {t}^ {{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - ( p - 1)\left( \frac{n}{2} - \sigma \right) + 1 \right) }} &{}\text {if} \quad \frac{1}{1 - \sigma } \left( -\frac{n}{2} (p-1) + p \sigma \right) > -2 \\ {t}^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} - 1 + 2 \sigma \right) }} \log (\langle t \rangle + 1) \quad &{}\text {if} \quad \frac{1}{1 - \sigma } \left( -\frac{n}{2} (p-1) + p \sigma \right) = -2 \\ {t}^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} -1 + 2 \sigma \right) }} \quad &{}\text {if} \quad \frac{1}{1 - \sigma } \left( -\frac{n}{2} (p-1) + p \sigma \right) < - 2. \end{array}\right. } \nonumber \\&\le {t}^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - \min \{ ( p - 1)\left( \frac{n}{2} - \sigma \right) - 1, 1 - 2 \sigma \} \right) }}. \end{aligned}$$

(7.40)

Last we estimate $L_4$. Since $n \ge 2$, the assumptions (2.14) and (2.13) imply

$$\begin{aligned} \frac{2n}{n + 2\delta } \le \frac{2n}{2n - 2} = 1 + \frac{1}{n-1} \le 1 + \frac{2}{n - 2\sigma } < p. \end{aligned}$$

By this inequality and (2.7), we can apply (6.8) with $r = 1$ and $q = p$ to obtain

$$\begin{aligned} \Vert u(\tau ,\cdot ) \Vert _p \lesssim \langle \tau \rangle ^{{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(1 - \frac{1}{p}) + \sigma \right) }}\Vert {u}\Vert _{X}. \end{aligned}$$

This together with (7.9) yields

$$\begin{aligned} \int \nolimits _{t/2}^\infty \left| \int \nolimits _{{{\mathbb {R}}}^n}f(u(\tau ,y))dy \right| d\tau&\lesssim \int \nolimits _{t/2}^\infty \langle \tau \rangle ^{{\frac{1}{1 - \sigma } \left( -\frac{n}{2}(p - 1) + p \sigma \right) }} d\tau \Vert {u}\Vert _{X}^p \nonumber \\&\sim \langle t \rangle ^{{\frac{1}{1 - \sigma } \left( -(\frac{n}{2} - \sigma )(p - 1) + 1 \right) }} \Vert {u}\Vert _{X}^p. \end{aligned}$$

(7.41)

Taking the product of (2.4) and (7.41), we obtain

$$\begin{aligned} L_4 \lesssim {t}^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{4} + \sigma - ( p - 1)\left( \frac{n}{2} - \sigma \right) + 1 \right) }}. \end{aligned}$$

(7.42)

Substituting (7.28), (7.37), (7.40) and (7.42) into (7.25), we obtain (7.24). $\square$

8 Proof of Proposition and Theorems

8.1 Proof of Proposition 1

Let $\varepsilon > 0$, and

$$\begin{aligned} X(\varepsilon ) = X_{r,\delta , \bar{s}}(\varepsilon ) := \{ \varphi \in X_{r,\delta , \bar{s}}; \Vert \varphi \Vert _{{X_{r,\delta , \bar{s}}}} < \varepsilon \}, \end{aligned}$$

where $X_{r,\delta , \bar{s}}$ and $\Vert \varphi \Vert _{{X_{r,\delta , \bar{s}}}}$ are defined by (6.1) and (6.2), respectively. We put $X = X_{r,\delta , \bar{s}}$ and $\Vert \cdot \Vert _X =\Vert \varphi \Vert _{{X_{r,\delta , \bar{s}}}}$, throughout this subsection.

If u is a solution of (1.1), then Duhamel’s principles implies

$$\begin{aligned} u(t,x)=K_0(t,\cdot ) *u_0 +K_1(t,\cdot ) *u_1 + \int _0 ^t K_1(t- \tau ,\cdot )*f(u( \tau ,\cdot ) ) d\tau , \end{aligned}$$

where $K_0$ and $K_1$ are defined by (4.2) and (4.3). Taking account of the formula above, we define the mapping $\Phi$ on $X(\varepsilon )$ by

$$\begin{aligned} (\Phi u)(t) := K_0(t,\cdot )*u_0+K_1(t,\cdot )*u_1 + \int \nolimits _0^t K_1(t - \tau ,\cdot )* f(u(\tau ,\cdot )) d \tau .\nonumber \\ \end{aligned}$$

(8.1)

We prove that $\Phi$ is a contraction mapping on $X(\varepsilon )$ provided $\varepsilon$ and initial data are sufficiently small.

First we estimate $K_1(t,\dot{)}*u_1$. By (2.14), we see that assumtion (4.43) of Lemma 2 is satisfied for $\vartheta = 0$ and $\delta$, $s_1 = s_2 = 0$, $q_1 = r$ and $q_2 = \frac{nr}{n - r \vartheta }(\in [r,2))$ (that is, $\frac{1}{q_2} = \frac{1}{r} - \frac{\vartheta }{n}$). Then, (4.46) gives estimate of low frequency part. The high and middle frequency parts are given by (4.64) and (4.70). Then we have

$$\begin{aligned} {\left\| \langle \cdot \rangle ^\delta \left( K_{1}(t,\cdot )*u_1 \right) \right\| _{2}}\lesssim & {} \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} (\Vert u_1 \Vert _{r}^\prime + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{{ \frac{nr}{n - r \delta }, 2}} }}) \nonumber \\&\quad + e^{-\varepsilon _{\sigma } t} \Vert (1 - \Delta )^{-\frac{1}{2}} \langle \cdot \rangle ^\delta u_1 \Vert _{{{{2}} }}. \end{aligned}$$

(8.2)

Assumption (2.14) implies $\frac{ n - r \delta }{r} - 1 \le \frac{n}{2}$, and therefore, sharp Sobolev’s embedding theorem (Lemma E) yields

$$\begin{aligned} \Vert (1 - \Delta )^{-\frac{1}{2}} \langle \cdot \rangle ^\delta u_1 \Vert _{{2}} \le \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{{ \frac{nr}{n - r \delta }},2 }}}. \end{aligned}$$

Substituting this inequality into (8.2), we obtain

$$\begin{aligned} \begin{aligned}&{\left\| \langle \cdot \rangle ^\delta \left( K_{1}(t,\cdot )*u_1 \right) \right\| _{2}} \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} (\Vert u_1 \Vert _{r}^\prime + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{ \frac{nr}{n - r \delta }, 2}} ). \end{aligned}\nonumber \\ \end{aligned}$$

(8.3)

Inequality (4.46) with $\vartheta = 0$, $s_1 = \bar{s}$, $s_2 = 0$, $q_1 =q_2 = r$, and inequalities (4.63) and (4.69) with $s = \bar{s}$ yield

$$\begin{aligned}&{\left\| (-\Delta )^{\frac{\bar{s}}{2}} \left( K_{1}(t,\cdot ) * u_1 \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{\bar{s}}{2} + \sigma \right) }} \Vert u_1\Vert _{{r}}^\prime + e^{-\varepsilon _{\sigma } t} \Vert (-\Delta )^{\frac{\bar{s}}{2}} (1 - \Delta )^{-\frac{1}{2}} u_1 \Vert _{{2}}. \end{aligned}$$

(8.4)

Next we estimate $K_0(t,\dot{)}*u_0$. By (2.14), we see that the assumption of Lemma 2 is satisfied for

$$\begin{aligned} \vartheta = 0 \;\text {and}\; \delta , \quad s_1 = s_2 = 0, \quad q_3 = \frac{nr}{n - 2 r \sigma }, \quad q_4 = \frac{nr}{n - r(\vartheta + 2\sigma )}. \end{aligned}$$

Then, (4.49) gives estimate of low frequency part. Inequalities (4.66) and (4.72) give estimate of high and middle frequency parts. Then we obtain

$$\begin{aligned} \begin{aligned}&{\left\| \langle \cdot \rangle ^\delta \left( K_{0}(t,\cdot ) *u_0 \right) \right\| _{2}} \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} (\Vert u_0 \Vert _{{ \frac{nr}{n -2 r \sigma }, 2}} + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\frac{nr}{n - r(\delta + 2\sigma )},2}} }}) + e^{-\varepsilon _{\sigma } t} \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{2}} }}. \end{aligned} \end{aligned}$$

By Corollary A, we have

$$\begin{aligned} \Vert u_0 \Vert _{{ \frac{nr}{n -2 r \sigma },2 }} \lesssim \Vert | x |^{-\delta }\Vert _{{{\frac{n}{\delta }, \infty } }} \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\frac{nr}{n - r(\delta + 2\sigma )},2}} }} \lesssim \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\frac{nr}{n - r(\delta + 2\sigma )},2}} }}. \end{aligned}$$

(8.5)

Hence, we have

$$\begin{aligned} \begin{aligned}&{\left\| \langle \cdot \rangle ^\delta \left( K_{0}(t,\cdot ) *u_0 \right) \right\| _{2}} \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) + \frac{\delta }{2} + \sigma \right) }} \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\frac{nr}{n - r(\delta + 2\sigma )},2}} }} + e^{-\varepsilon _{\sigma } t} \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{2}} }}. \end{aligned} \end{aligned}$$

(8.6)

Inequality (4.49) with $\vartheta = 0$, $s_1 = \bar{s}$, $s_2 = 0$ and $q_3 = q_4 =\frac{nr}{n - 2 r \sigma }$ and inequalities (4.65) and (4.71) with $s = \bar{s}$ imply

$$\begin{aligned}&{\left\| (-\Delta )^{\frac{\bar{s}}{2}} \left( K_{0}(t,\cdot ) * u_0 \right) \right\| _{2}} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{\bar{s}}{2} + \sigma \right) }} {\left\| u_0\right\| _{{\frac{nr}{n - 2 r \sigma } },2 }} + e^{-\varepsilon _{\sigma } t} \Vert u_0 \Vert _{{H^{\bar{s}} }} \nonumber \\&\quad \lesssim \langle t \rangle ^{{ \frac{1}{1 - \sigma } \left( -\frac{n}{2}(\frac{1}{r} - \frac{1}{2}) - \frac{\bar{s}}{2} + \sigma \right) }} \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\frac{nr}{n - r(\delta + 2\sigma )},2}} }} + e^{-\varepsilon _{\sigma } t} \Vert u_0 \Vert _{{H^{\bar{s}} }}. \end{aligned}$$

(8.7)

For the last inequality, we used (8.5).

By (8.1), (8.3), (8.4), (8.6), (8.7), (7.2) and (7.14) with $s = 0,\bar{s}$, we have

$$\begin{aligned} \begin{aligned} \Vert \Phi u \Vert _X&\lesssim \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\frac{nr}{n - r(\delta + 2\sigma )}},2} }} + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{2}} }} + \Vert u_0 \Vert _{{H^{\bar{s}} }} \\&\quad + \Vert u_1\Vert _{r}^\prime + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{{ \frac{nr}{n - r \delta },2}} }} + \Vert (-\Delta )^{\frac{\bar{s}}{2}} (1 - \Delta )^{-\frac{1}{2}} u_1 \Vert _{{2}} + \Vert u \Vert _X^p. \end{aligned} \end{aligned}$$

(8.8)

(i) First we consider the case $r = 1$. By (8.8), there is a positive constant $C_1$ independent of initial data such that

$$\begin{aligned} \begin{aligned} \Vert \Phi u \Vert _X&\le C_1 \Big ( \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{\frac{n}{n - (\delta + 2\sigma )}, 2}} + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{2}} }} + \Vert u_0 \Vert _{{H^{\bar{s}} }}\\&\quad + \Vert u_1\Vert _{{1}} + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{ \frac{n}{n - \delta },2 }} + \Vert (-\Delta )^{\frac{\bar{s}}{2}} (1 - \Delta )^{-\frac{1}{2}} u_1 \Vert _{{2}} + \Vert u \Vert _X^p \Big ). \end{aligned} \end{aligned}$$

Hence, taking $\varepsilon _1 > 0$ such that $C_1 \varepsilon _1^{p-1} \le \frac{1}{2}$, and assuming $u_0$ and $u_1$ satisfy

$$\begin{aligned} \begin{aligned}&C_1 \Big ( \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{\frac{n}{n - (\delta + 2\sigma )},2 }} + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{2} + \Vert (-\Delta )^{\frac{\bar{s}}{2}} u_0 \Vert _{{2}} \\&\quad + \Vert u_1\Vert _{{{1}}} + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{ \frac{n}{n - \delta }, 2}} + \Vert (-\Delta )^{\frac{\bar{s}}{2}} (1 - \Delta )^{-\frac{1}{2}} u_1 \Vert _{{2}} \Big ) \le \frac{\varepsilon _1}{2}, \end{aligned} \end{aligned}$$

we see that $\Phi$ is a mapping from $X(\varepsilon _1)$ to $X(\varepsilon _1)$.

By (7.3) and (7.15), there is a positive constant $C_2$ independent of initial data such that

$$\begin{aligned} \Vert \Phi u - \Phi v \Vert _X \le C_2 (\Vert u \Vert _X + \Vert v \Vert _X)^{p-1} \Vert u - v \Vert _X \end{aligned}$$

(8.9)

for every $u, v \in X$. We take $\varepsilon \in (0,\varepsilon _1)$ such that

$$\begin{aligned} C_2 (2 \varepsilon )^{p-1} < 1. \end{aligned}$$

Then we see that $\Phi$ is a contraction mapping from $X(\varepsilon )$ to $X(\varepsilon )$, and therefore $\Phi$ has the only one fixed point u, which is the unique solution.

(ii) Next we consider the case $r \in (1,\frac{2n}{n + 2 \sigma }]$. By Corollary A, we have

$$\begin{aligned} \Vert u_1 \Vert _{{{r}}}^\prime = \Vert u_1 \Vert _{{{r,2}}} \lesssim \Vert | x |^{-\delta }\Vert _{{{\frac{n}{\delta }, \infty } }} \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{{ \frac{nr}{n - r \delta },2}} }} \lesssim \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{{ \frac{nr}{n - r \delta }},2} }}. \end{aligned}$$

Substituting this inequality into (8.8), we obtain

$$\begin{aligned} \Vert \Phi u \Vert _X\le & {} C_3 \Big ( \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\frac{nr}{n - r(\delta + 2\sigma )}},2} }} + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{2}} }} + \Vert u_0 \Vert _{{H^{\bar{s}} }} \nonumber \\&\quad + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{{ \frac{nr}{n - r \delta },2}} }} + \Vert (-\Delta )^{\frac{\bar{s}}{2}} (1 - \Delta )^{-\frac{1}{2}} u_1 \Vert _{{2}} + \Vert u \Vert _X^p \Big ), \end{aligned}$$

(8.10)

for a positive constant $C_3$ independent of initial data. Hence, taking $\varepsilon _2 > 0$ such that $C_3 \varepsilon _2^{p-1} \le \frac{1}{2}$, and assuming $u_0$ and $u_1$ satisfy

$$\begin{aligned} \begin{aligned}&C_3 \Big ( \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{\frac{nr}{n - r(\delta + 2\sigma )}},2} }} + \Vert \langle \cdot \rangle ^\delta u_0 \Vert _{{{{2}} }} + \Vert u_0 \Vert _{{H^{\bar{s}} }} \\&\quad + \Vert \langle \cdot \rangle ^\delta u_1 \Vert _{{{{ \frac{nr}{n - r \delta },2}} }} + \Vert (-\Delta )^{\frac{\bar{s}}{2}} (1 - \Delta )^{-\frac{1}{2}} u_1 \Vert _{{2}} \Big ) \le \frac{\varepsilon _2}{2}, \end{aligned} \end{aligned}$$

we see that $\Phi$ is a mapping from $X(\varepsilon _2)$ to $X(\varepsilon _2)$. In the same way as (8.9), we see that there is a positive constant $C_4$ independent of initial data satisfying

$$\begin{aligned} \Vert \Phi u - \Phi v \Vert _X \le C_4 (\Vert u \Vert + \Vert v \Vert )^{p-1} \Vert u - v \Vert _X \end{aligned}$$

for every $u, v \in X$. We take $\varepsilon \in (0,\varepsilon _2)$ such that $C_4 (2 \varepsilon )^{p-1} < 1$. Then $\Phi$ becomes a contraction mapping $X(\varepsilon )$ to $X(\varepsilon )$, and therefore $\Phi$ has the only one fixed point u, which is the unique solution.

8.2 Proof of Theorems 2 and 3

Proof of Theorems 2

We prove Theorem 2 by using Proposition 1.

(Case 1) In the case $p_\sigma < p \le 1 + \frac{4}{n}$, we can $\eta > 0$ sufficiently small such that r defined by

$$\begin{aligned} \frac{1}{r} = \frac{2}{n} \left( \frac{1}{p-1} + \sigma \right) + \eta \end{aligned}$$

(8.11)

satisfies

$$\begin{aligned} \frac{1}{2} + \frac{2 \sigma }{n}< \frac{1}{r} < 1, \quad \text {that is}, \quad r \in \Bigg (1, \frac{2n}{n + 4 \sigma }\Bigg ). \end{aligned}$$

(8.12)

Then $\frac{1}{r} > \frac{2}{n} \left( \frac{1}{p-1} + \sigma \right)$ implies assumtion (2.13).

(Case 1-1) First we consider the case $p_\sigma < p \le 1 + \frac{4}{n + 2 - 4 \sigma }$. We put

$$\begin{aligned} \delta ^\prime = 2 \left( \frac{1}{p-1} + \sigma \right) - \frac{n}{2} - 1 + n \eta . \end{aligned}$$

(8.13)

By definition (8.11), we have

$$\begin{aligned} \delta ^\prime = n \left( \frac{1}{r} - \frac{1}{2} \right) - 1. \end{aligned}$$

(8.14)

Comparing (2.9) and (8.13), we see that

$$\begin{aligned} \delta ^\prime \le \delta \end{aligned}$$

(8.15)

if $\eta >0$ is sufficiently small. Hence, taking $\eta > 0$ sufficiently small, we can assume that r and $\delta ^\prime$ defined above satisfy (8.12) and (8.15). We check that the assumptions (2.14) and (2.15) of Proposition 1 are satisfied with $\delta$ replaced $\delta ^\prime$. Since $2 \sigma < 1$, (2.14) is trivial by (8.14). Since $n \ge 2$ and $2 \sigma < 1$, we have

$$\begin{aligned} p \le 1 + \frac{4}{n + 2 - 4 \sigma } < 1 + \frac{1}{1 - 2\sigma } = \frac{2 - 2\sigma }{1 - 2\sigma }, \end{aligned}$$

from which it follows that

$$\begin{aligned} 2\left( \frac{1}{p - 1} + \sigma \right) - 1 > \frac{2}{p} \left( \frac{1}{p - 1} + \sigma \right) . \end{aligned}$$

From this and the definition of $\delta ^\prime$ and r, it follows that

$$\begin{aligned} \delta ^\prime = 2 \left( \frac{1}{p-1} + \sigma \right) -1 - \frac{n}{2} + n \eta > \frac{2}{p} \left( \frac{1}{p - 1} + \sigma \right) + n \eta - \frac{n}{2} = n\left( \frac{1}{pr} - \frac{1}{2} \right) ,\!\!\!\!\!\!\nonumber \\ \end{aligned}$$

(8.16)

that is, (2.15) is satisfied with $\delta$ replaced by $\delta ^\prime$. Hence the assumption of Proposition 1 is satisfied. Let ${\hat{q}}_j$ $(j = 0,1)$ be the constants defined by (2.16) with $\delta = \delta ^\prime$ and r defined above. Then

$$\begin{aligned} \frac{1}{{\hat{q}}_0 } = \frac{n - r(\delta ^\prime + 2 \sigma )}{nr} = \frac{n + 2 - 4 \sigma }{2 n}, \quad \frac{1}{{\hat{q}}_1 } = \frac{n - r \delta ^\prime }{nr} = \frac{n+2}{2n}, \end{aligned}$$

(8.17)

that is, ${\hat{q}}_j = q_j$ $(j = 0,1)$. Since $\delta ^\prime \le \delta$, the assumptions (2.10) and (2.11) imply (2.17) and (2.19) with $\delta$ replaced by $\delta ^\prime$. Thus, Proposition 1 guarantees the existence of the solution $u \in C^1([0,\infty ), H^{\bar{s}}) \cap C([0,\infty ), H^{\bar{s} - 1})$ if $\varepsilon$ is sufficiently small. By the standard argument, the uniqueness holds in the class $C^1([0,\infty ), H^{\bar{s}}) \cap C([0,\infty ), H^{\bar{s} - 1})$.

(Case 1-2) Next we consider the case

$$\begin{aligned} 1 + \frac{4}{n + 2 - 4\sigma } < p \le 1 + \frac{4}{n}. \end{aligned}$$

(8.18)

We show that $\delta = 0$ satisfies the assumptions (2.14) and (2.15), that is,

$$\begin{aligned}&2 \left( \frac{1}{p-1} + \sigma \right) + n \eta - \frac{n}{2} - 1 = n\Bigg (\frac{1}{r} - \frac{1}{2}\Bigg ) - 1 \le 0 < n\Bigg (\frac{1}{r} - \frac{1}{2}\Bigg ) - 2\sigma , \end{aligned}$$

(8.19)

$$\begin{aligned}&\frac{2}{p}\left( \frac{1}{p-1} + \sigma \right) + \frac{n \eta }{p} - \frac{n}{2} = n\Bigg (\frac{1}{pr} - \frac{1}{2}\Bigg ) \le 0, \end{aligned}$$

(8.20)

if $\eta > 0$ is sufficiently small.

The assumption (8.12) implies $n(\frac{1}{r} - \frac{1}{2}) - 2\sigma > 0$. The assumption $1~+~\frac{4}{n + 2 - 4\sigma } < p$ is equivalent to

$$\begin{aligned} 2 \left( \frac{1}{p-1} + \sigma \right) - \frac{n}{2} - 1 < 0. \end{aligned}$$

(8.21)

Hence (8.19) holds if $\eta$ is sufficiently small.

From the assumption that $p \ge 1 + \frac{4}{n + 2 - 4 \sigma }$, it follows that

$$\begin{aligned} \frac{2}{p}\left( \frac{1}{p-1} + \sigma \right) - \frac{n}{2}&\le \frac{2(n + 2 - 4\sigma )}{n + 6 - 4\sigma }\left( \frac{n + 2 - 4\sigma }{4} + \sigma \right) - \frac{n}{2} \\&= -\frac{n+ 4 \sigma - 2}{n + 6 - 4\sigma } < 0, \end{aligned}$$

and thus assumtion (8.20) holds if $\eta > 0$ is sufficiently small. Hence the assumption of Proposition 1 is satisfied with $\delta = 0$. Let ${\hat{q}}_j$ $(j = 0,1)$ be the constants defined by (2.16) with $\delta = 0$, and r be defined by (8.11) for sufficiently small $\eta$ satisfying conditions described above. Then

$$\begin{aligned} \frac{1}{{\hat{q}}_0 } = \frac{1}{r} - \frac{2 \sigma }{n} = \frac{2}{ n(p-1)} + \eta , \quad \frac{1}{{\hat{q}}_1 } = \frac{1}{r} = = \frac{2}{n(p-1)} + \frac{2 \sigma }{n} + \eta .\nonumber \\ \end{aligned}$$

(8.22)

Then by the assumption of $q_j$ ($j = 0,1$), we have

$$\begin{aligned} q_j < {\hat{q}}_j \quad (j = 0,1), \end{aligned}$$

(8.23)

if $\eta > 0$ is sufficiently small. Since $p \le 1 + \frac{4}{n}$,

$$\begin{aligned} \frac{1}{{\hat{q}}_j} > \frac{2}{n(p-1)} \ge \frac{1}{2}, \quad \text {that is,} \quad {\hat{q}}_j < 2 \quad (j = 0,1). \end{aligned}$$

(8.24)

By (8.23) and (8.24),

$$\begin{aligned} L_{{\hat{q}}_j,2 } \subset L_{q_j,2} \cap L_{2}. \end{aligned}$$

Hence (2.11) implies (2.19). Thus the conclusion holds by Proposition 1 in the same way as above.

(Case 2) Last we consider the case $p \ge 1 + \frac{4}{n}$. We define r by

$$\begin{aligned} \frac{1}{r} = \frac{1}{2} + \frac{2\sigma }{n} + \eta \quad (\eta > 0). \end{aligned}$$

(8.25)

Since $2 \sigma < 1$, (2.14) holds for $\delta = 0$ if $\eta > 0$ is sufficiently small. The assumption

$$\begin{aligned} n\left( \frac{1}{pr} - \frac{1}{2}\right) = \frac{n}{p}\left( \frac{1}{2} + \frac{2\sigma }{n} + \eta \right) - \frac{n}{2} < 0 \end{aligned}$$

is equivalent to

$$\begin{aligned} 1 + \frac{4\sigma }{n} + 2 \eta < p, \end{aligned}$$

which holds if $\eta > 0$ is sufficiently small, since $\sigma < 1$ and $p \ge 1 + \frac{4}{n}$. Hence, defining r by (8.25) with sufficiently small $\eta > 0$, we can take $\delta = 0$ in Proposition 1. Let ${\hat{q}}_j$ $(j = 0,1)$ be the constants defined by (2.16) with $\delta = 0$ and r defined above. Then, considering the asumption of $q_j$ ($j = 0,1$), we see that

$$\begin{aligned} \frac{1}{{\hat{q}}_0 } = \frac{1}{r} - \frac{2 \sigma }{n} = \frac{1}{2} + \eta< \frac{1}{q_0}, \quad \frac{1}{{\hat{q}}_1 } = \frac{1}{r} = \frac{1}{2} + \frac{2 \sigma }{n} + \eta < \frac{1}{q_1}, \end{aligned}$$

(8.26)

if $\eta > 0$ is sufficiently small. This imply that $q_j< {\hat{q}}_j < 2$ $(j = 0,1$). Hence (2.12) implies (2.19) with $\delta = 0$, and the conclusion holds by Proposition 1.

$\square$

Proof of Theorem 3

Since $u =\Phi u$ ($\Phi$ is defined by (8.1)), we can write

$$\begin{aligned}&u(t,\cdot ) - \varTheta G_{\sigma }(t,x) = \left( K_0(t,\cdot )*u_0 + K_1(t,\cdot )*u_1 - \vartheta _1 G_{\sigma }(t,x) \right) \nonumber \\&\quad + \left( \int \nolimits _0^t K_1(t - \tau ,\cdot )* f(u(\tau ,\cdot )) d \tau - G_{\sigma }(t,x) \int \nolimits _0^\infty \int \nolimits _{{{\mathbb {R}}}^n}f(u(\tau ,y))dy d\tau \right) , \end{aligned}$$

(8.27)

where $\vartheta _1$ is defined by (2.3). Since $K_0(t,\cdot )*u_0 + K_1(t,\cdot )*u_1$ is a solution of the linear equation (1.3), the first term of the right-hand side of (8.27) is estimated by Theorem 1. The second term is estimated in (7.24). Combining these estimates, we obtain the assertion. $\square$