Asymptotic behaviors of global solutions for a semilinear diffusion equation in the de Sitter spacetime

Abstract

A semilinear diffusion equation is considered in the de Sitter spacetime with the spatially flat curvature. Global solutions for small initial data and their asymptotic behaviors in high order are obtained. Some effects of the spatial expansion and contraction are studied through the problem.

Keywords

Semilinear diffusion equation global solution asymptotic behavior de Sitter spacetime

1. Introduction

In this paper, we consider the Cauchy problem for a semilinear diffusion equation $\begin{matrix} (1.1) & \{\begin{matrix} \partial_{t} u (t, x) - Δ u (t, x) = A (t) f (u) (t, x), \\ u (0, x) = u_{0} (x) \end{matrix} \end{matrix}$ for $0 ⩽ t < T_{0}$ and $x \in R^{n}$ , where $u = u (t, x)$ , $n ⩾ 1$ , $H \in R$ is the Hubble constant, $1 < p < \infty$ , $A (\cdot)$ is a weight function defined by $\begin{matrix} (1.2) & A (t) : = {(1 - 2 H t)}^{n (p - 1) / 4 - 1}, \end{matrix}$ $f (u)$ is a semilinear term defined by $\begin{matrix} (1.3) & f (u) : = λ | u |^{p - 1} u or λ | u |^{p} \end{matrix}$ for $λ \in R$ , $u_{0}$ is a given initial datum, and $T_{0}$ is defined by $\begin{matrix} T_{0} : = \{\begin{matrix} \infty & if H ⩽ 0, \\ \frac{1}{2 H} & if H > 0 . \end{matrix} \end{matrix}$ The de Sitter spacetime is the solution of the Einstein equations with the cosmological constant in the vacuum. We consider the de Sitter spacetime with the spatially flat curvature whose metric is given by $- c^{2} {(d τ)}^{2} = - c^{2} {(d t)}^{2} + e^{2 H t} \sum_{j = 1}^{n} {(d x^{j})}^{2}$ , which describes the spatial expansion and contraction when $H > 0$ and $H < 0$ , respectively, where c denotes the speed of the light, and τ denotes the proper time. We note that the case $H = 0$ yields the Minkowski spacetime. The derivation of the first equation in (1.1) by the nonrelativistic limit of a semilinear field equation has been shown in [18]. In this paper, we study the effects of the spatial expansion and contraction through the Cauchy problem of the semilinear diffusion equation in the de Sitter spacetime. Our main purpose is to study the large time behavior of the solution of (1.1).

The corresponding Cauchy problem for the semilinear Schrödinger equation in the de Sitter spacetime has been considered in [17]. The global solutions for small data for the Klein–Gordon equation have been shown in asymptotically de Sitter spacetime in [3]. The wave equation with quadratic nonlinear term in asymptotically de Sitter and Kerr–de Sitter spacetimes has been considered in [11]. The global solutions for semilinear wave equations have been shown on the manifold with the time slices being real hyperbolic spaces in [1,2,13,14]. The Klein-Gordon equation in the de Sitter spacetime has been considered in [4] for the tachyonic field, in [27] for the Huygens principle, and in [26] for the Higgs scalar field. The Cauchy problem of the semilinear Klein–Gordon equations has been considered in [8,16,25] in the de Sitter spacetime, and in [7] in the Friedmann–Lemaître–Robertson–Walker spacetime.

We denote the Lebesgue space of order $1 ⩽ r ⩽ \infty$ by $L^{r}$ with the norm $‖ \cdot ‖_{L^{r}}$ . Put $⟨ x ⟩ : = {(1 + | x |^{2})}^{1 / 2}$ for $x \in R^{n}$ . For any real number $k ⩾ 0$ , we define the weighted Lebesgue space $L_{k}^{1} (R^{n})$ by $\begin{matrix} L_{k}^{1} (R^{n}) : = {f \in L^{1} (R^{n}); {⟨ \cdot ⟩}^{k} f (\cdot) \in L^{1} (R^{n})} \end{matrix}$ with the norm $‖ f ‖_{L_{k}^{1} (R^{n})} : = ‖ {⟨ | \cdot | ⟩}^{k} f (\cdot) ‖_{L^{1} (R^{n})}$ . For $m ⩾ 0$ and $1 ⩽ r ⩽ \infty$ , we denote the Sobolev space by $\begin{matrix} W^{m, r} (R^{n}) : = {f \in L^{r} (R^{n}); \partial^{α} f \in L^{r} (R^{n}) for | α | ⩽ m} \end{matrix}$ with the norm $‖ f ‖_{W^{m, r} (R^{n})} : = {\sum_{| α | ⩽ m} ‖ \partial^{α} f ‖_{L^{r} (R^{n})}^{r}}^{1 / r}$ . We also denote the homogeneous Sobolev space by $\begin{matrix} {\dot{W}}^{m, r} (R^{n}) : = {f \in L_{loc}^{1} (R^{n}); \partial^{α} f \in L^{r} (R^{n}) for | α | = m} \end{matrix}$ with $‖ f ‖_{{\dot{W}}^{m, r} (R^{n})} : = {\sum_{| α | = m} ‖ \partial^{α} f ‖_{L^{r} (R^{n})}^{r}}^{1 / r}$ .

We denote the set of natural numbers by $N : = {1, 2, 3, \dots}$ . For any real number $r > 0$ , we denote the largest integer which is less than r or equal to r by $[r]$ . For $k = 0, 1, 2, \dots$ , we define the function space defined by $\begin{matrix} (1.4) & X_{k} : = {u \in L^{\infty} ((0, \infty), L_{loc}^{1} (R^{n})); ‖ u ‖_{X^{\infty}} < \infty, ‖ u ‖_{X_{0}^{1}} < \infty, ‖ u ‖_{X_{k}^{1}} < \infty} \end{matrix}$ with the metric $d (u, v) : = ‖ u - v ‖_{X_{k}}$ , where we have put $\begin{array}{l} ‖ u ‖_{X^{\infty}} : = sup_{t > 0} {(1 + t)}^{n / 2} {‖ u (t, \cdot) ‖}_{L^{\infty} (R^{n})}, \\ ‖ u ‖_{X_{k}^{1}} : = sup_{t > 0} {(1 + t)}^{- k / 2} {‖ | x |^{k} u (t, x) ‖}_{L_{x}^{1} (R^{n})}, \\ ‖ u ‖_{X_{k}} : = max {‖ u ‖_{X^{\infty}}, ‖ u ‖_{X_{0}^{1}}, ‖ u ‖_{X_{k}^{1}}} . \end{array}$ For real numbers $R_{\infty} > 0$ and $R_{1} > 0$ , we put $\begin{matrix} X_{k} (R_{\infty}, R_{1}) : = {u \in X_{k}; ‖ u ‖_{X^{\infty}} ⩽ R_{\infty}, ‖ u ‖_{X_{0}^{1}} ⩽ R_{1}, ‖ u ‖_{X_{k}^{1}} ⩽ R_{1}} . \end{matrix}$

We denote the Gauss kernel $G (t, x)$ for the diffusion equation by $\begin{matrix} (1.5) & G (t, x) : = \frac{1}{{(4 π t)}^{n / 2}} e^{- | x |^{2} / 4 t} \end{matrix}$ for $t > 0$ and $x \in R^{n}$ , and we define the operator $e^{t Δ}$ by $e^{t Δ} : = G (t, x) *_{x}$ , where $*_{x}$ denotes the convolution for the variables x. We regard the solution of the Cauchy problem (1.1) as the fixed point of the mapping Φ defined by $\begin{matrix} (1.6) & Φ (u) (t) : = e^{t Δ} u_{0} + \int_{0}^{t} e^{(t - s) Δ} A (s) f (u) (s) d s . \end{matrix}$

First, we show the existence of the solutions of the Cauchy problem (1.1) for $H ⩽ 0$ .

Theorem 1.1 (Global solutions for $H ⩽ 0$ ).

Let $n \in N$ , $H ⩽ 0$ , $k \in N \cup {0}$ . Let p satisfy $\begin{matrix} (1.7) & \{\begin{matrix} 1 + \frac{2}{n} < p < \infty & if H = 0, \\ 1 < p < \infty & if H < 0 . \end{matrix} \end{matrix}$ Let $u_{0} \in L_{k}^{1} (R^{n}) \cap L^{\infty} (R^{n})$ . If $‖ u_{0} ‖_{L_{k}^{1} (R^{n}) \cap L^{\infty} (R^{n})}$ is sufficiently small, then there exist $R_{\infty} > 0$ , $R_{1} > 0$ and a unique global solution u of ( 1.1 ) in $X_{k} (R_{\infty}, R_{1}) \cap C ([0, \infty), L_{k}^{1} (R^{n}))$ satisfying the estimates $\begin{array}{l} (1.8) & {‖ u (t) ‖}_{L^{\infty} (R^{n})} ⩽ C {(1 + t)}^{- n / 2} ‖ u_{0} ‖_{L^{1} (R^{n}) \cap L^{\infty} (R^{n})}, \\ (1.9) & {‖ u (t) ‖}_{L_{k}^{1} (R^{n})} ⩽ C {(1 + t)}^{k / 2} ‖ u_{0} ‖_{L_{k}^{1} (R^{n})} \end{array}$ for $t ⩾ 0$ .

The global solutions in Theorem 1.1 have the following asymptotic behaviors.

Theorem 1.2 (Asymptotic behaviors for $H ⩽ 0$ ).

Under the assumption in Theorem 1.1 , moreover, assume $\begin{matrix} (1.10) & p > \{\begin{matrix} 1 + \frac{2 + k}{n} & if H = 0 or p = 1 + \frac{4}{n}, \\ 1 + \frac{2 k}{n} & if H < 0 and p \neq 1 + \frac{4}{n} . \end{matrix} \end{matrix}$ Let $1 ⩽ q ⩽ \infty$ . The solution u obtained in Theorem 1.1 satisfies $\begin{matrix} (1.11) & {‖ u (t, \cdot) - v (t, \cdot) ‖}_{L^{q} (R^{n})} = o (t^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ as $t \to \infty$ , where $\begin{array}{l} (1.12) & m_{α} : = \frac{1}{α!} \int_{R^{n}} {(- y)}^{α} u_{0} (y) d y, \\ (1.13) & M_{j, α} : = \frac{1}{j! α!} \int_{0}^{\infty} {(- s)}^{j} A (s) \int_{R^{n}} {(- y)}^{α} f (u) (s, y) d y d s \end{array}$ and $\begin{matrix} (1.14) & v (t, \cdot) : = \sum_{| α | ⩽ k} m_{α} \partial_{x}^{α} G (t, \cdot) + \sum_{\begin{array}{c} 0 ⩽ j ⩽ [k / 2] \\ 2 j + | α | ⩽ k \end{array}} M_{j, α} \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot) . \end{matrix}$ Here, $| m_{α} | < \infty$ and $| M_{j, α} | < \infty$ hold for j and α with $2 j + | α | ⩽ k$ .

The corresponding result in Theorem 1.2 in the case $H = 0$ has been shown in [12] with a different expression of the asymptotic function v, and the result in Theorem 1.1 in the case $H = 0$ follows from the estimates in it (see [21]). Our results extend the results in the case $H = 0$ into $H ⩽ 0$ , and our proofs show how the spatial contraction by $H < 0$ affects the problem compared with the case $H = 0$ . The basic idea of our proofs is based on the method in [5] and [15] (see also [10,22]), which obtained a higher order expansion of the solutions of the Navier–Stoke equations in terms of Gaussian-like functions. We can find similar argument in [23], which dealt with nonlinear damped wave equations. The exponent $1 + 2 / n$ in (1.7) is called the Fujita exponent. When $H = 0$ , it is well-known for (1.1) that any nontrivial and nonnegative solution blows up in finite time when $p ⩽ 1 + 2 / n$ , while the solution u exists globally for small data when $p > 1 + 2 / n$ [6,9,24]. Theorem 1.1 shows that the Fujita exponent disappears if the space is contracting (i.e., $H < 0$ ). We note that $1 + 2 k / n < 1 + (2 + k) / n$ holds if and only if $k = 0, 1$ .

Next, we show the existence of solutions of the Cauchy problem (1.1) for $H > 0$ . We say that the solution u of (1.1) is global if u exists on $[0, 1 / 2 H)$ . Let $1 ⩽ r ⩽ \infty$ . For integers $N ⩾ 0$ and $l ⩾ 0$ , we define the function spaces D and Y by $\begin{matrix} D : = \{\begin{matrix} L^{r} (R^{n}) \cap L^{\infty} (R^{n}) & if N = l = 0, \\ L^{r} (R^{n}) \cap L^{\infty} (R^{n}) \cap ⋂_{m = 1}^{2 N + l} W^{m, r_{m}} (R^{n}) & if N \neq 0 or l \neq 0, \end{matrix} \end{matrix}$ and $Y : = L^{\infty} ((0, 1 / 2 H), D)$ with the metric $\begin{matrix} d (u, v) : = ‖ u - v ‖_{L^{\infty} ((0, 1 / 2 H), L^{r} (R^{n}) \cap L^{\infty} (R^{n}))}, \end{matrix}$ where $r_{m}$ is defined by $\begin{matrix} (1.15) & \frac{1}{r_{m}} = \frac{m}{(2 N + l) r} \end{matrix}$ for $1 ⩽ m ⩽ 2 N + l$ . For $R > 0$ , put $\begin{matrix} Y (R) : = {u \in Y; ‖ u ‖_{Y} ⩽ R} . \end{matrix}$

Theorem 1.3 (Global solutions for $H > 0$ ).

Let $n \in N$ , $H > 0$ , $1 ⩽ r ⩽ \infty$ , $N ⩾ 0$ , $l ⩾ 0$ . Let p satisfy $max {1, 2 N + l} < p < \infty$ . Let $u_{0} \in D$ . Then the following results hold.

There exists constants $C_{0} > 0$ and $C > 0$ such that if $u_{0}$ satisfies $‖ u_{0} ‖_{D}^{p - 1} < C H$ , then there exists a unique global solution u of ( 1.1 ) in $Y (C_{0} ‖ u_{0} ‖_{D})$ .

The solution u in (1) satisfies $u \in C ([0, 1 / 2 H), D^{'})$ , where $D^{'}$ is defined by $\begin{matrix} D^{'} : = \{\begin{matrix} W^{2 N + l, r} (R^{n}) & if r \neq \infty, \\ W^{2 N + l - 1, \infty} (R^{n}) & if r = \infty and “ N \neq 0 or l \neq 0 . ” \end{matrix} \end{matrix}$

The solution u in (1) is unique in $C ([0, 1 / 2 H), D^{'}) \cap L^{\infty} ((0, 1 / 2 H) \times R^{n})$ .

The solutions in Theorem 1.3 have the following asymptotic behaviors.

Theorem 1.4 (Asymptotic behaviors for $H > 0$ ).

Under the assumption in Theorem 1.3 , let $l = 2$ . Put $\begin{array}{l} v_{0} (t) : = \sum_{j = 0}^{N} \frac{1}{j!} {(\frac{1}{2 H} - t)}^{j} {(- Δ)}^{j} e^{Δ / 2 H} u_{0}, \\ v_{1} (t) : = \sum_{j = 0}^{N} \frac{1}{j!} {(\frac{1}{2 H} - t)}^{j} {(- Δ)}^{j} \int_{0}^{t} e^{(1 / 2 H - s) Δ} A (s) f (u) (s) d s, \\ v (t) = v_{0} (t) + v_{1} (t) \end{array}$ for $0 ⩽ t < 1 / 2 H$ . Then the following results hold. $\begin{array}{l} (1) v \in L^{\infty} ((0, \frac{1}{2 H}), W^{2, r} (R^{n})), \\ (2) {‖ u (t) - v (t) ‖}_{L^{r} (R^{n})} ≲ \frac{1}{(N + 1)!} {(\frac{1}{2 H} - t)}^{N + 1} {\frac{‖ u_{0} ‖_{L^{r} (R^{n})}}{t^{N + 1}} + \frac{‖ u ‖_{Y}^{p}}{(p - 1) H}}, \end{array}$ for $0 < t < 1 / 2 H$ . Especially, $‖ u (t) - v (t) ‖_{L^{r} (R^{n})} = O ({(1 / 2 H - t)}^{N + 1})$ as $t ↗ 1 / 2 H$ .

Remark 1.5.
One might consider to obtain the same result in Theorem 1.4 with $v_{1}$ replaced by $\tilde{v_{1}}$ defined by $\begin{matrix} \tilde{v_{1}} (t) : = \sum_{j = 0}^{N} \frac{1}{j!} {(\frac{1}{2 H} - t)}^{j} {(- Δ)}^{j} \int_{0}^{1 / 2 H} e^{(1 / 2 H - s) Δ} A (s) f (u) (s) d s . \end{matrix}$ In this case, we need to show $‖ \tilde{v_{1}} (t) - v_{1} (t) ‖_{L^{r} (R^{n})} = O ({(1 / 2 H - t)}^{N + 1})$ as $t ↗ 1 / 2 H$ . When $N = 0$ , since we have $\begin{array}{rcl} {‖ \tilde{v_{1}} (t) - v_{1} (t) ‖}_{L^{r} (R^{n})} & ⩽ & \int_{t}^{1 / 2 H} A (s) {‖ f (u) (s) ‖}_{L^{r} (R^{n})} d s \\ = & \frac{4 {(2 H)}^{n (p - 1) / 4 - 1}}{n (p - 1)} {(\frac{1}{2 H} - t)}^{n (p - 1) / 4} ‖ u ‖_{Y}^{p} \end{array}$ by Lemma 2.1, below, the additional condition $1 + 4 / n ⩽ p$ is required to show $‖ \tilde{v_{1}} (t) - v_{1} (t) ‖_{L^{r} (R^{n})} = O (1 / 2 H - t)$ , while this condition is not necessary in the theorem by the use of $v_{1}$ .

The global solutions in Theorem 1.1 when $k = 0$ and in Theorem 1.3 when $N = l = 0$ have been shown in [19] with their asymptotic behaviors based on the different method in [20]. Our results extend those results to the general order $k ⩾ 0$ , $N ⩾ 0$ and $l ⩾ 0$ , which require the estimates for the higher order terms in the Taylor expansion of the Gauss kernel and the heat operator given by Lemma 2.3 and (6.2).

For A and B, the notation $A ≲ B$ denotes the inequality $A ⩽ C B$ for some constant $C > 0$ which is not essential for the argument throughout the paper.
2. Preliminaries

In this section, we prepare several estimates to prove our theorems. We recall three fundamental lemmas known as the $L^{p} - L^{q}$ estimate, the expansion of the Gauss kernel, and the asymptotic behavior for the free solution of the diffusion equation. We prove them for the convenience of readers.

Lemma 2.1.
Let $j ⩾ 0$ , $1 ⩽ r ⩽ q ⩽ \infty$ , and let α denote arbitrary multi-index. Then $\begin{array}{l} (2.1) & (1) {‖ \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ t^{- n (1 - 1 / q) / 2 - j - | α | / 2}, \\ (2.2) & (2) {‖ \partial_{t}^{j} \partial_{x}^{α} e^{t Δ} g ‖}_{L^{q} (R^{n})} ≲ t^{- n (1 / r - 1 / q) / 2 - j - | α | / 2} ‖ g ‖_{L^{r} (R^{n})} \end{array}$ hold for $t > 0$ .
Proof.
(1) Put $η (x) : = G (1, x) = {(4 π)}^{- n / 2} e^{- | x |^{2} / 4}$ , $r : = | x | = {\sum_{k = 1}^{n} {(x_{k})}^{2}}^{1 / 2}$ , $r \partial_{r} η (x) : = \sum_{k = 1}^{n} x_{k} \partial_{k} η (x)$ . We define functions $η_{1}, η_{2}, \dots$ inductively by $\begin{matrix} η_{1} (x) : = - \frac{n}{2} η (x) - \frac{1}{2} r \partial_{r} η (x) \end{matrix}$ and $\begin{matrix} η_{j} (x) : = - \frac{n + 2 j - 2}{2} η_{j - 1} (x) - \frac{1}{2} r \partial_{r} η_{j - 1} (x) \end{matrix}$ for $j ⩾ 2$ . By $G (t, x) = t^{- n / 2} η (t^{- 1 / 2} x)$ , we have $\partial_{t} G (t, x) = t^{- n / 2 - 1} η_{1} (t^{- 1 / 2} x)$ . We are able to show $\partial_{t}^{j} G (t, x) = t^{- n / 2 - j} η_{j} (t^{- 1 / 2} x)$ inductively for $j ⩾ 2$ . We obtain $\begin{matrix} \partial_{t}^{j} \partial_{x}^{α} G (t, x) = t^{- n / 2 - j} \partial_{x}^{α} (η_{j} (t^{- 1 / 2} x)) = t^{- n / 2 - j - | α | / 2} \partial_{x}^{α} η_{j} (t^{- 1 / 2} x) \end{matrix}$ for $j ⩾ 0$ and any multi-index α. So that, we have $\begin{array}{rcl} {‖ \partial_{t}^{j} \partial_{x}^{α} G (t, x) ‖}_{L_{x}^{q} (R^{n})} & = & t^{- n / 2 - j - | α | / 2} {‖ \partial_{x}^{α} η_{j} (t^{- 1 / 2} x) ‖}_{L_{x}^{q} (R^{n})} \\ = & t^{- n (1 - 1 / q) / 2 - j - | α | / 2} {‖ \partial_{x}^{α} η_{j} ‖}_{L^{q} (R^{n})}, \end{array}$ which yields the required inequality.

(2) By $\partial_{t}^{j} \partial_{x}^{α} e^{t Δ} g = (\partial_{t}^{j} \partial_{x}^{α} G (t, \cdot)) * g$ , the Young inequality and (1), we have $\begin{array}{rcl} {‖ \partial_{t}^{j} \partial_{x}^{α} e^{t Δ} g ‖}_{L^{q} (R^{n})} & ⩽ & {‖ \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot) ‖}_{L^{s} (R^{n})} ‖ g ‖_{L^{r} (R^{n})} \\ ≲ & t^{- n (1 - 1 / s) / 2 - j - | α | / 2} ‖ g ‖_{L^{r} (R^{n})}, \end{array}$ where s is defined by $1 / q = 1 / s + 1 / r - 1$ , which is the required result. □
Lemma 2.2.
Let $N ⩾ 0$ . The expansion $\begin{matrix} e^{(t - s) Δ} = \sum_{j = 0}^{N} \frac{{(- s)}^{j}}{j!} \frac{d^{j}}{d t^{j}} e^{t Δ} + \frac{{(- 1)}^{N + 1}}{N!} \int_{0}^{s} {(s - τ)}^{N} \frac{d^{N + 1}}{d t^{N + 1}} e^{(t - τ) Δ} d τ \end{matrix}$ holds for $0 ⩽ s ⩽ t$ .
Proof.
For $t ⩾ 0$ and $ξ \in R^{n}$ , put $h (s) : = e^{- (t - s) | ξ |^{2}}$ . Since h has the expansion $\begin{matrix} h (s) = \sum_{j = 0}^{N} \frac{s^{j}}{j!} h^{(j)} (0) + \frac{1}{N!} \int_{0}^{s} {(s - τ)}^{N} h^{(N + 1)} (τ) d τ, \end{matrix}$ we have $\begin{matrix} e^{- (t - s) | ξ |^{2}} = \sum_{j = 0}^{N} \frac{{(- s)}^{j}}{j!} \frac{d^{j}}{d t^{j}} e^{- t | ξ |^{2}} + \frac{{(- 1)}^{N + 1}}{N!} \int_{0}^{s} {(s - τ)}^{N} \frac{d^{N + 1}}{d t^{N + 1}} e^{- (t - τ) | ξ |^{2}} d τ \end{matrix}$ by $h^{(j)} (s) = {(- 1)}^{j} (d^{j} / d t^{j}) (e^{- (t - s) | ξ |^{2}})$ . Taking the Fourier transform and its inverse for the both sides, we obtain the required result. □
Lemma 2.3.
For any $k \in N \cup {0}$ , the equation $\begin{array}{rcl} G (t, x - y) & = & \sum_{| α | = 0}^{k} \frac{{(- y)}^{α}}{α!} \partial_{x}^{α} G (t, x) \\ + (k + 1) \sum_{| α | = k + 1} \frac{{(- y)}^{α}}{α!} \int_{0}^{1} \partial_{x}^{α} G (t, x - θ y) {(1 - θ)}^{k} d θ \end{array}$ holds for any $t > 0$ , $x, y \in R^{n}$ .
Proof.
The required result follows from the fundamental theorem of calculus. □

For $H \in R$ and $1 ⩽ p < \infty$ , put $\begin{matrix} (2.3) & A_{} : = \{\begin{matrix} {(- 2 H)}^{n (p - 1) / 4 - 1} & if H \neq 0 and (2 H + 1) (\frac{n (p - 1)}{4} - 1) < 0, \\ 1 & otherwise . \end{matrix} \end{matrix}$
Lemma 2.4.
Let* $H < 0$ and $p \neq 1 + 4 / n$ . Put $g (s) : = (1 - 2 H s) / (1 + s)$ . Then the following inequalities hold for $0 < s < \infty$ .
$1 ⩽ {(g (s))}^{n (p - 1) / 4 - 1} < {(- 2 H)}^{n (p - 1) / 4 - 1}$ if $(2 H + 1) (\frac{n (p - 1)}{4} - 1) < 0$ .

$1 ⩾ {(g (s))}^{n (p - 1) / 4 - 1} > {(- 2 H)}^{n (p - 1) / 4 - 1}$ if $(2 H + 1) (\frac{n (p - 1)}{4} - 1) > 0$ .

$A (s) ⩽ A_{} {(1 + s)}^{n (p - 1) / 4 - 1}$ .

Proof.
We have $g (s) \to - 2 H$ as $s \to \infty$ . By $g^{'} (s) = - (1 + 2 H) / {(1 + s)}^{2}$ , the function g is increasing if $2 H < - 1$ , g is constant if $2 H = - 1$ , and g is decreasing if $2 H > - 1$ . Thus, we have $1 ⩽ g (s) < - 2 H$ if $2 H < - 1$ , and $1 ⩾ g (s) > - 2 H$ if $- 1 < 2 H < 0$ . We note that $(2 H + 1) (n (p - 1) / 4 - 1) < 0$ is equivalent to “ $2 H < - 1$ and $n (p - 1) / 4 - 1 > 0$ ” or “ $- 1 < 2 H < 0$ and $n (p - 1) / 4 - 1 < 0$ ,” and $(2 H + 1) (n (p - 1) / 4 - 1) > 0$ is equivalent to “ $2 H < - 1$ and $n (p - 1) / 4 - 1 < 0$ ” or “ $- 1 < 2 H < 0$ and $n (p - 1) / 4 - 1 > 0$ ” under $H < 0$ . So that, we obtain the required results in (1) and (2). The result in (3) follows from $A (s) = g {(s)}^{n (p - 1) / 4 - 1} {(1 + s)}^{n (p - 1) / 4 - 1}$ and $g {(s)}^{n (p - 1) / 4 - 1} ⩽ A_{}$ by the results in (1), (2) and the definition of $A_{*}$ . □
Lemma 2.5 (Nonlinear estimates).

For $k ⩾ 0$ , α with $| α | ⩽ k$ and $f (u)$ defined by ( 1.3 ), the estimate $\begin{matrix} \int_{R^{n}} | y |^{| α |} | f (u) (s, y) | d y ≲ | λ | {(1 + s)}^{- n (p - 1) / 2 + | α | / 2} ‖ u ‖_{X_{k}}^{p} \end{matrix}$ holds for any $s ⩾ 0$ .

Proof.
Since we have $\begin{matrix} | y |^{| α |} | f (u) (s, y) | ⩽ | λ | | y |^{| α |} {| u (s, y) |}^{p} \end{matrix}$ and $| u (s, y) | = {(1 + s)}^{- n / 2} {(1 + s)}^{n / 2} | u (s, y) | ⩽ {(1 + s)}^{- n / 2} ‖ u ‖_{X_{k}}$ by the definition of $f (u)$ and $X_{k}$ , we have $\begin{matrix} | y |^{| α |} | f (u) (s, y) | ⩽ | λ | | y |^{| α |} | u (s, y) | {(1 + s)}^{- n (p - 1) / 2} ‖ u ‖_{X_{k}}^{p - 1} \end{matrix}$ for any $s ⩾ 0$ . So that, we obtain $\begin{array}{rcl} \int_{R^{n}} | y |^{k} | f (u) (s, y) | d y & ⩽ & | λ | \int_{R^{n}} | y |^{k} | u (s, y) | d y {(1 + s)}^{- n (p - 1) / 2} ‖ u ‖_{X_{k}}^{p - 1} \\ ⩽ & | λ | {(1 + s)}^{- n (p - 1) / 2 + | α | / 2} ‖ u ‖_{X_{k}}^{p}, \end{array}$ where we have used $\int_{R^{n}} | y |^{k} | u (s, y) | d y ⩽ {(1 + s)}^{| α | / 2} ‖ u ‖_{X_{k}}$ by the definition of $X_{k}$ . □
Lemma 2.6.
Let $m = 0, 1, 2, \dots$ , and let $1 ⩽ r ⩽ \infty$ , $T > 0$ . Let $u_{0} \in W^{m, r} (R^{n})$ , $h \in L^{1} ((0, T), W^{m, r} (R^{n}))$ . Let $u_{0} \in W^{m + 1, \infty} (R^{n})$ , $h (s) \in W^{m + 1, \infty} (R^{n})$ for a.e. $s \in (0, T)$ if $r = \infty$ . Then the function u defined by $\begin{matrix} u (t) : = e^{t Δ} u_{0} + \int_{0}^{t} e^{(t - s) Δ} h (s) d s \end{matrix}$ for $0 ⩽ t < T$ satisfies $u \in C ([0, T), W^{m, r} (R^{n}))$ .
Proof.
For $0 ⩽ t ⩽ t^{'} < T$ , we have $\begin{matrix} u (t) - u (t^{'}) = I + II + III, \end{matrix}$ where we have put $\begin{array}{l} I : = e^{t Δ} (1 - e^{(t^{'} - t) Δ}) u_{0}, II : = \int_{0}^{t} e^{(t - s) Δ} (1 - e^{(t^{'} - t) Δ}) h (s) d s, \\ III : = - \int_{t}^{t^{'}} e^{(t^{'} - s) Δ} h (s) d s . \end{array}$ Since the function $u \in W^{1, \infty} (R^{n})$ is uniformly continuous by $\begin{matrix} | u (t, x) - u (t, y) | ⩽ \int_{0}^{1} | \nabla u (t, y + θ (x - y)) \cdot (x - y) | d θ ⩽ {‖ \nabla u (t, \cdot) ‖}_{L^{\infty} (R^{n})} \cdot | x - y | \end{matrix}$ for $x, y \in R^{n}$ , the function $\partial^{α} u$ is uniformly continuous for any α with $| α | ⩽ m$ if $u \in W^{m + 1, \infty} (R^{n})$ . We have $\begin{matrix} (2.4) & {‖ u (t) - u (t^{'}) ‖}_{W^{m, r} (R^{n})} ⩽ ‖ I ‖_{W^{m, r} (R^{n})} + ‖ II ‖_{W^{m, r} (R^{n})} + ‖ III ‖_{W^{m, r} (R^{n})} . \end{matrix}$ Since we have $\begin{matrix} ‖ I ‖_{W^{m, r} (R^{n})} ⩽ {‖ (1 - e^{(t^{'} - t) Δ}) u_{0} ‖}_{W^{m, r} (R^{n})} \end{matrix}$ by Lemma 2.1, we have $\begin{matrix} (2.5) & ‖ I ‖_{W^{m, r} (R^{n})} \to 0 \end{matrix}$ as $t^{'} ↘ t$ or $t ↗ t^{'}$ , where we have used that ${\partial^{α} u_{0}}_{| α | ⩽ m}$ are uniformly continuous when $r = \infty$ . Similarly, we have $\begin{matrix} ‖ II ‖_{W^{m, r} (R^{n})} ⩽ \int_{0}^{T} χ_{(0, t)} (s) {‖ (1 - e^{(t^{'} - t) Δ}) h (s) ‖}_{W^{m, r} (R^{n})} d s \end{matrix}$ by Lemma 2.1, where $χ_{(0, t)}$ is the characteristic function on the interval $(0, t)$ . Since we have $\begin{matrix} {‖ (1 - e^{(t^{'} - t) Δ}) h (s) ‖}_{W^{m, r} (R^{n})} ⩽ 2 {‖ h (s) ‖}_{W^{m, r} (R^{n})} \end{matrix}$ and $\begin{matrix} {‖ (1 - e^{(t^{'} - t) Δ}) h (s) ‖}_{W^{m, r} (R^{n})} \to 0 \end{matrix}$ as $t^{'} ↘ t$ or $t ↗ t^{'}$ , where we use that ${\partial^{α} h (s)}_{| α | ⩽ m}$ are uniformly continuous for a.e. $s \in (0, t)$ when $r = \infty$ , we obtain $\begin{matrix} (2.6) & ‖ II ‖_{W^{m, r} (R^{n})} \to 0 \end{matrix}$ as $t^{'} ↘ t$ or $t ↗ t^{'}$ by the Lebesgue convergence theorem. We have $\begin{matrix} (2.7) & ‖ III ‖_{W^{m, r} (R^{n})} ⩽ \int_{t}^{t^{'}} {‖ h (s) ‖}_{W^{m, r} (R^{n})} d s \to 0 \end{matrix}$ as $t^{'} ↘ t$ or $t ↗ t^{'}$ by Lemma 2.1 and $h \in L^{1} ((0, T), W^{m, r} (R^{n}))$ . By (2.4), (2.5), (2.6) and (2.7), the function u is continuous. □

We show some weighted estimates for the operator $e^{t Δ}$ as follows. Lemma 2.7.
Let $n ⩾ 1$ . Let $1 ⩽ r ⩽ q ⩽ \infty$ . For any multi-index α, the following results hold. $\begin{array}{rcl} (1) x^{α} e^{t Δ} h (x) & = & e^{t Δ} (x^{α} h (x)) \\ + \sum_{\begin{array}{c} 0 \neq β ⩽ α \\ 0 ⩽ j ⩽ | β | / 2 \end{array}} \sum_{\begin{array}{c} γ ⩽ β \\ | γ | = | β | - 2 j \end{array}} C (α, β, γ, j) t^{| β | - j} \partial_{x}^{γ} e^{t Δ} (x^{α - β} h (x)), \end{array}$ where $C (α, β, γ, j)$ is a constant dependent on α, β, γ and j. $\begin{array}{rcl} (2) {‖ x^{α} e^{t Δ} h (x) ‖}_{L_{x}^{q} (R^{n})} & ≲ & \sum_{β ⩽ α} t^{| β | / 2 - n (1 / r - 1 / q) / 2} {‖ x^{α - β} h (x) ‖}_{L_{x}^{r} (R^{n})} \\ ≲ & t^{- n (1 / r - 1 / q) / 2} ({‖ | x |^{| α |} h (x) ‖}_{L_{x}^{r} (R^{n})} + t^{| α | / 2} ‖ h ‖_{L^{r} (R^{n})}) . \end{array}$
Proof.
(1) The required result follows from $\begin{matrix} (2.8) & \partial_{ξ}^{α} e^{- t | ξ |^{2}} = t^{| α | / 2} \sum_{0 ⩽ j ⩽ | α | / 2} \sum_{\begin{array}{c} β ⩽ α \\ | β | = | α | - 2 j \end{array}} C_{j, β} {(t^{1 / 2} ξ)}^{β} e^{- t | ξ |^{2}}, \end{matrix}$ where $C_{j, β}$ is a constant dependent on j and β, and we note $C_{0, α} = {(- 2)}^{| α |}$ .

(2) By the result (1) and (2.2), we have $\begin{array}{rcl} {‖ x^{α} e^{t Δ} h (x) ‖}_{L_{x}^{q} (R^{n})} & ≲ & \sum_{\begin{array}{c} β ⩽ α \\ 0 ⩽ j ⩽ | β | / 2 \end{array}} \sum_{\begin{array}{c} γ ⩽ β \\ | γ | = | β | - 2 j \end{array}} t^{| β | - j} {‖ \partial^{γ} e^{t Δ} (x^{α - β} h) (x) ‖}_{L^{q} (R^{n})} \\ ≲ & \sum_{β ⩽ α} t^{| β | / 2 - n (1 / r - 1 / q) / 2} {‖ x^{α - β} h (x) ‖}_{L_{x}^{r} (R^{n})} . \end{array}$ Since we have $\begin{array}{rcl} t^{| β | / 2} {‖ x^{α - β} h (x) ‖}_{L_{x}^{r} (R^{n})} & ⩽ & {‖ | x |^{| α |} h (x) ‖}_{L_{x}^{r} (R^{n})}^{1 - | β | / | α |} {(t^{| α | / 2} ‖ h ‖_{L^{r} (R^{n})})}^{| β | / | α |} \\ ⩽ & {‖ | x |^{| α |} h (x) ‖}_{L_{x}^{r} (R^{n})} + t^{| α | / 2} ‖ h ‖_{L^{r} (R^{n})} \end{array}$ by the interpolation, we obtain the last result. □
Corollary 2.8.
Let k be a nonnegative integer. If h satisfies ${⟨ \cdot ⟩}^{k} h (\cdot) \in L^{1} (R^{n})$ , then $\begin{matrix} {‖ | x |^{k} (1 - e^{t Δ}) h (x) ‖}_{L_{x}^{1} (R^{n})} \to 0 \end{matrix}$ as $t ↘ 0$ .
Proof.
For any α with $| α | = k$ , we have $\begin{matrix} x^{α} e^{t Δ} h = e^{t Δ} (x^{α} h) + I \end{matrix}$ by (1) in Lemma 2.7, where we have put $\begin{matrix} I : = \sum_{\begin{array}{c} 0 \neq β ⩽ α \\ 0 ⩽ j ⩽ | β | / 2 \end{array}} \sum_{\begin{array}{c} γ ⩽ β \\ | γ | = | β | - 2 j \end{array}} C (α, β, γ, j) t^{| β | - j} \partial^{γ} e^{t Δ} (x^{α - β} h) . \end{matrix}$ Since we have $\begin{array}{rcl} ‖ I ‖_{L^{1} (R^{n})} & ≲ & \sum_{\begin{array}{c} 0 \neq β ⩽ α \\ 0 ⩽ j ⩽ | β | / 2 \end{array}} \sum_{\begin{array}{c} γ ⩽ β \\ | γ | = | β | - 2 j \end{array}} t^{| β | - j} {‖ \partial^{γ} e^{t Δ} (x^{α - β} h) ‖}_{L^{1} (R^{n})} \\ ≲ & \sum_{\begin{array}{c} β ⩽ α \\ β \neq 0 \end{array}} t^{| β | / 2} {‖ x^{α - β} h ‖}_{L^{1} (R^{n})} \end{array}$ by (2.2), we obtain $‖ I ‖_{L^{1} (R^{n})} \to 0$ as $t ↘ 0$ . Since we have $\begin{matrix} x^{α} (1 - e^{t Δ}) h = x^{α} h - e^{t Δ} x^{α} h - I, \end{matrix}$ we obtain $\begin{matrix} {‖ x^{α} (1 - e^{t Δ}) h ‖}_{L^{1} (R^{n})} ⩽ {‖ x^{α} h - e^{t Δ} x^{α} h ‖}_{L^{1} (R^{n})} + ‖ I ‖_{L^{1} (R^{n})} \to 0 \end{matrix}$ as $t ↘ 0$ by $⟨ \cdot ⟩ h (\cdot) \in L^{1} (R^{n})$ as required. □
Corollary 2.9.
Let k be a nonnegative integer. Let $s ⩾ 0$ . The following results hold for any rapidly decaying function h on $[0, s) \times R^{n}$ . $\begin{array}{l} (1) \int_{0}^{s} {‖ e^{(s - τ) Δ} h (τ, \cdot) ‖}_{L^{\infty} (R^{n})} d τ \\ ≲ {(1 + s)}^{- n / 2} {\int_{0}^{s} {‖ h (τ, \cdot) ‖}_{L^{1} (R^{n})} d τ + \int_{0}^{s} {(1 + τ)}^{n / 2} {‖ h (τ, \cdot) ‖}_{L^{\infty} (R^{n})} d τ} . \\ (2) \int_{0}^{s} {‖ | x |^{k} e^{(s - τ) Δ} h (τ, x) ‖}_{L_{x}^{1} (R^{n})} d τ \\ ≲ {(1 + s)}^{k / 2} {\int_{0}^{s} {‖ h (τ, \cdot) ‖}_{L^{1} (R^{n})} d τ + \int_{0}^{s} {(1 + τ)}^{- k / 2} {‖ | x |^{k} h (τ, x) ‖}_{L_{x}^{1} (R^{n})} d τ} . \end{array}$
Proof.
(1) When $0 < s ⩽ 1$ , we have $\begin{matrix} \int_{0}^{s} {‖ e^{(s - τ) Δ} h (τ, \cdot) ‖}_{L^{\infty} (R^{n})} d τ ≲ {(1 + s)}^{- n / 2} \int_{0}^{s} {(1 + τ)}^{n / 2} {‖ h (τ, \cdot) ‖}_{L^{\infty} (R^{n})} d τ \end{matrix}$ by (2.2). When $s ⩾ 1$ , we divide the interval $[0, s]$ into $[0, s / 2]$ and $[s / 2, s]$ . For each interval, we have $\begin{matrix} \int_{0}^{s / 2} {‖ e^{(s - τ) Δ} h (τ, \cdot) ‖}_{L^{\infty} (R^{n})} d τ ≲ {(1 + s)}^{- n / 2} \int_{0}^{s / 2} {‖ h (τ, \cdot) ‖}_{L^{1} (R^{n})} d τ \end{matrix}$ and $\begin{matrix} \int_{s / 2}^{s} {‖ e^{(s - τ) Δ} h (τ, \cdot) ‖}_{L^{\infty} (R^{n})} d τ ≲ {(1 + s)}^{- n / 2} \int_{s / 2}^{s} {(1 + τ)}^{n / 2} {‖ h (τ, \cdot) ‖}_{L^{\infty} (R^{n})} d τ \end{matrix}$ by (2.2). So that, we obtain the required result.

(2) By Lemma 2.7, we have $\begin{array}{l} \int_{0}^{s} {‖ | x |^{k} e^{(s - τ) Δ} h (τ, x) ‖}_{L_{x}^{1} (R^{n})} d τ \\ ≲ \int_{0}^{s} {(s - τ)}^{k / 2} {‖ h (τ, \cdot) ‖}_{L^{1} (R^{n})} + {‖ | x |^{k} h (τ, x) ‖}_{L_{x}^{1} (R^{n})} d τ \\ ≲ {(1 + s)}^{k / 2} {\int_{0}^{s} {‖ h (τ, \cdot) ‖}_{L^{1} (R^{n})} d τ + \int_{0}^{s} {(1 + τ)}^{- k / 2} {‖ | x |^{k} h (τ, x) ‖}_{L_{x}^{1} (R^{n})} d τ} \end{array}$ as required. □
Lemma 2.10.
Let $1 < p < \infty$ , $λ \in C$ . Let $f (u)$ be defined by ( 1.3 ). For $T > 0$ and any continuous function $A (\cdot) ⩾ 0$ on $[0, T)$ . Put $\begin{matrix} (2.9) & h (u) : = A f (u), I (t) : = \int_{0}^{t} A (τ) {(1 + τ)}^{- n (p - 1) / 2} d τ . \end{matrix}$ For nonnegative integer k, put $\begin{matrix} ‖ u ‖_{Y^{\infty} (t)} : = {(1 + t)}^{n / 2} {‖ u (t) ‖}_{L^{\infty} (R^{n})}, ‖ u ‖_{Y_{k}^{1} (t)} : = {(1 + t)}^{- k / 2} {‖ | x |^{k} u (t) ‖}_{L^{1} (R^{n})}, \\ ‖ u ‖_{X^{\infty} (t)} : = sup_{0 < s < t} ‖ u ‖_{Y^{\infty} (s)}, ‖ u ‖_{X_{k}^{1} (t)} : = sup_{0 < s < t} ‖ u ‖_{Y_{k}^{1} (s)} . \end{matrix}$ Then the following results hold.
$‖ f (u) (t) ‖_{L^{\infty} (R^{n})} ⩽ | λ | {(1 + t)}^{- n (p - 1) / 2 - n / 2} ‖ u ‖_{Y^{\infty} (t)}^{p}$ .

$‖ | x |^{k} f (u) (t) ‖_{L^{1} (R^{n})} ⩽ | λ | {(1 + t)}^{- n (p - 1) / 2 + k / 2} ‖ u ‖_{Y^{\infty} (t)}^{p - 1} ‖ u ‖_{Y_{k}^{1} (t)}$ .

$\int_{0}^{t} {(1 + τ)}^{n / 2} ‖ h (u) (τ) ‖_{L^{\infty} (R^{n})} d τ ⩽ | λ | I (t) ‖ u ‖_{X^{\infty} (t)}^{p}$ .

$\int_{0}^{t} {(1 + τ)}^{- k / 2} ‖ | x |^{k} h (u) (τ) ‖_{L^{1} (R^{n})} d τ ⩽ | λ | I (t) ‖ u ‖_{X^{\infty} (t)}^{p - 1} ‖ u ‖_{X_{k}^{1} (t)}$ .

$‖ f (u) (t) - f (v) (t) ‖_{L^{\infty} (R^{n})} ≲ | λ | {(1 + t)}^{- n (p - 1) / 2 - n / 2} {max}_{w = u, v} ‖ w ‖_{Y^{\infty} (t)}^{p - 1} ‖ u - v ‖_{Y^{\infty} (t)}$ .

$‖ | x |^{k} (f (u) (t) - f (v) (t)) ‖_{L^{1} (R^{n})} ≲ | λ | {(1 + t)}^{- n (p - 1) / 2 + k / 2} {max}_{w = u, v} ‖ w ‖_{Y^{\infty} (t)}^{p - 1} ‖ u - v ‖_{Y_{k}^{1} (t)}$ .

$\int_{0}^{t} {(1 + τ)}^{n / 2} ‖ h (u) (τ) - h (v) (τ) ‖_{L^{\infty} (R^{n})} d τ ≲ | λ | I (t) {max}_{w = u, v} ‖ w ‖_{X^{\infty} (t)}^{p - 1} ‖ u - v ‖_{X^{\infty} (t)}$ .

$\int_{0}^{t} {(1 + τ)}^{- k / 2} ‖ | x |^{k} (h (u) (τ) - h (v) (τ)) ‖_{L^{1} (R^{n})} d τ ≲ | λ | I (t) {max}_{w = u, v} ‖ w ‖_{X^{\infty} (t)}^{p - 1} ‖ u - v ‖_{X_{k}^{1} (t)}$ .

Proof.
(1) Since we have $‖ u (t) ‖_{L^{\infty} (R^{n})} = {(1 + t)}^{- n / 2} ‖ u ‖_{Y^{\infty} (t)}$ , we have $\begin{matrix} {‖ f (u) (t) ‖}_{L^{\infty} (R^{n})} ⩽ | λ | {(1 + t)}^{- n (p - 1) / 2 - n / 2} ‖ u ‖_{Y^{\infty} (t)}^{p} . \end{matrix}$

(2) Since we have $‖ | x |^{k} u (t) ‖_{L^{1} (R^{n})} = {(1 + t)}^{k / 2} ‖ u ‖_{Y_{k}^{1} (t)}$ , we have $\begin{matrix} {‖ | x |^{k} f (u) (t) ‖}_{L^{1} (R^{n})} ⩽ | λ | {(1 + t)}^{- n (p - 1) / 2 + k / 2} ‖ u ‖_{Y^{\infty} (t)}^{p - 1} ‖ u ‖_{Y_{k}^{1} (t)} . \end{matrix}$

(3) By (1), we have $\begin{matrix} {‖ h (u) (τ) ‖}_{L^{\infty} (R^{n})} ⩽ A (τ) {‖ f (u) (τ) ‖}_{L^{\infty} (R^{n})} ⩽ | λ | A (τ) {(1 + τ)}^{- n (p - 1) / 2 - n / 2} ‖ u ‖_{Y^{\infty} (τ)}^{p} \end{matrix}$ for $0 < τ < t$ . So that, we obtain $\begin{array}{rcl} \int_{0}^{t} {(1 + τ)}^{n / 2} {‖ h (u) (τ) ‖}_{L^{\infty} (R^{n})} d τ & ⩽ & | λ | \int_{0}^{t} A (τ) {(1 + τ)}^{- n (p - 1) / 2} d τ ‖ u ‖_{X^{\infty} (t)}^{p} \\ = & | λ | I (t) ‖ u ‖_{X^{\infty} (t)}^{p} . \end{array}$

(4) By (2), we have $\begin{array}{rcl} {‖ | x |^{k} h (u) (τ) ‖}_{L^{1} (R^{n})} & ⩽ & A (τ) {‖ | x |^{k} f (u) (τ) ‖}_{L^{1} (R^{n})} \\ ⩽ & | λ | A (τ) {(1 + τ)}^{- n (p - 1) / 2 + k / 2} ‖ u ‖_{Y^{\infty} (τ)}^{p - 1} ‖ u ‖_{Y_{k}^{1} (τ)} . \end{array}$ So that, we obtain $\begin{array}{l} \int_{0}^{t} {(1 + τ)}^{- k / 2} {‖ | x |^{k} h (u) (τ) ‖}_{L^{1} (R^{n})} d τ \\ ⩽ | λ | \int_{0}^{t} A (τ) {(1 + τ)}^{- n (p - 1) / 2} d τ ‖ u ‖_{X^{\infty} (t)}^{p - 1} ‖ u ‖_{X_{k}^{1} (t)} \\ = | λ | I (t) ‖ u ‖_{X^{\infty} (t)}^{p - 1} ‖ u ‖_{X_{k}^{1} (t)} . \end{array}$

(5) Since we have $| f (u) - f (v) | ≲ | λ | (| u |^{p - 1} + | v |^{p - 1}) | u - v |$ , we have $\begin{matrix} {‖ f (u) - f (v) ‖}_{L^{\infty} (R^{n})} ≲ | λ | max_{w = u, v} ‖ w ‖_{L^{\infty} (R^{n})}^{p - 1} ‖ u - v ‖_{L^{\infty} (R^{n})} . \end{matrix}$ We obtain the required result by the definition of $‖ \cdot ‖_{Y^{\infty} (t)}$ .

(6) Since we have $| x |^{k} | f (u) - f (v) | ≲ | λ | (| u |^{p - 1} + | v |^{p - 1}) | x |^{k} | u - v |$ , we have $\begin{matrix} {‖ | x |^{k} (f (u) - f (v)) ‖}_{L^{1} (R^{n})} ≲ | λ | max_{w = u, v} ‖ w ‖_{L^{\infty} (R^{n})}^{p - 1} {‖ | x |^{k} (u - v) ‖}_{L^{1} (R^{n})} . \end{matrix}$ We obtain the required result by the definitions of $‖ \cdot ‖_{Y^{\infty} (t)}$ and $‖ \cdot ‖_{Y_{k}^{1} (t)}$ .

(7) By (5), we have $\begin{array}{l} {‖ h (u) (τ) - h (v) (τ) ‖}_{L^{\infty} (R^{n})} \\ ⩽ A (τ) {‖ f (u) (τ) - f (v) (τ) ‖}_{L^{\infty} (R^{n})} \\ ≲ | λ | A (τ) {(1 + τ)}^{- n (p - 1) / 2 - n / 2} max_{w = u, v} ‖ w ‖_{Y^{\infty} (τ)}^{p - 1} ‖ u - v ‖_{Y^{\infty} (τ)} \end{array}$ for $0 < τ < t$ . We obtain the required result by $\begin{array}{l} \int_{0}^{t} {(1 + τ)}^{n / 2} {‖ h (u) (τ) - h (v) (τ) ‖}_{L^{\infty} (R^{n})} d τ \\ ≲ | λ | \int_{0}^{t} A (τ) {(1 + τ)}^{- n (p - 1) / 2} d τ max_{w = u, v} ‖ w ‖_{X^{\infty} (t)}^{p - 1} ‖ u - v ‖_{X^{\infty} (t)} \\ = | λ | I (t) max_{w = u, v} ‖ w ‖_{X^{\infty} (t)}^{p - 1} ‖ u - v ‖_{X^{\infty} (t)} \end{array}$ as required.

(8) By (6), we have $\begin{array}{l} {‖ | x |^{k} (h (u) (τ) - h (v) (τ)) ‖}_{L^{1} (R^{n})} \\ ⩽ A (τ) {‖ | x |^{k} (f (u) (τ) - f (v) (τ)) ‖}_{L^{1} (R^{n})} \\ ≲ | λ | A (τ) {(1 + τ)}^{- n (p - 1) / 2 + k / 2} max_{w = u, v} ‖ w ‖_{Y^{\infty} (τ)}^{p - 1} ‖ u - v ‖_{Y_{k}^{1} (τ)} \end{array}$ for $0 < τ < t$ . So that, we obtain $\begin{array}{l} \int_{0}^{t} {(1 + τ)}^{- k / 2} {‖ | x |^{k} (h (u) (τ) - h (v) (τ)) ‖}_{L^{1} (R^{n})} d τ \\ ≲ | λ | \int_{0}^{t} A (τ) {(1 + τ)}^{- n (p - 1) / 2} d τ max_{w = u, v} ‖ w ‖_{X^{\infty} (t)}^{p - 1} ‖ u - v ‖_{X_{k}^{1} (t)} \\ = | λ | I (t) max_{w = u, v} ‖ w ‖_{X^{\infty} (t)}^{p - 1} ‖ u - v ‖_{X_{k}^{1} (t)} \end{array}$ as required. □

3. Proof of Theorem 1.1

For $t ⩾ 0$ , we have $\begin{matrix} (3.1) & {‖ e^{t Δ} u_{0} ‖}_{L^{\infty} (R^{n})} ≲ {(1 + t)}^{- n / 2} ‖ u_{0} ‖_{L^{1} (R^{n}) \cap L^{\infty} (R^{n})} \end{matrix}$ by (2.2). We also have $\begin{array}{rcl} {‖ | x |^{k} e^{t Δ} u_{0} ‖}_{L^{1} (R^{n})} & ≲ & t^{k / 2} ‖ u_{0} ‖_{L^{1} (R^{n})} + {‖ | x |^{k} u_{0} ‖}_{L^{1} (R^{n})} \\ (3.2) & ≲ & {(1 + t)}^{k / 2} {‖ {⟨ x ⟩}^{k} u_{0} ‖}_{L^{1} (R^{n})} \end{array}$ by Lemma 2.7. Put $h : = A f (u)$ . We have $\begin{matrix} (3.3) & {‖ Φ (u) (t) ‖}_{L^{\infty} (R^{n})} ⩽ {‖ e^{t Δ} u_{0} ‖}_{L^{\infty} (R^{n})} + \int_{0}^{t} {‖ e^{(t - τ) Δ} h (τ) ‖}_{L^{\infty} (R^{n})} d τ \end{matrix}$ by (1.6). Put $\begin{matrix} H_{\infty} (t) : = \int_{0}^{t} {‖ e^{(t - τ) Δ} h (τ) ‖}_{L^{\infty} (R^{n})} d τ . \end{matrix}$ By Corollary 2.9, we have $\begin{matrix} H_{\infty} (t) ≲ {(1 + t)}^{- n / 2} {\int_{0}^{t} {‖ h (τ) ‖}_{L^{1} (R^{n})} d τ + \int_{0}^{t} {(1 + τ)}^{n / 2} {‖ h (τ) ‖}_{L^{\infty} (R^{n})} d τ}, \end{matrix}$ which yields $\begin{matrix} H_{\infty} (t) ≲ {(1 + t)}^{- n / 2} | λ | I (t) ‖ u ‖_{X^{\infty} (t)}^{p - 1} ‖ u ‖_{X_{0}^{1} (t) \cap X^{\infty} (t)} \end{matrix}$ by (3) and (4) in Lemma 2.10. Thus, we obtain $\begin{array}{l} {(1 + t)}^{n / 2} {‖ Φ (u) (t) ‖}_{L^{\infty} (R^{n})} \\ ⩽ C_{0} ‖ u_{0} ‖_{L^{1} (R^{n}) \cap L^{\infty} (R^{n})} + C | λ | I (t) ‖ u ‖_{X^{\infty} (t)}^{p - 1} ‖ u ‖_{X_{0}^{1} (t) \cap X^{\infty} (t)} \\ ⩽ C_{0} ‖ u_{0} ‖_{L^{1} (R^{n}) \cap L^{\infty} (R^{n})} + C | λ | sup_{s ⩾ 0} I (s) R_{\infty}^{p - 1} max {R_{\infty}, R_{1}} \\ ⩽ R_{\infty} \end{array}$ for $t ⩾ 0$ under the conditions $\begin{matrix} (3.4) & R_{\infty} ⩾ 2 C_{0} ‖ u_{0} ‖_{L^{1} (R^{n}) \cap L^{\infty} (R^{n})}, 2 C | λ | sup_{s ⩾ 0} I (s) R_{\infty}^{p - 1} max {R_{\infty}, R_{1}} ⩽ R_{\infty} . \end{matrix}$

Similarly to (3.3), we have $\begin{matrix} {‖ | x |^{k} Φ (u) (t) ‖}_{L^{1} (R^{n})} ⩽ {‖ | x |^{k} e^{t Δ} u_{0} ‖}_{L^{1} (R^{n})} + \int_{0}^{t} {‖ | x |^{k} e^{(t - τ) Δ} h (τ) ‖}_{L^{1} (R^{n})} d τ . \end{matrix}$ Put $\begin{matrix} H_{1}^{k} (t) : = \int_{0}^{t} {‖ | x |^{k} e^{(t - τ) Δ} h (τ) ‖}_{L^{1} (R^{n})} d τ . \end{matrix}$ By Corollary 2.9, we have $\begin{matrix} H_{1}^{k} (t) ≲ {(1 + t)}^{k / 2} {\int_{0}^{t} {‖ h (τ) ‖}_{L^{1} (R^{n})} d τ + \int_{0}^{t} {(1 + τ)}^{- k / 2} {‖ | x |^{k} h (τ) ‖}_{L^{1} (R^{n})} d τ}, \end{matrix}$ which yields $\begin{matrix} H_{1}^{k} (t) ≲ {(1 + t)}^{k / 2} | λ | I (t) ‖ u ‖_{X^{\infty} (t)}^{p - 1} ‖ u ‖_{X_{0}^{1} (t) \cap X_{k}^{1} (t)} \end{matrix}$ by (4) in Lemma 2.10. Thus, we obtain $\begin{array}{l} {(1 + t)}^{- k / 2} {‖ | x |^{k} Φ (u) (t) ‖}_{L^{1} (R^{n})} \\ ⩽ C_{0} {‖ {⟨ \cdot ⟩}^{k} u_{0} (\cdot) ‖}_{L^{1} (R^{n})} + C | λ | I (t) ‖ u ‖_{X^{\infty} (t)}^{p - 1} ‖ u ‖_{X_{0}^{1} (t) \cap X_{k}^{1} (t)} \\ ⩽ C_{0} {‖ {⟨ \cdot ⟩}^{k} u_{0} (\cdot) ‖}_{L^{1} (R^{n})} + C | λ | sup_{s ⩾ 0} I (s) R_{\infty}^{p - 1} R_{1} \\ ⩽ R_{1} \end{array}$ under the conditions $\begin{matrix} (3.5) & R_{1} ⩾ 2 C_{0} {‖ {⟨ \cdot ⟩}^{k} u_{0} (\cdot) ‖}_{L^{1} (R^{n})}, 2 C | λ | sup_{s ⩾ 0} I (s) R_{\infty}^{p - 1} ⩽ 1 . \end{matrix}$ Especially, putting $k = 0$ , we obtain $\begin{array}{rcl} {‖ Φ (u) (t) ‖}_{L^{1} (R^{n})} & ⩽ & R_{1} \end{array}$ under the same condition (3.5). So that, Φ is a mapping from $X_{k} (R_{\infty}, R_{1})$ into itself under the conditions (3.4) and (3.5).

Next, we consider the metric. Put $h (u) : = A f (u)$ . Since we have $\begin{matrix} Φ (u) (t) - Φ (v) (t) = \int_{0}^{t} e^{(t - τ) Δ} (h (u) - h (v)) (τ) d τ, \end{matrix}$ we have $\begin{array}{l} {‖ Φ (u) (t) - Φ (v) (t) ‖}_{L^{\infty} (R^{n})} \\ ⩽ \int_{0}^{t} {‖ e^{(t - τ) Δ} (h (u) - h (v)) (τ) ‖}_{L^{\infty} (R^{n})} d τ \\ ⩽ C {(1 + t)}^{- n / 2} {\int_{0}^{t} {(1 + τ)}^{n / 2} {‖ (h (u) - h (v)) (τ) ‖}_{L^{\infty} (R^{n})} d τ \\ + \int_{0}^{t} {‖ (h (u) - h (v)) (τ) ‖}_{L^{1} (R^{n})} d τ} \\ ⩽ C {(1 + t)}^{- n / 2} | λ | I (t) max_{w = u, v} ‖ w ‖_{X^{\infty} (t)}^{p - 1} ‖ u - v ‖_{X^{\infty} (t) \cap X_{0}^{1} (t)} \end{array}$ by Corollary 2.9 and Lemma 2.10. So that, we obtain $\begin{array}{l} {(1 + t)}^{n / 2} {‖ Φ (u) (t) - Φ (v) (t) ‖}_{L^{\infty} (R^{n})} \\ ⩽ C | λ | I (t) R_{\infty}^{p - 1} ‖ u - v ‖_{X^{\infty} (t) \cap X_{0}^{1} (t)} \\ ⩽ \frac{1}{2} ‖ u - v ‖_{X^{\infty} (t) \cap X_{0}^{1} (t)} \end{array}$ for any $u, v \in X_{k} (R_{\infty}, R_{1})$ under the second condition in (3.5). Similarly, we also have $\begin{array}{l} {‖ | x |^{k} (Φ (u) - Φ (v)) (t) ‖}_{L^{1} (R^{n})} \\ ⩽ \int_{0}^{t} {‖ | x |^{k} e^{(t - τ) Δ} (h (u) - h (v)) (τ) ‖}_{L^{1} (R^{n})} d τ \\ ⩽ C {(1 + t)}^{k / 2} {\int_{0}^{t} {‖ (h (u) - h (v)) (τ) ‖}_{L^{1} (R^{n})} d τ \\ + \int_{0}^{t} {(1 + τ)}^{- k / 2} {‖ | x |^{k} (h (u) - h (v)) (τ) ‖}_{L^{1} (R^{n})} d τ} \\ ⩽ C {(1 + t)}^{k / 2} | λ | I (t) max_{w = u, v} ‖ w ‖_{X^{\infty} (t)}^{p - 1} ‖ u - v ‖_{X_{0}^{1} (t) \cap X_{k}^{1} (t)} \end{array}$ by Corollary 2.9 and Lemma 2.10. So that, we obtain $\begin{array}{l} {(1 + t)}^{- k / 2} {‖ | x |^{k} (Φ (u) - Φ (v)) (t) ‖}_{L^{1} (R^{n})} \\ ⩽ C | λ | I (t) R_{\infty}^{p - 1} ‖ u - v ‖_{X_{0}^{1} (t) \cap X_{k}^{1} (t)} \\ ⩽ \frac{1}{2} ‖ u - v ‖_{X_{0}^{1} (t) \cap X_{k}^{1} (t)} \end{array}$ for any $u, v \in X_{k} (R_{\infty}, R_{1})$ under the second condition in (3.5). Especially, taking $k = 0$ , we also obtain $\begin{matrix} {‖ Φ (u) (t) - Φ (v) (t) ‖}_{L^{1} (R^{n})} ⩽ \frac{1}{2} ‖ u - v ‖_{X_{0}^{1} (t)} . \end{matrix}$ Therefore, Φ is a contraction mapping on $X_{k} (R_{\infty}, R_{1})$ under the conditions (3.4) and (3.5). We obtain the fixed point $u \in X_{k} (R_{\infty}, R_{1})$ of Φ by the Banach fixed point theorem.

Since we have $\begin{matrix} A (t) ≲ \{\begin{matrix} {(1 + t)}^{n (p - 1) / 4 - 1} & if H < 0, \\ 1 & if H = 0 \end{matrix} \end{matrix}$ for $t ⩾ 0$ by the definition of $A (\cdot)$ in (1.2), we obtain $\begin{matrix} I (t) ≲ \{\begin{matrix} \int_{0}^{t} {(1 + τ)}^{- 1 - n (p - 1) / 4} d τ < \infty & if H < 0 and p > 1, \\ \int_{0}^{t} {(1 + τ)}^{- n (p - 1) / 2} d τ < \infty & if H = 0 and p > 1 + \frac{2}{n}, \end{matrix} \end{matrix}$ for $t ⩾ 0$ by the definition of $I (\cdot)$ in (2.9). So that, the conditions (3.4) and (3.5) are satisfied if $R_{\infty} > 0$ and $R_{1} > 0$ are sufficiently small, which require the smallness of $‖ u_{0} ‖_{L_{k}^{1} (R^{n}) \cap L^{\infty} (R^{n})}$ . Especially, taking $R_{\infty} = 2 C_{0} ‖ u_{0} ‖_{L^{1} (R^{n}) \cap L^{\infty} (R^{n})}$ and $R_{1} = 2 C_{0} ‖ {⟨ \cdot ⟩}^{k} u_{0} (\cdot) ‖_{L^{1} (R^{n})}$ , we obtain the solution $u \in X_{k} (R_{\infty}, R_{1})$ , which shows $\begin{array}{l} {‖ u (t) ‖}_{L^{\infty} (R^{n})} ⩽ {(1 + t)}^{- n / 2} R_{\infty} ⩽ 2 C_{0} {(1 + t)}^{- n / 2} ‖ u_{0} ‖_{L^{1} (R^{n}) \cap L^{\infty} (R^{n})}, \\ {‖ | x |^{k} u (t) ‖}_{L^{1} (R^{n})} ⩽ {(1 + t)}^{k / 2} R_{1} ⩽ 2 C_{0} {(1 + t)}^{k / 2} {‖ {⟨ \cdot ⟩}^{k} u_{0} (\cdot) ‖}_{L^{1} (R^{n})} \end{array}$ for $t ⩾ 0$ by the definition (1.4) as required in (1.8) and (1.9).

Next, let us show $u \in C ([0, \infty), L_{k}^{1} (R^{n}))$ . Let $0 ⩽ t < t^{'} < T < \infty$ . Since u satisfies $\begin{matrix} u (t) = e^{t Δ} u_{0} + \int_{0}^{t} e^{(t - τ) Δ} A (τ) f (u) (τ) d τ, \end{matrix}$ we have $\begin{array}{rcl} u (t) - u (t^{'}) & = & e^{t Δ} (1 - e^{(t^{'} - t) Δ}) u_{0} + \int_{0}^{t} e^{(t - τ) Δ} (1 - e^{(t^{'} - t) Δ}) A (τ) f (u) (τ) d τ \\ - \int_{t}^{t^{'}} e^{(t^{'} - τ) Δ} A (τ) f (u) (τ) d τ . \end{array}$ Thus, we have $\begin{matrix} (3.6) & {‖ | x |^{k} (u (t) - u (t^{'})) ‖}_{L^{1} (R^{n})} ⩽ ‖ I ‖_{L^{1} (R^{n})} + ‖ II ‖_{L^{1} (R^{n})} + ‖ III ‖_{L^{1} (R^{n})}, \end{matrix}$ where we have put $\begin{array}{l} I : = | x |^{k} e^{t Δ} (1 - e^{(t^{'} - t) Δ}) u_{0}, \\ II : = \int_{0}^{t} | x |^{k} e^{(t - τ) Δ} (1 - e^{(t^{'} - t) Δ}) A (τ) f (u) (τ) d τ, \\ III : = \int_{t}^{t^{'}} | x |^{k} e^{(t^{'} - τ) Δ} A (τ) f (u) (τ) d τ . \end{array}$ We have $\begin{matrix} (3.7) & ‖ I ‖_{L^{1} (R^{n})} ≲ t^{k / 2} {‖ (1 - e^{(t^{'} - τ) Δ}) u_{0} ‖}_{L^{1} (R^{n})} + {‖ | x |^{k} (1 - e^{(t^{'} - t) Δ}) u_{0} ‖}_{L^{1} (R^{n})} \to 0 \end{matrix}$ as $t ↗ t^{'}$ or $t^{'} ↘ t$ if ${⟨ \cdot ⟩}^{k} u_{0} (\cdot) \in L^{1} (R^{n})$ by Lemma 2.7 and Corollary 2.8. We consider $II$ for $t > 0$ since $II = 0$ when $t = 0$ . Since $II$ is rewritten as $\begin{matrix} II = \int_{0}^{T} χ_{(0, t)} (τ) | x |^{k} e^{(t - τ) Δ} (1 - e^{(t^{'} - t) Δ}) h (u) (τ) d τ, \end{matrix}$ where $χ_{(0, T)}$ is the characteristic function on the interval $(0, T)$ , we have $\begin{matrix} ‖ II ‖_{L^{1} (R^{n})} ⩽ \int_{0}^{T} χ_{(0, t)} (τ) {‖ | x |^{k} e^{(t - τ) Δ} (1 - e^{(t^{'} - t) Δ}) h (u) (τ) ‖}_{L^{1} (R^{n})} d τ . \end{matrix}$ We have $\begin{array}{l} {‖ | x |^{k} e^{(t - τ) Δ} (1 - e^{(t^{'} - t) Δ}) h (u) (τ) ‖}_{L^{1} (R^{n})} \\ ⩽ {‖ | x |^{k} (1 - e^{(t^{'} - t) Δ}) h (u) (τ) ‖}_{L^{1} (R^{n})} + {(t - τ)}^{k / 2} {‖ (1 - e^{(t^{'} - t) Δ}) h (u) (τ) ‖}_{L^{1} (R^{n})} \\ (3.8) & \to 0 \end{array}$ as $t ↗ t^{'}$ or $t^{'} ↘ t$ if ${⟨ \cdot ⟩}^{k} h (u) (τ) \in L^{1} (R^{n})$ for $0 < τ < t$ by Lemma 2.7 and Corollary 2.8. We also have $\begin{array}{l} {‖ | x |^{k} e^{(t - τ) Δ} (1 - e^{(t^{'} - t) Δ}) h (u) (τ) ‖}_{L^{1} (R^{n})} \\ ⩽ {‖ | x |^{k} e^{(t - τ) Δ} h (u) (τ) ‖}_{L^{1} (R^{n})} + {‖ | x |^{k} e^{(t^{'} - τ) Δ} h (u) (τ) ‖}_{L^{1} (R^{n})} \\ ≲ {‖ | x |^{k} h (u) (τ) ‖}_{L^{1} (R^{n})} + {(t - τ)}^{k / 2} {‖ h (u) (τ) ‖}_{L^{1} (R^{n})} + {(t^{'} - τ)}^{k / 2} {‖ h (u) (τ) ‖}_{L^{1} (R^{n})} \\ ≲ {(1 + T)}^{k / 2} {{(1 + τ)}^{- k / 2} {‖ | x |^{k} h (u) (τ) ‖}_{L^{1} (R^{n})} + {‖ h (u) (τ) ‖}_{L^{1} (R^{n})}} \end{array}$ by which we obtain $\begin{array}{l} \int_{0}^{T} χ_{(0, t)} (τ) {‖ | x |^{k} e^{(t - τ) Δ} (1 - e^{(t^{'} - t) Δ}) h (u) (τ) ‖}_{L^{1} (R^{n})} d τ \\ ≲ {(1 + T)}^{k / 2} \int_{0}^{T} {(1 + τ)}^{- k / 2} {‖ | x |^{k} h (u) (τ) ‖}_{L^{1} (R^{n})} + {‖ h (u) (τ) ‖}_{L^{1} (R^{n})} d τ \\ (3.9) & < \infty \end{array}$ by Lemma 2.10. Thus, we obtain $\begin{matrix} (3.10) & ‖ II ‖_{L^{1} (R^{n})} \to 0 \end{matrix}$ as $t ↗ t^{'}$ or $t^{'} ↘ t$ by (3.8) and (3.9) and the Lebesgue convergence theorem. We have $\begin{array}{l} ‖ III ‖_{L^{1} (R^{n})} \\ ⩽ \int_{t}^{t^{'}} {‖ | x |^{k} e^{(t^{'} - τ) Δ} h (u) (τ) ‖}_{L^{1} (R^{n})} d τ \\ ≲ \int_{t}^{t^{'}} {(t^{'} - τ)}^{k / 2} {‖ h (u) (τ) ‖}_{L^{1} (R^{n})} + {‖ | x |^{k} h (u) (τ) ‖}_{L^{1} (R^{n})} d τ \\ ⩽ {(1 + t^{'})}^{k / 2} {\int_{t}^{t^{'}} {‖ h (u) (τ) ‖}_{L^{1} (R^{n})} + {(1 + τ)}^{- k / 2} {‖ | x |^{k} h (u) (τ) ‖}_{L^{1} (R^{n})} d τ} \\ (3.11) & \to 0 \end{array}$ as $t ↗ t^{'}$ or $t^{'} ↘ t$ by Lemma 2.7 and Lemma 2.10. By (3.6), (3.7), (3.10) and (3.11), we obtain $| \cdot |^{k} u (t, \cdot)$ is continuous in $L^{1} (R^{n})$ for $t ⩾ 0$ . Since the above argument is also valid for the case $k = 0$ , we have $u \in C ([0, \infty), L^{1} (R^{n}))$ . So that, we obtain $u \in C ([0, \infty), L_{k}^{1} (R^{n}))$ as required.

4. Proof of Theorem 1.2

For any integer $k ⩾ 0$ , let v be the function defined by (1.14).

Lemma 4.1.
For the solution u obtained in Theorem 1.1 , we have $\begin{matrix} u (t) - v (t) = J_{1} (t) + J_{2} (t) + J_{3} (t) + J_{4} (t) + J_{5} (t), \end{matrix}$ where we have put $\begin{array}{l} J_{1} (t) : = e^{t Δ} u_{0} - \sum_{| α | ⩽ k} m_{α} \partial_{x}^{α} G (t, \cdot), \\ J_{2} (t) : = \int_{0}^{\frac{t}{2}} e^{(t - s) Δ} A (s) f (u) (s) d s - \sum_{j = 0}^{[k / 2]} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} A (s) f (u) (s) d s, \\ J_{3} (t) : = \sum_{j = 0}^{[k / 2]} {\int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} A (s) f (u) (s) d s \\ - \sum_{| α | + 2 j ⩽ k} \frac{1}{j! α!} (\int_{0}^{t / 2} {(- s)}^{j} A (s) \int_{R^{n}} {(- y)}^{α} f (u) (s, y) d y d s) \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot)}, \\ J_{4} (t) : = \int_{t / 2}^{t} e^{(t - s) Δ} A (s) f (u) (s) d s, \\ J_{5} (t) : = - \sum_{\begin{array}{c} 0 ⩽ j ⩽ [k / 2] \\ | α | + 2 j ⩽ k \end{array}} (\int_{t / 2}^{\infty} \frac{{(- s)}^{j}}{j!} A (s) \int_{R^{n}} \frac{{(- y)}^{α}}{α!} f (u) (s, y) d y d s) \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot) . \end{array}$
Proof.
Put $h : = A f (u)$ . Since u is the fixed point of Φ defined by (1.6), and we have $\begin{array}{rcl} \int_{0}^{t} e^{(t - s) Δ} h (s) d s & = & \int_{0}^{t / 2} e^{(t - s) Δ} h (s) d s + \int_{t / 2}^{t} e^{(t - s) Δ} h (s) d s \\ - \sum_{2 j ⩽ k} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} h (s) d s + \sum_{2 j ⩽ k} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} h (s) d s, \end{array}$ we obtain $\begin{array}{rcl} u (t) & = & e^{t Δ} u_{0} + \int_{0}^{t / 2} (e^{(t - s) Δ} - \sum_{2 j ⩽ k} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ}) h (s) d s \\ + \sum_{2 j ⩽ k} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} h (s) d s + \int_{t / 2}^{t} e^{(t - s) Δ} h (s) d s \\ = & e^{t Δ} u_{0} + J_{2} (t) + \sum_{2 j ⩽ k} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} h (s) d s + J_{4} (t) . \end{array}$ On the other hand, since v is defined by (1.14), we have $\begin{array}{rcl} v (t) & = & \sum_{| α | ⩽ k} m_{α} \partial_{x}^{α} G (t, \cdot) - J_{5} (t) \\ + \sum_{\begin{array}{c} 0 ⩽ j ⩽ [k / 2] \\ | α | + 2 j ⩽ k \end{array}} \int_{0}^{t / 2} \int_{R^{n}} \frac{{(- s)}^{j}}{j!} \cdot \frac{{(- y)}^{α}}{α!} h (s, y) d y d s \cdot \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot) \end{array}$ by dividing the interval $(0, \infty)$ into $(0, τ / 2)$ and $(τ / 2, \infty)$ . So that, we have $\begin{array}{rcl} u (t) - v (t) & = & e^{t Δ} u_{0} - \sum_{| α | ⩽ k} m_{α} \partial_{x}^{α} G (t, \cdot) + J_{2} (t) + \sum_{0 ⩽ j ⩽ [k / 2]} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} h (s) d s \\ - \sum_{\begin{array}{c} 0 ⩽ j ⩽ [k / 2] \\ | α | + 2 j ⩽ k \end{array}} \int_{0}^{t / 2} \int_{R^{n}} \frac{{(- s)}^{j}}{j!} \cdot \frac{{(- y)}^{α}}{α!} h (s, y) d y d s \cdot \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot) \\ + J_{4} (t) + J_{5} (t) \\ = & J_{1} (t) + J_{2} (t) + J_{3} (t) + J_{4} (t) + J_{5} (t) \end{array}$ as required. □

In the following, we estimate $J_{1}, \dots, J_{5}$ in Lemma 4.1, respectively.
Lemma 4.2 (Estimate for $J_{1}$ ).

Let $k ⩾ N \cup {0}$ and $1 ⩽ q ⩽ \infty$ . Let $m_{α}$ and $M_{k + 1, α}^{*}$ be defined by ( 1.12 ) and $\begin{matrix} M_{k + 1, α}^{*} (t, x) : = \int_{0}^{1} \int_{R^{n}} \frac{{(- y)}^{α}}{α!} u_{0} (y) \partial_{x}^{α} G (t, x - θ y) d y {(1 - θ)}^{k} d θ . \end{matrix}$ Then the following estimates hold.

$e^{t Δ} u_{0} (x) = \sum_{| α | ⩽ k} m_{α} \partial_{x}^{α} G (t, x) + (k + 1) \sum_{| α | = k + 1} M_{k + 1, α}^{*} (t, x)$ .

If $u_{0} \in L_{k + 1}^{1} (R^{n})$ , then $\begin{matrix} {‖ e^{t Δ} u_{0} (\cdot) - \sum_{| α | ⩽ k} m_{α} \partial_{x}^{α} G (t, \cdot) ‖}_{L^{q} (R^{n})} = O (t^{- n (1 - 1 / q) / 2 - (k + 1) / 2}) as t \to \infty . \end{matrix}$

If $u_{0} \in L_{k}^{1} (R^{n})$ , then $\begin{matrix} {‖ e^{t Δ} u_{0} (\cdot) - \sum_{| α | ⩽ k} m_{α} \partial_{x}^{α} G (t, \cdot) ‖}_{L^{q} (R^{n})} = o (t^{- n (1 - 1 / q) / 2 - k / 2}) as t \to \infty . \end{matrix}$

Proof.
(1) By Lemma 2.3, we have $\begin{array}{rcl} e^{t Δ} u_{0} (x) & = & \int_{R^{n}} G (t, x - y) u_{0} (y) d y \\ (4.1) & = & \sum_{| α | ⩽ k} m_{α} \partial_{x}^{α} G (t, x) + (k + 1) \sum_{| α | = k + 1} M_{k + 1, α}^{} (t, x) . \end{array}$

(2) Since we have $\begin{matrix} {‖ M_{k + 1, α}^{} (t, x) ‖}_{L_{x}^{q} (R^{n})} ⩽ \int_{0}^{1} \int_{R^{n}} \frac{| y |^{| α |}}{α!} | u_{0} (y) | {‖ \partial_{x}^{α} G (t, x - θ y) ‖}_{L_{x}^{q} (R^{n})} d y {(1 - θ)}^{k} d θ \end{matrix}$ and $\begin{matrix} {‖ \partial_{x}^{α} G (t, x - θ y) ‖}_{L_{x}^{q} (R^{n})} = {‖ \partial_{x}^{α} G (t, x) ‖}_{L_{x}^{q} (R^{n})} ≲ t^{- n (1 - 1 / q) / 2 - | α | / 2} \end{matrix}$ by Lemma 2.1, we have $\begin{matrix} (4.2) & {‖ M_{k + 1, α}^{*} (t, x) ‖}_{L_{x}^{q} (R^{n})} ≲ \frac{t^{- n (1 - 1 / q) / 2 - | α | / 2}}{k + 1} \int_{R^{n}} \frac{| y |^{α}}{α!} | u_{0} (y) | d y \end{matrix}$ for $t > 0$ . By (4.1) and (4.2), we obtain the required result.

(3) Put $I : = e^{t Δ} u_{0} (x) - \sum_{| α | ⩽ k} m_{α} \partial_{x}^{α} G (t, x)$ and $\begin{matrix} II : = G (t, x - y) u_{0} (y) - \sum_{| α | ⩽ k} \frac{{(- y)}^{α}}{α!} \partial_{x}^{α} G (t, x) u_{0} (y) . \end{matrix}$ Then we have $\begin{matrix} (4.3) & I = \int_{R^{n}} II d y = \int_{| y | < t^{ε}} II d y + \int_{| y | > t^{ε}} II d y = : I_{1} + I_{2}, \end{matrix}$ where we have decomposed $R^{n}$ into the regions $| y | < t^{ε}$ and $| y | > t^{ε}$ for any $ε > 0$ . Since $II$ is rewritten as $\begin{matrix} II = (k + 1) \sum_{| α | = k + 1} \frac{{(- y)}^{α}}{α!} \int_{0}^{1} \partial_{x}^{α} G (t, x - θ y) {(1 - θ)}^{k} d θ \cdot u_{0} (y) \end{matrix}$ by Lemma 2.3, we have $\begin{matrix} ‖ I_{1} ‖_{L_{x}^{q} (R^{n})} ⩽ (k + 1) \sum_{| α | = k + 1} \int_{| y | < t^{ε}} \frac{| y |^{k + 1}}{α!} | u_{0} (y) | \int_{0}^{1} {‖ \partial_{x}^{α} G (t, x - θ y) ‖}_{L_{x}^{q} (R^{n})} {(1 - θ)}^{k} d θ d y . \end{matrix}$ By $| y |^{k + 1} < | y |^{k} t^{ε}$ and $\begin{matrix} (4.4) & {‖ \partial_{x}^{α} G (t, x - θ y) ‖}_{L_{x}^{q} (R^{n})} ≲ {‖ \partial_{x}^{α} G (t, x) ‖}_{L_{x}^{q} (R^{n})} ≲ t^{- n (1 - 1 / q) / 2 - | α | / 2}, \end{matrix}$ we have $\begin{matrix} (4.5) & ‖ I_{1} ‖_{L_{x}^{q} (R^{n})} ≲ \sum_{| α | = k + 1} \frac{t^{- n (1 - 1 / q) / 2 - | α | / 2 + ε}}{α!} \int_{R^{n}} | y |^{k} | u_{0} (y) | d y . \end{matrix}$ Next, we estimate $I_{2}$ . Since $II$ is rewritten as $\begin{array}{rcl} II & = & G (t, x - y) u_{0} (y) - \sum_{| α | ⩽ k - 1} \frac{{(- y)}^{α}}{α!} u_{0} (y) \partial_{x}^{α} G (t, x) \\ - \sum_{| α | = k} \frac{{(- y)}^{α}}{α!} u_{0} (y) \partial_{x}^{α} G (t, x) \\ = & k \sum_{| α | = k} \int_{0}^{1} \frac{{(- y)}^{α}}{α!} u_{0} (y) \partial_{x}^{α} G (t, x - θ y) {(1 - θ)}^{k - 1} d θ \\ - \sum_{| α | = k} \frac{{(- y)}^{α}}{α!} u_{0} (y) \partial_{x}^{α} G (t, x), \end{array}$ where we have used Lemma 2.3 for the last equality. Thus, we have $\begin{array}{rcl} ‖ II ‖_{L_{x}^{q} (R^{n})} & ⩽ & k \sum_{| α | = k} \int_{0}^{1} \frac{| y |^{| α |}}{α!} | u_{0} (y) | {‖ \partial_{x}^{α} G (t, x - θ y) ‖}_{L_{x}^{q} (R^{n})} {(1 - θ)}^{k - 1} d θ \\ + \sum_{| α | = k} \frac{| y |^{| α |}}{α!} | u_{0} (y) | {‖ \partial_{x}^{α} G (t, x) ‖}_{L_{x}^{q} (R^{n})} \\ ≲ & \sum_{| α | = k} \frac{| y |^{k}}{α!} | u_{0} (y) | t^{- n (1 - 1 / q) / 2 - k / 2} \end{array}$ by (4.4), which yields $\begin{array}{rcl} ‖ I_{2} ‖_{L_{x}^{q} (R^{n})} & ⩽ & \int_{| y | > t^{ε}} ‖ II ‖_{L_{x}^{q} (R^{n})} d y \\ (4.6) & ≲ & \sum_{| α | = k} \frac{t^{- n (1 - 1 / q) / 2 - k / 2}}{α!} \int_{| y | > t^{ε}} | y |^{k} | u_{0} (y) | d y . \end{array}$ By (4.3), (4.5) and (4.6), we have $\begin{array}{rcl} ‖ I ‖_{L^{q} (R^{n})} & ⩽ & ‖ I_{1} ‖_{L^{q} (R^{n})} + ‖ I_{2} ‖_{L^{q} (R^{n})} \\ ≲ & \sum_{| α | = k + 1} \frac{t^{- n (1 - 1 / q) / 2 - | α | / 2 + ε}}{α!} \int_{R^{n}} | y |^{k} | u_{0} (y) | d y \\ + \sum_{| α | = k} \frac{t^{- n (1 - 1 / q) / 2 - k / 2}}{α!} \int_{| y | > t^{ε}} | y |^{k} | u_{0} (y) | d y \\ = & o (t^{- n (1 - 1 / q) / 2 - k / 2}), \end{array}$ which is the required estimate. □
Lemma 4.3 (Estimate for $J_{2}$ ).

Let $N ⩾ 0$ be an integer. Let $1 ⩽ q ⩽ \infty$ . Let u be the global solution of ( 1.1 ) in Theorem 1.1 . Then the following results hold.

The solution u satisfies $\begin{array}{rcl} \int_{0}^{t / 2} e^{(t - s) Δ} A (s) f (u) (s) d s & = & \sum_{j = 0}^{N} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} A (s) f (u) (s) d s \\ + \frac{{(- 1)}^{N + 1}}{N!} \int_{0}^{t / 2} \int_{0}^{s} {(s - τ)}^{N} \partial_{t}^{N + 1} e^{(t - τ) Δ} A (s) f (u) (s) d τ d s \end{array}$ for $t ⩾ 0$ .

If $H = 0$ or $p = 1 + 4 / n$ , then $\begin{matrix} (4.7) & \begin{matrix} {‖ \int_{0}^{t / 2} e^{(t - s) Δ} A (s) f (u) (s) d s - \sum_{j = 0}^{N} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} A (s) f (u) (s) d s ‖}_{L^{q} (R^{n})} \\ ≲ ‖ u ‖_{X_{0}}^{p} \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 - N - 1} & if p > 1 + \frac{2 (N + 2)}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - N - 1} log (2 + t) & if p = 1 + \frac{2 (N + 2)}{n}, \\ {(1 + t)}^{1 - n (1 - 1 / q) / 2 - n (p - 1) / 2} & if 1 < p < 1 + \frac{2 (N + 2)}{n} \end{matrix} \end{matrix} \end{matrix}$ for $t > 0$ . Especially, $\begin{matrix} ‖ J_{2} ‖_{L^{q} (R^{n})} ≲ ‖ u ‖_{X_{0}}^{p} o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ for any integer $k ⩾ 0$ if $p > 1 + (2 + k) / n$ by putting $N : = [k / 2]$ .

Let $A_{*}$ be the constant defined by ( 2.3 ). If $H < 0$ and $p \neq 1 + 4 / n$ , then $\begin{matrix} (4.8) & \begin{matrix} {‖ \int_{0}^{t / 2} e^{(t - s) Δ} A (s) f (u) (s) d s - \sum_{j = 0}^{N} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} A (s) f (u) (s) d s ‖}_{L^{q} (R^{n})} \\ ≲ A_{*} ‖ u ‖_{X_{0}}^{p} \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 - N - 1} & if p > 1 + \frac{4 (N + 1)}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - N - 1} log (2 + t) & if p = 1 + \frac{4 (N + 1)}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - n (p - 1) / 4} & if 1 < p < 1 + \frac{4 (N + 1)}{n} \end{matrix} \end{matrix} \end{matrix}$ for $t > 0$ . Especially, $\begin{matrix} ‖ J_{2} ‖_{q} ≲ A_{*} ‖ u ‖_{X_{0}}^{p} o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ for any integer $k ⩾ 0$ if $p > 1 + 2 k / n$ by putting $N : = [k / 2]$ .

Proof.
(1) By Lemma 2.2, we have $\begin{array}{rcl} \int_{0}^{t / 2} e^{(t - s) Δ} A (s) f (u) (s) d s & = & \sum_{j = 0}^{N} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} A (s) f (u) (s) d s \\ + \frac{{(- 1)}^{N + 1}}{N!} \int_{0}^{t / 2} \int_{0}^{s} {(s - τ)}^{N} \partial_{t}^{N + 1} e^{(t - τ) Δ} A (s) f (u) (s) d τ d s \end{array}$ for $t > 0$ as required.

(2) By the result (1), we have $\begin{array}{l} {‖ \int_{0}^{t / 2} e^{(t - s) Δ} A (s) f (u) (s) d s - \sum_{j = 0}^{N} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} A (s) f (u) (s) d s ‖}_{L^{q} (R^{n})} \\ ⩽ \frac{1}{N!} \int_{0}^{t / 2} \int_{0}^{s} {(s - τ)}^{N} A (s) {‖ \partial_{t}^{N + 1} e^{(t - τ) Δ} f (u) (s) ‖}_{L^{q} (R^{n})} d τ d s \\ ≲ t^{- n (1 - 1 / q) / 2 - N - 1} \int_{0}^{t / 2} s^{N + 1} A (s) {‖ f (u) (s) ‖}_{L^{1} (R^{n})} d s \\ (4.9) & ≲ t^{- n (1 - 1 / q) / 2 - N - 1} \int_{0}^{t / 2} s^{N + 1} {(1 + s)}^{- n (p - 1) / 2} A (s) d s \cdot ‖ u ‖_{X_{0}}^{p}, \end{array}$ where we have used $\begin{array}{rcl} {‖ \partial_{t}^{N + 1} e^{(t - τ) Δ} f (u) (s) ‖}_{L^{q} (R^{n})} & ≲ & {(t - τ)}^{- n (1 - 1 / q) / 2 - N - 1} {‖ f (u) (s) ‖}_{L^{1} (R^{n})} \\ ≲ & t^{- n (1 - 1 / q) / 2 - N - 1} {‖ f (u) (s) ‖}_{L^{1} (R^{n})} \end{array}$ for $0 ⩽ τ ⩽ s ⩽ t / 2$ by Lemma 2.1, and $\begin{matrix} {‖ f (u) (s) ‖}_{L^{1} (R^{n})} ⩽ {‖ u (s) ‖}_{L^{\infty} (R^{n})}^{p - 1} {‖ u (s) ‖}_{L^{1} (R^{n})} ≲ {(1 + s)}^{- n (p - 1) / 2} ‖ u ‖_{X_{0}}^{p} \end{matrix}$ by the definition of $X_{0}$ . Since $A (\cdot) = 1$ when $H = 0$ or $p = 1 + 4 / n$ , we have $\begin{matrix} (4.10) & \int_{0}^{t / 2} s^{N + 1} {(1 + s)}^{- n (p - 1) / 2} A (s) d s ≲ \{\begin{matrix} 1 & if p > 1 + \frac{2 (N + 2)}{n}, \\ log (t + 2) & if p = 1 + \frac{2 (N + 2)}{n}, \\ {(1 + t)}^{N + 2 - n (p - 1) / 2} & if 1 < p < 1 + \frac{2 (N + 2)}{n} . \end{matrix} \end{matrix}$ We obtain the required result by (4.9) and (4.10).

(3) Since we have $A (s) ≲ A_{} {(1 + s)}^{n (p - 1) / 4 - 1}$ when $H < 0$ and $p \neq 1 + 4 / n$ , we have $\begin{array}{l} \int_{0}^{t / 2} s^{N + 1} {(1 + s)}^{- n (p - 1) / 2} A (s) d s \\ (4.11) & ≲ A_{} \cdot \{\begin{matrix} 1 & if p > 1 + \frac{4 (N + 1)}{n}, \\ log (t + 2) & if p = 1 + \frac{4 (N + 1)}{n}, \\ {(1 + t)}^{N + 1 - n (p - 1) / 4} & if 1 < p < 1 + \frac{4 (N + 1)}{n} . \end{matrix} \end{array}$ We obtain the required result by (4.9) and (4.11). □
Lemma 4.4 (Decomposition of $J_{3}$ ).

Put $h : = A f (u)$ . $J_{3}$ is decomposed as $\begin{matrix} J_{3} = J_{31} + J_{32}, \end{matrix}$ where $J_{31}$ and $J_{32}$ are defined by $\begin{matrix} J_{31} : = \int_{0}^{t / 2} e^{t Δ} h (s) d s - \sum_{| α | ⩽ k} \int_{0}^{t / 2} \int_{R^{n}} \frac{{(- y)}^{α}}{α!} h (s) d y d s \cdot \partial_{x}^{α} G (t, x) \end{matrix}$ and $\begin{array}{rcl} J_{32} & : = & \sum_{1 ⩽ j ⩽ [k / 2]} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} \partial_{t}^{j} e^{t Δ} h (s) d s \\ - \sum_{\begin{array}{c} 1 ⩽ j ⩽ [k / 2] \\ 2 j + | α | ⩽ k \end{array}} \int_{0}^{t / 2} \int_{R^{n}} \frac{{(- s)}^{j} {(- y)}^{α}}{j! α!} h (s) d y d s \cdot \partial_{t}^{j} \partial_{x}^{α} G (t, x) . \end{array}$ Here, $J_{32}$ is ignored if $k ⩽ 1$ .

Proof.
The decomposition follows directly from the definition of $J_{3}$ since $J_{31}$ is the case of $j = 0$ in $J_{3}$ , and $J_{32}$ is the case of $j ⩾ 1$ in $J_{3}$ . □
Lemma 4.5 (Estimate for $J_{31}$ ).

Let $1 ⩽ q ⩽ \infty$ , and let $k = 0, 1, 2, \dots$ . The following estimates hold.

If $H = 0$ or $p = 1 + 4 / n$ , then $\begin{matrix} {‖ J_{31} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ ‖ u ‖_{X_{k}}^{p} \cdot \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 + 1 - n (p - 1) / 2} & if p < 1 + \frac{k + 2}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2} log (2 + t) & if p = 1 + \frac{k + 2}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2} & if p > 1 + \frac{k + 2}{n} . \end{matrix} \end{matrix}$ Moreover, $‖ J_{31} (t, \cdot) ‖_{L^{q} (R^{n})} = o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2})$ if $p > 1 + \frac{k + 2}{n}$ .

If $H < 0$ and $p \neq 1 + 4 / n$ , then $\begin{matrix} {‖ J_{31} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ A_{*} ‖ u ‖_{X_{k}}^{p} \cdot \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 - n (p - 1) / 4} & if p < 1 + \frac{2 k}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2} log (2 + t) & if p = 1 + \frac{2 k}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2} & if p > 1 + \frac{2 k}{n} . \end{matrix} \end{matrix}$ Moreover, $‖ J_{31} (t, \cdot) ‖_{L^{q} (R^{n})} = o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2})$ if $p > 1 + \frac{2 k}{n}$ .

Proof.
We rewrite $J_{31}$ as $\begin{matrix} J_{31} (t, x) = \int_{0}^{t / 2} \int_{R^{n}} A (s) {G (t, x - y) - \sum_{| α | ⩽ k} \frac{{(- y)}^{α}}{α!} \partial_{x}^{α} G (t, x)} f (u) (s, y) d y d s \end{matrix}$ by $\begin{array}{rcl} e^{t Δ} A (s) f (u) (s, \cdot) & = & G (t, \cdot) * (A (s) f (u) (s, \cdot)) \\ = & \int_{R^{n}} G (t, \cdot - y) A (s) f (u) (s, y) d y . \end{array}$ For any $ε > 0$ , we decompose $J_{31}$ into three parts given by $\begin{matrix} (4.12) & J_{31} = I + II + III, \end{matrix}$ where we have put $\begin{array}{l} I (t, x) : = \int_{0}^{t / 2} A (s) \int_{| y | < t^{ε}} {G (t, x - y) - \sum_{| α | ⩽ k} \frac{{(- y)}^{α}}{α!} \partial_{x}^{α} G (t, x)} f (u) (s, y) d y d s, \\ II (t, x) : = \int_{0}^{t / 2} A (s) \int_{| y | > t^{ε}} {G (t, x - y) - \sum_{| α | ⩽ k - 1} \frac{{(- y)}^{α}}{α!} \partial_{x}^{α} G (t, x)} f (u) (s, y) d y d s \end{array}$ and $\begin{matrix} III (t, x) : = \int_{0}^{t / 2} A (s) \int_{| y | > t^{ε}} \sum_{| α | = k} \frac{{(- y)}^{α}}{α!} \partial_{x}^{α} G (t, x) f (u) (s, y) d y d s . \end{matrix}$ Here, $II$ is ignored when $k = 0$ . We estimate I. Since we have $\begin{array}{l} G (t, x - y) - \sum_{| α | ⩽ k} \frac{{(- y)}^{α}}{α!} \partial_{x}^{α} G (t, x) \\ = (k + 1) \sum_{| α | = k + 1} \frac{{(- y)}^{α}}{α!} \int_{0}^{1} \partial_{x}^{α} G (t, x - θ y) \cdot {(1 - θ)}^{k} d θ \end{array}$ by Lemma 2.3, we have $\begin{array}{rcl} {‖ I (t, x) ‖}_{L_{x}^{q} (R^{n})} & ⩽ & (k + 1) \sum_{| α | = k + 1} \int_{0}^{t / 2} A (s) \int_{| y | < t^{ε}} \frac{| y |^{k + 1}}{α!} \\ \cdot \int_{0}^{1} {‖ \partial_{x}^{α} G (t, x - θ y) ‖}_{L_{x}^{q} (R^{n})} d θ | f (u) (s, y) | d y d s \\ ≲ & I^{} (t) t^{- n (1 - 1 / q) / 2 - (k + 1) / 2 + ε} ‖ u ‖_{X_{k}}^{p}, \end{array}$ where we have used $| y |^{k + 1} ⩽ t^{ε} | y |^{k}$ , $\begin{matrix} {‖ \partial_{x}^{α} G (t, x - θ y) ‖}_{L_{x}^{q} (R^{n})} ≲ {‖ \partial_{x}^{α} G (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ t^{- n (1 - 1 / q) / 2 - (k + 1) / 2} \end{matrix}$ for $| α | = k + 1$ by Lemma 2.1, $\begin{matrix} \int_{| y | < t^{ε}} | y |^{k} | f (u) (s, y) | d y ⩽ | λ | {(1 + s)}^{- n (p - 1) / 2 + k / 2} ‖ u ‖_{X_{k}}^{p} \end{matrix}$ by Lemma 2.5, and we have put $\begin{matrix} I^{} (t) : = \int_{0}^{t / 2} A (s) {(1 + s)}^{- n (p - 1) / 2 + k / 2} d s . \end{matrix}$ We have $\begin{matrix} (4.13) & A (s) \{\begin{matrix} = 1 & if H = 0 or p = 1 + \frac{4}{n}, \\ ≲ A_{} {(1 + s)}^{n (p - 1) / 4 - 1} & if H \neq 0 and p \neq 1 + \frac{4}{n} \end{matrix} \end{matrix}$ by Lemma 2.4. When $H = 0$ or $p = 1 + 4 / n$ , we have $\begin{matrix} (4.14) & I^{} (t) ≲ \{\begin{matrix} {(1 + t)}^{1 - n (p - 1) / 2 + k / 2} & if p < 1 + \frac{k + 2}{n}, \\ log (2 + t) & if p = 1 + \frac{k + 2}{n}, \\ 1 & if p > 1 + \frac{k + 2}{n}, \end{matrix} \end{matrix}$ by which we obtain $\begin{matrix} (4.15) & {‖ I (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ ‖ u ‖_{X_{k}}^{p} \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 + 1 / 2 - n (p - 1) / 2 + ε} & if p < 1 + \frac{k + 2}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - (k + 1) / 2 + ε} log (2 + t) & if p = 1 + \frac{k + 2}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - (k + 1) / 2 + ε} & if p > 1 + \frac{k + 2}{n} . \end{matrix} \end{matrix}$ When $H \neq 0$ and $p \neq 1 + 4 / n$ , we have $\begin{matrix} (4.16) & I^{} (t) ≲ A_{} \{\begin{matrix} {(1 + t)}^{- n (p - 1) / 4 + k / 2} & if p < 1 + \frac{2 k}{n}, \\ log (2 + t) & if p = 1 + \frac{2 k}{n}, \\ 1 & if p > 1 + \frac{2 k}{n}, \end{matrix} \end{matrix}$ by which we obtain $\begin{matrix} (4.17) & {‖ I (t, x) ‖}_{L_{x}^{q} (R^{n})} ≲ A_{} ‖ u ‖_{X_{k}}^{p} \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 - 1 / 2 - n (p - 1) / 4 + ε} & if p < 1 + \frac{2 k}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - (k + 1) / 2 + ε} log (2 + t) & if p = 1 + \frac{2 k}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - (k + 1) / 2 + ε} & if p > 1 + \frac{2 k}{n} . \end{matrix} \end{matrix}$

We estimate $II$ . By Lemma 2.3, we have $\begin{array}{rcl} {‖ II (t, x) ‖}_{L_{x}^{q} (R^{n})} & ⩽ & \sum_{| α | = k} \int_{0}^{t / 2} A (s) \int_{| y | > t^{ε}} \frac{| y |^{k}}{α!} \int_{0}^{1} {‖ \partial_{x}^{α} G (t, x - θ y) ‖}_{L_{x}^{q} (R^{n})} d θ \\ (4.18) & \cdot | f (u) (s, y) | d y d s . \end{array}$ Thus, we obtain $\begin{array}{rcl} {‖ II (t, x) ‖}_{L_{x}^{q} (R^{n})} & ≲ & \sum_{| α | = k} \frac{1}{α!} \int_{0}^{t / 2} A (s) \int_{| y | > t^{ε}} | y |^{k} | f (u) (s, y) | d y d s \cdot t^{- n (1 - 1 / q) / 2 - k / 2} \\ (4.19) & ≲ & | λ | I^{} (t) t^{- n (1 - 1 / q) / 2 - k / 2} ‖ u ‖_{X_{k}}^{p}, \end{array}$ where we have used Lemma 2.1, and $\begin{matrix} \int_{| y | > t^{ε}} | y |^{k} | f (u) (s, y) | d y ⩽ | λ | {(1 + s)}^{- n (p - 1) / 2 + k / 2} ‖ u ‖_{X_{k}}^{p} \end{matrix}$ by Lemma 2.5. When $H = 0$ or $p = 1 + 4 / n$ , we obtain $\begin{matrix} (4.20) & {‖ II (t, x) ‖}_{L_{x}^{q} (R^{n})} ≲ | λ | ‖ u ‖_{X_{k}}^{p} \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 + 1 - n (p - 1) / 2} & if p < 1 + \frac{k + 2}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2} log (2 + t) & if p = 1 + \frac{k + 2}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2} & if p > 1 + \frac{k + 2}{n} \end{matrix} \end{matrix}$ by (4.14). Especially, we have $\begin{matrix} (4.21) & {‖ II (t, x) ‖}_{L_{x}^{q} (R^{n})} ≲ o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ if $p > 1 + (k + 2) / n$ by the Lebesgue convergence theorem for (4.19). When $H \neq 0$ and $p \neq 1 + 4 / n$ , we obtain $\begin{matrix} (4.22) & {‖ II (t, x) ‖}_{L_{x}^{q} (R^{n})} ≲ A_{*} ‖ u ‖_{X_{k}}^{p} \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 - n (p - 1) / 4} & if p < 1 + \frac{2 k}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2} log (2 + t) & if p = 1 + \frac{2 k}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2} & if p > 1 + \frac{2 k}{n} \end{matrix} \end{matrix}$ by (4.16). Especially, we have $\begin{matrix} (4.23) & {‖ II (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ if $p > 1 + 2 k / n$ by the Lebesgue convergence theorem for (4.19).

We estimate $III$ . We have $\begin{matrix} {‖ III (t, x) ‖}_{L_{x}^{q} (R^{n})} ⩽ \sum_{| α | = k} \int_{0}^{t / 2} A (s) \int_{| y | > t^{ε}} \frac{| y |^{k}}{α!} {‖ \partial_{x}^{α} G (t, \cdot) ‖}_{L^{q} (R^{n})} | f (u) (s, y) | d y d s \end{matrix}$ directly from the definition of $III$ . Analogously to the argument starting from (4.18), we obtain the same estimates (4.20), (4.21), (4.22), (4.23) with $‖ II (t, \cdot) ‖_{L^{q} (R^{n})}$ replaced by $‖ III (t, \cdot) ‖_{L^{q} (R^{n})}$ .

The result in (1) follows from (4.12), (4.15), (4.20) with the same estimate for $III$ , and (4.21). The result in (2) follows from (4.12), (4.17), (4.22) with the same estimate for $III$ , and (4.23). □
Lemma 4.6 (Estimate for $J_{32}$ ).

Let $1 ⩽ q ⩽ \infty$ and $k ⩾ 2$ . The following estimates hold.

If $H = 0$ or $p = 1 + 4 / n$ , then $\begin{matrix} {‖ J_{32} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ ‖ u ‖_{X_{k}}^{p} \cdot \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 + 1 - n (p - 1) / 2} & if p < 1 + \frac{k + 3}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - (k + 1) / 2} log (2 + t) & if p = 1 + \frac{k + 3}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - (k + 1) / 2} & if p > 1 + \frac{k + 3}{n} . \end{matrix} \end{matrix}$ Moreover, $‖ J_{32} (t, \cdot) ‖_{L^{q} (R^{n})} = ‖ u ‖_{X_{k}}^{p} o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2})$ if $p > 1 + (k + 2) / n$ .

If $H < 0$ and $p \neq 1 + 4 / n$ , then $\begin{matrix} {‖ J_{32} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ A_{*} ‖ u ‖_{X_{k}}^{p} \cdot \{\begin{matrix} {(1 + t)}^{- n (1 - 1 / q) / 2 + 1 - n (p - 1) / 2} & if p < 1 + \frac{k + 3}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - (k + 1) / 2} log (2 + t) & if p = 1 + \frac{k + 3}{n}, \\ {(1 + t)}^{- n (1 - 1 / q) / 2 - (k + 1) / 2} & if p > 1 + \frac{k + 3}{n} . \end{matrix} \end{matrix}$ Moreover, $‖ J_{32} (t, \cdot) ‖_{L^{q} (R^{n})} = A_{*} ‖ u ‖_{X_{k}}^{p} o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2})$ if $p > 1 + (k + 2) / n$ .

Proof.
We rewrite $J_{32}$ as $\begin{array}{rcl} J_{32} (t, x) & = & \sum_{\begin{array}{c} 2 j ⩽ k \\ j ⩾ 1 \end{array}} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} A (s) \int_{R^{n}} {\partial_{t}^{j} G (t, x - y) - \sum_{| α | ⩽ k - 2 j} \frac{{(- y)}^{α}}{α!} \partial_{t}^{j} \partial_{x}^{α} G (t, x)} \\ \cdot f (u) (s, y) d y d s \end{array}$ by $\begin{matrix} \partial_{t}^{j} e^{t Δ} A (s) f (u) (s, x) = \int_{R^{n}} \partial_{t}^{j} G (t, x - y) A (s) f (u) (s, y) d y . \end{matrix}$ Since we have $\begin{array}{l} \partial_{t}^{j} G (t, x - y) - \sum_{| α | ⩽ k - 2 j} \frac{{(- y)}^{α}}{α!} \partial_{t}^{j} \partial_{x}^{α} G (t, x) \\ = (k - 2 j + 1) \sum_{| α | = k - 2 j + 1} \frac{{(- y)}^{α}}{α!} \int_{0}^{1} \partial_{t}^{j} \partial_{x}^{α} G (t, x - θ y) \cdot {(1 - θ)}^{k - 2 j} d θ \end{array}$ by Lemma 2.3, we obtain $\begin{array}{rcl} J_{32} (t, x) & = & \sum_{\begin{array}{c} 2 j ⩽ k \\ j ⩾ 1 \end{array}} \sum_{| α | = k - 2 j + 1} (k - 2 j + 1) \int_{0}^{1} \int_{0}^{t / 2} \frac{{(- s)}^{j}}{j!} A (s) \\ \cdot \int_{R^{n}} \frac{{(- y)}^{α}}{α!} \partial_{t}^{j} \partial_{x}^{α} G (t, x - θ y) f (u) (s, y) d y d s {(1 - θ)}^{k - 2 j} d θ . \end{array}$ Taking $L^{q} (R^{n})$ -norm of the both sides, we have $\begin{array}{rcl} {‖ J_{32} (t, \cdot) ‖}_{L^{q} (R^{n})} & ⩽ & \sum_{\begin{array}{c} 2 j ⩽ k \\ j ⩾ 1 \end{array}} \sum_{| α | = k - 2 j + 1} (k - 2 j + 1) \int_{0}^{1} \int_{0}^{t / 2} \frac{s^{j}}{j!} A (s) \\ \cdot \int_{R^{n}} \frac{| y |^{| α |}}{α!} {‖ \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot - θ y) ‖}_{L^{q} (R^{n})} | f (u) (s, y) | d y d s {(1 - θ)}^{k - 2 j} d θ . \end{array}$ Here, we have $\begin{matrix} {‖ \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot - θ y) ‖}_{L^{q} (R^{n})} ≲ t^{- n (1 - 1 / q) / 2 - j - | α | / 2} \end{matrix}$ by Lemma 2.1, $\begin{matrix} \int_{R^{n}} | y |^{| α |} | f (u) (s, y) | d y ⩽ | λ | {(1 + s)}^{- n (p - 1) / 2 + | α | / 2} ‖ u ‖_{X_{k}}^{p} \end{matrix}$ by Lemma 2.5. So that, we obtain $\begin{matrix} {‖ J_{32} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ ‖ u ‖_{X_{k}}^{p} \sum_{\begin{array}{c} 2 j ⩽ k \\ j ⩾ 1 \end{array}} \sum_{| α | = k - 2 j + 1} \frac{| λ |}{j! α!} J_{32}^{} (t) t^{- n (1 - 1 / q) / 2 - (k + 1) / 2}, \end{matrix}$ where we have used (4.13) and we have put $\begin{matrix} J_{32}^{} (t) = \{\begin{matrix} \int_{0}^{t / 2} {(1 + s)}^{- n (p - 1) / 2 + j + | α | / 2} d s & if H = 0 or p = 1 + \frac{4}{n}, \\ A_{} \int_{0}^{t / 2} {(1 + s)}^{- n (p - 1) / 4 + j + | α | / 2 - 1} d s & if H < 0 and p \neq 1 + \frac{4}{n} . \end{matrix} \end{matrix}$

(1) When $H = 0$ or $p = 1 + 4 / n$ , we have $\begin{matrix} J_{32}^{} (t) ≲ \{\begin{matrix} {(1 + t)}^{1 - n (p - 1) / 2 + (k + 1) / 2} & if p < 1 + \frac{k + 3}{n}, \\ log (2 + t) & if p = 1 + \frac{k + 3}{n}, \\ 1 & if p > 1 + \frac{k + 3}{n}, \end{matrix} \end{matrix}$ by which we obtain the required result. Since $1 - n (p - 1) / 2 < - k / 2$ holds when $p > 1 + (k + 2) / n$ , we obtain $\begin{matrix} {‖ J_{32} (t, \cdot) ‖}_{L^{q} (R^{n})} = A_{} ‖ u ‖_{X_{k}}^{p} o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ if $p > 1 + (k + 2) / n$ .

(2) When $H < 0$ and $p \neq 1 + 4 / n$ , we have $\begin{matrix} J_{32}^{} (t) ≲ \{\begin{matrix} {(1 + t)}^{- n (p - 1) / 4 + (k + 1) / 2} & if p < 1 + \frac{2 (k + 1)}{n}, \\ log (2 + t) & if p = 1 + \frac{2 (k + 1)}{n}, \\ 1 & if p > 1 + \frac{2 (k + 1)}{n}, \end{matrix} \end{matrix}$ by which we obtain the required result. Since $n (p - 1) / 4 > k / 2$ holds when $p > 1 + 2 k / n$ , we obtain $‖ J_{32} (t, \cdot) ‖_{L^{q} (R^{n})} = A_{*} ‖ u ‖_{X_{k}}^{p} o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2})$ if $p > 1 + (k + 2) / n$ . □
Lemma 4.7 (Estimate for $J_{4}$ ).

Let $1 ⩽ q ⩽ \infty$ and $k ⩾ 0$ . The following inequalities hold.

If $H = 0$ or $p = 1 + 4 / n$ , then $\begin{matrix} (4.24) & {‖ J_{4} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ ‖ u ‖_{X_{0}}^{p} {(1 + t)}^{- n (1 - 1 / q) / 2 + 1 - n (p - 1) / 2} \end{matrix}$ for $t ⩾ 0$ . Especially, $\begin{matrix} (4.25) & {‖ J_{4} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ ‖ u ‖_{X_{0}}^{p} o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ if $p > 1 + (k + 2) / n$ .

If $H < 0$ and $p \neq 1 + 4 / n$ , then $\begin{matrix} (4.26) & {‖ J_{4} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ A_{*} ‖ u ‖_{X_{0}}^{p} {(1 + t)}^{- n (1 - 1 / q) / 2 - n (p - 1) / 4} \end{matrix}$ for $t ⩾ 0$ . Especially, $\begin{matrix} (4.27) & {‖ J_{4} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ A_{*} ‖ u ‖_{X_{0}}^{p} o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ if $p > 1 + 2 k / n$ .

Proof.
We have $\begin{matrix} {‖ J_{4} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ \int_{t / 2}^{t} A (s) {‖ e^{(t - s) Δ} f (u) (s) ‖}_{L^{q} (R^{n})} d s \end{matrix}$ and $\begin{matrix} {‖ e^{(t - s) Δ} f (u) (s) ‖}_{L^{q} (R^{n})} ≲ {‖ f (u) (s) ‖}_{L^{q} (R^{n})} ≲ ‖ u ‖_{L^{\infty} (R^{n})}^{p - 1} ‖ u ‖_{L^{q} (R^{n})} \end{matrix}$ by the Hölder inequality. By the interpolation $‖ u ‖_{L^{q} (R^{n})} ⩽ ‖ u ‖_{L^{\infty} (R^{n})}^{1 - 1 / q} ‖ u ‖_{L^{1} (R^{n})}^{1 / q}$ and the definition of $‖ u ‖_{X_{0}}$ , we have $\begin{matrix} (4.28) & {‖ J_{4} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ {(1 + t)}^{- n (p - 1 / q) / 2} \int_{t / 2}^{t} A (s) d s ‖ u ‖_{X_{0}}^{p} . \end{matrix}$ By (4.13), we obtain $\begin{matrix} (4.29) & \int_{t / 2}^{t} A (s) d s ≲ \{\begin{matrix} t & if H = 0 or p = 1 + \frac{4}{n}, \\ A_{*} \cdot {(1 + t)}^{n (p - 1) / 4} & if H < 0 and p \neq 1 + \frac{4}{n} . \end{matrix} \end{matrix}$ By (4.28) and (4.29), we have the required results (4.24) and (4.26). The results (4.25) and (4.27) follow directly from (4.24) and (4.26), respectively, since $k / 2 + 1 - n (p - 1) / 2 < 0$ is rewritten as $p > 1 + (k + 2) / n$ , and $k / 2 - n (p - 1) / 4 < 0$ is rewritten as $p > 1 + 2 k / n$ . □
Lemma 4.8 (Estimate for $J_{5}$ ).

Let $1 ⩽ q ⩽ \infty$ and $k ⩾ 0$ . The following results hold.

Let $H = 0$ or $p = 1 + 4 / n$ . Let p satisfy $1 + (k + 2) / n < p$ . Then $\begin{matrix} (4.30) & {‖ J_{5} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ {(1 + t)}^{- n (1 - 1 / q) / 2 - n (p - 1) / 2 + 1} ‖ u ‖_{X_{k}}^{p} . \end{matrix}$ Especially, $\begin{matrix} (4.31) & {‖ J_{5} (t, \cdot) ‖}_{L^{q} (R^{n})} = o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2}) ‖ u ‖_{X_{k}}^{p} . \end{matrix}$

Let $H < 0$ and $p \neq 1 + 4 / n$ . Let p satisfy $1 + 2 k / n < p$ . Then $\begin{matrix} (4.32) & {‖ J_{5} (t, \cdot) ‖}_{L^{q} (R^{n})} ≲ A_{*} {(1 + t)}^{- n (1 - 1 / q) / 2 - n (p - 1) / 4} ‖ u ‖_{X_{k}}^{p} . \end{matrix}$ Especially, $\begin{matrix} (4.33) & {‖ J_{5} (t, \cdot) ‖}_{L^{q} (R^{n})} = A_{*} o ({(1 + t)}^{- n (1 - 1 / q) / 2 - k / 2}) ‖ u ‖_{X_{k}}^{p} . \end{matrix}$

Proof.
By the definition of $J_{5}$ , we have $\begin{array}{l} (4.34) & {‖ J_{5} (t, \cdot) ‖}_{L^{q} (R^{n})} \\ ⩽ | λ | \sum_{\begin{array}{c} 0 ⩽ j ⩽ [k / 2] \\ | α | + 2 j ⩽ k \end{array}} \frac{1}{j! α!} \int_{t / 2}^{\infty} s^{j} A (s) \int_{R^{n}} | y |^{| α |} {| u (s, y) |}^{p} d y d s {‖ \partial_{t}^{j} \partial_{x}^{α} G (t, \cdot) ‖}_{L^{q} (R^{n})} \\ ⩽ | λ | \sum_{\begin{array}{c} 0 ⩽ j ⩽ [k / 2] \\ | α | + 2 j ⩽ k \end{array}} \frac{1}{j! α!} \int_{t / 2}^{\infty} s^{j} A (s) {(1 + s)}^{- n (p - 1) / 2 + | α | / 2} d s \\ (4.35) & \cdot ‖ u ‖_{X_{k}}^{p} t^{- n (1 - 1 / q) / 2 - j - | α | / 2}, \end{array}$ where we have used (2.1) and Lemma 2.5.

(1) Since $A (\cdot) = 1$ by Lemma 2.4 when $H = 0$ or $p = 1 + 4 / n$ , we have $\begin{matrix} (4.36) & \int_{t / 2}^{\infty} s^{j} A (s) {(1 + s)}^{- n (p - 1) / 2 + | α | / 2} d s ≲ {(1 + t)}^{- n (p - 1) / 2 + j + | α | / 2 + 1} \end{matrix}$ if $p > 1 + (k + 2) / n$ . So that, we obtain (4.30) by (4.35) and (4.36). The result (4.31) follows from (4.30) since $- n (p - 1) / 2 + 1 < - k / 2$ holds by $p > 1 + (k + 2) / n$ .

(2) Since $A (\cdot) ⩽ A_{} {(1 + s)}^{n (p - 1) / 4 - 1}$ by Lemma 2.4 when $H < 0$ and $p \neq 1 + 4 / n$ , we have $\begin{matrix} (4.37) & \int_{t / 2}^{\infty} s^{j} A (s) {(1 + s)}^{- n (p - 1) / 2 + | α | / 2} d s ≲ A_{} {(1 + t)}^{- n (p - 1) / 4 + j + | α | / 2} \end{matrix}$ if $p > 1 + 2 k / n$ . So that, we obtain (4.32) by (4.35) and (4.37). The result (4.33) follows from (4.32) since $n (p - 1) / 4 > k / 2$ holds by $p > 1 + 2 k / n$ . □

Now, we prove Theorem 1.2. Let v be the function defined by (1.14). By Lemma 4.1, we have $\begin{matrix} {‖ u (t) - v (t) ‖}_{L^{q} (R^{n})} ⩽ \sum_{l = 1}^{5} {‖ J_{l} (t) ‖}_{L^{q} (R^{n})} . \end{matrix}$ We have $\begin{matrix} {‖ J_{1} (t) ‖}_{L^{q} (R^{n})} = o (t^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ as $t \to \infty$ if $\int_{R^{n}} | y |^{k} | u_{0} (y) | d y < \infty$ by (3) in Lemma 4.2. We have $\begin{matrix} {‖ J_{2} (t) ‖}_{L^{q} (R^{n})} + {‖ J_{4} (t) ‖}_{L^{q} (R^{n})} + {‖ J_{5} (t) ‖}_{L^{q} (R^{n})} = A_{*} ‖ u ‖_{X_{k}}^{p} o (t^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ under the condition (1.10) by Lemmas 4.3, 4.7 and 4.8. We have $\begin{matrix} {‖ J_{3} (t) ‖}_{L^{q} (R^{n})} ⩽ {‖ J_{31} (t) ‖}_{L^{q} (R^{n})} + {‖ J_{32} (t) ‖}_{L^{q} (R^{n})} = o (t^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ under the condition (1.10) by Lemmas 4.4, 4.5 and 4.6. So that, we obtain $\begin{matrix} {‖ u (t) - v (t) ‖}_{L^{q} (R^{n})} = o (t^{- n (1 - 1 / q) / 2 - k / 2}) \end{matrix}$ as $t \to \infty$ , which is the required result in the theorem.
5. Proof of Theorem 1.3

For any multi-index α with $| α | ⩾ 1$ , we have $\begin{matrix} \partial_{x}^{α} f (u) = \sum_{\begin{array}{c} 1 ⩽ k ⩽ | α | \\ β (1) + \dots + β (k) = α \end{array}} C (α, β (1), \dots, β (k)) f^{(k)} (u) \partial_{x}^{β (1)} u \dots \partial_{x}^{β (k)} u, \end{matrix}$ where $C (α, β (1), \dots, β (k))$ are constants dependent on α, $β (1), \dots, β (k)$ . Thus, when $1 ⩽ 2 N + l$ , we have $\begin{array}{l} \sum_{| α | = 2 N + l} {‖ \partial_{x}^{α} f (u) ‖}_{L^{r} (R^{n})} \\ ≲ \sum_{| α | = 2 N + l} \sum_{\begin{array}{c} 1 ⩽ k ⩽ | α | \\ β (1) + \dots + β (k) = α \end{array}} {‖ f^{(k)} (u) \partial_{x}^{β (1)} u \dots \partial_{x}^{β (k)} u ‖}_{L^{r} (R^{n})} \\ ≲ \sum_{| α | = 2 N + l} \sum_{\begin{array}{c} 1 ⩽ k ⩽ | α | \\ β (1) + \dots + β (k) = α \end{array}} ‖ u ‖_{L^{\infty} (R^{n})}^{p - k} {‖ \partial_{x}^{β (1)} u ‖}_{L^{r_{| β (1) |}} (R^{n})} \dots {‖ \partial_{x}^{β (k)} u ‖}_{L^{r_{| β (k) |}} (R^{n})} \\ (5.1) & ≲ ‖ u ‖_{Y}^{p}, \end{array}$ where $r_{m}$ is defined by (1.15) for $1 ⩽ m ⩽ 2 N + l$ and we have used $\begin{matrix} \frac{1}{r} = \frac{| β (1) |}{2 N + l} \cdot \frac{1}{r} + \dots + \frac{| β (k) |}{2 N + l} \cdot \frac{1}{r} = \frac{1}{r_{| β (1) |}} + \dots + \frac{1}{r_{| β (k) |}} . \end{matrix}$ Since we also have $\begin{matrix} (5.2) & {‖ f (u) ‖}_{L^{r_{*}} (R^{n})} ≲ ‖ u ‖_{L^{\infty} (R^{n})}^{p - 1} ‖ u ‖_{L^{r_{*}} (R^{n})} ≲ ‖ u ‖_{Y}^{p} \end{matrix}$ for $r_{*} = r, \infty, r_{| β (1) |}, \dots, r_{| β (k) |}$ , we obtain $\begin{matrix} (5.3) & {‖ f (u) ‖}_{W^{2 N + l, r} (R^{n})} ≲ {‖ f (u) ‖}_{L^{r} (R^{n})} + \sum_{| α | = 2 N + l} {‖ \partial_{x}^{α} f (u) ‖}_{L^{r} (R^{n})} ≲ ‖ u ‖_{Y}^{p} . \end{matrix}$ Similarly, we have $\begin{matrix} (5.4) & {‖ f (u) ‖}_{W^{m, r_{m}} (R^{n})} ≲ ‖ u ‖_{Y}^{p} \end{matrix}$ for $1 ⩽ m ⩽ 2 N + l$ by (5.2), where we note $r = r_{2 N + l}$ .

Let Φ be the operator defined by (1.6). By (2.2) and (5.3), we have $\begin{array}{rcl} {‖ Φ (u) ‖}_{W^{m, r_{m}} (R^{n})} & ⩽ & {‖ e^{t Δ} u_{0} ‖}_{W^{m, r_{m}} (R^{n})} + \int_{0}^{t} A (s) {‖ e^{(t - s) Δ} f (u) (s) ‖}_{W^{m, r_{m}} (R^{n})} d s \\ ⩽ & ‖ u_{0} ‖_{W^{m, r_{m}} (R^{n})} + \frac{C}{H} ‖ u ‖_{Y}^{p} \end{array}$ for some constant $C > 0$ , where we have used $\begin{matrix} (5.5) & \int_{0}^{t} A (s) d s ⩽ \frac{2}{n (p - 1) H} . \end{matrix}$ Thus, we have $\begin{array}{rcl} {‖ Φ (u) ‖}_{Y} & ⩽ & max_{0 ⩽ m ⩽ 2 N + l} {‖ Φ (u) ‖}_{L^{\infty} ((0, T), (W^{m, r_{m}} \cap L^{r}) (R^{n}))} \\ (5.6) & ⩽ & ‖ u_{0} ‖_{D} + \frac{C R^{p}}{H} \end{array}$ for any $u \in Y (R)$ . We also have $\begin{array}{rcl} {‖ Φ (u) - Φ (v) ‖}_{L^{r_{*}} (R^{n})} & = & \int_{0}^{t} A (s) {‖ (f (u) - f (v)) (s) ‖}_{L^{r_{*}} (R^{n})} d s \\ ≲ & \frac{1}{H} \cdot max {‖ u ‖_{Y}^{p - 1}, ‖ v ‖_{Y}^{p - 1}} \cdot ‖ u - v ‖_{L^{r_{*}} (R^{n})} \end{array}$ for $r_{*} = r, \infty$ by (1.6), (5.5) and $\begin{matrix} {‖ f (u) - f (v) ‖}_{L^{r_{*}} (R^{n})} ≲ max {‖ u ‖_{L^{\infty} (R^{n})}^{p - 1}, ‖ v ‖_{L^{\infty} (R^{n})}^{p - 1}} \cdot ‖ u - v ‖_{L^{r_{*}} (R^{n})} \end{matrix}$ for any u, v. Thus, we have $\begin{array}{rcl} d (Φ (u), Φ (v)) & = & {‖ Φ (u) - Φ (v) ‖}_{L^{\infty} ((0, 1 / 2 H), L^{r} (R^{n}) \cap L^{\infty} (R^{n}))} \\ (5.7) & ⩽ & \frac{C R^{p - 1}}{H} d (u, v) \end{array}$ for any $u, v \in Y (R)$ . By (5.6) and (5.7), the operator Φ is a contraction mapping on the ball in Y if R satisfies $2 ‖ u_{0} ‖_{D} ⩽ R$ and $2 C R^{p - 1} ⩽ H$ .

(2) (i) Let us consider the case $r = \infty$ with $N \neq 0$ or $l \neq 0$ . In this case, we have $D = W^{2 N + l, \infty} (R^{n})$ and $D^{'} = W^{2 N + l - 1, \infty} (R^{n})$ by their definitions. We have $A (s) f (u) (s) \in W^{2 N + l, \infty} (R^{n})$ for a.e. $0 ⩽ s < 1 / 2 H$ by (5.4). By this result, $u_{0} \in W^{2 N + l, \infty} (R^{n})$ and Lemma 2.6, we have $u \in C ([0, 1 / 2 H), D^{'})$ .

(ii) Let us consider the case $r \neq \infty$ . Since $D^{'} = W^{2 N + l, r} (R^{n})$ , we have $u \in C ([0, 1 / 2 H), D^{'})$ by Lemma 2.6.

(3) Let $v \in C ([0, 1 / 2 H), D^{'}) \cap L^{\infty} ((0, 1 / 2 H) \times R^{n})$ be an another solution which has the initial data $u_{0}$ . Put $\begin{matrix} t_{0} : = sup_{0 ⩽ t < 1 / 2 H} {t; u (t) = v (t) \in L^{r} (R^{n})} . \end{matrix}$ We assume $t_{0} < 1 / 2 H$ and show a contradiction. Since $u, v \in C ([0, 1 / 2 H), D^{'})$ and $D^{'} \subset L^{r} (R^{n})$ , we have $u (t_{0}) = v (t_{0})$ . Since u and v are rewritten as $\begin{matrix} w (t) = e^{(t - t_{0}) Δ} w (t_{0}) + \int_{t_{0}}^{t} e^{(t - s) Δ} A (s) f (w) (s) d s \end{matrix}$ for $w = u, v$ , we have $\begin{matrix} u (t) - v (t) = \int_{t_{0}}^{t} e^{(t - s) Δ} A (s) (f (u) - f (v)) (s) d s \end{matrix}$ by $u (t_{0}) = v (t_{0})$ . By (2.2) and the Hölder inequality, we have $\begin{array}{rcl} {‖ u (t) - v (t) ‖}_{L^{r} (R^{n})} & ⩽ & \int_{t_{0}}^{t} A (s) {‖ e^{(t - s) Δ} (f (u) - f (v)) (s) ‖}_{L^{r} (R^{n})} d s \\ ⩽ & \int_{t_{0}}^{t} A (s) d s \cdot max_{w = u, v} ‖ w ‖_{L^{\infty} ((t_{0}, t) \times R^{n})}^{p - 1} \cdot ‖ u - v ‖_{L^{\infty} ((t_{0}, t), L^{r} (R^{n}))} . \end{array}$ Thus, we have $\begin{array}{l} {‖ u (t) - v (t) ‖}_{L^{\infty} ((t_{0}, t_{0} + ε), L^{r} (R^{n}))} \\ ⩽ \int_{t_{0}}^{t_{0} + ε} A (s) d s \cdot max_{w = u, v} ‖ w ‖_{L^{\infty} ((t_{0}, t_{0} + ε) \times R^{n})}^{p - 1} \cdot ‖ u - v ‖_{L^{\infty} ((t_{0}, t_{0} + ε), L^{r} (R^{n}))} . \end{array}$ Since $u, v \in L^{\infty} ((0, 1 / 2 H) \times R^{n})$ , we obtain $\begin{matrix} {‖ u (t) - v (t) ‖}_{L^{\infty} ((t_{0}, t_{0} + ε), L^{r} (R^{n}))} = 0 \end{matrix}$ for sufficiently small $ε > 0$ . This result shows $u (t) = v (t)$ for $t_{0} < t < t_{0} + ε$ , which is a contradiction to the definition of $t_{0}$ . So that, we have $t_{0} = 1 / 2 H$ , which yields the required uniqueness of the solution.

6. Proof of Theorem 1.4

This section is devoted to the proof of Theorem 1.4. The following lemma is useful to obtain the asymptotic profiles of the solutions of (1.1) when $H > 0$ .

Lemma 6.1.

Let $H > 0$ . Let $N, \tilde{N} \in N \cup {0}$ with $\tilde{N} ⩽ 2 (N + 1)$ . Let $1 ⩽ r ⩽ q ⩽ \infty$ . Then $\begin{matrix} (6.1) & \begin{matrix} {‖ e^{t Δ} u_{0} - \sum_{j = 0}^{N} \frac{1}{j!} {(\frac{1}{2 H} - t)}^{j} {(- Δ)}^{j} e^{Δ / 2 H} u_{0} ‖}_{L^{q} (R^{n})} \\ ≲ t^{- n (1 / r - 1 / q) / 2 - (N + 1) + \tilde{N} / 2} {(\frac{1}{2 H} - t)}^{N + 1} {‖ {(- Δ)}^{\tilde{N} / 2} u_{0} ‖}_{L^{r} (R^{n})} \end{matrix} \end{matrix}$ for $0 < t < 1 / 2 H$ .

Proof.

Put $T : = 1 / 2 H$ . For $N ⩾ 0$ , we use the expansion $\begin{array}{rcl} e^{t Δ} & = & \sum_{j = 0}^{N} \frac{{((T - t) (- Δ))}^{j}}{j!} e^{T Δ} \\ (6.2) & + \frac{{((T - t) (- Δ))}^{N + 1}}{N!} \int_{0}^{1} {(1 - θ)}^{N} e^{t Δ + (1 - θ) (T - t) Δ} d θ . \end{array}$ Defining the operator $\begin{matrix} P : = e^{t Δ} - \sum_{j = 0}^{N} \frac{{((T - t) (- Δ))}^{j}}{j!} e^{T Δ}, \end{matrix}$ we have $\begin{matrix} ‖ P u_{0} ‖_{L^{q} (R^{n})} ⩽ \frac{{(T - t)}^{N + 1}}{N!} \int_{0}^{1} {(1 - θ)}^{N} {‖ {(- Δ)}^{N + 1} e^{t Δ + (1 - θ) (T - t) Δ} u_{0} ‖}_{L^{q} (R^{n})} d θ . \end{matrix}$ Since we have $\begin{array}{l} {(- Δ)}^{N + 1} e^{t Δ + (1 - θ) (T - t) Δ} u_{0} \\ = {(- Δ)}^{N + 1 - \tilde{N} / 2} e^{t Δ + (1 - θ) (T - t) Δ} {(- Δ)}^{\tilde{N} / 2} u_{0} \\ = \frac{1}{{(2 π)}^{n / 2}} (F^{- 1} (| ξ |^{2 (N + 1 - \tilde{N} / 2)} e^{- t | ξ |^{2} - (1 - θ) (T - t) | ξ |^{2}})) * {(- Δ)}^{\tilde{N} / 2} u_{0}, \end{array}$ we have $\begin{matrix} (6.3) & ‖ P u_{0} ‖_{L^{q} (R^{n})} ⩽ \frac{{(T - t)}^{N + 1}}{{(2 π)}^{n / 2} N!} \int_{0}^{1} {(1 - θ)}^{N} I (q_{*}) d θ {‖ {(- Δ)}^{\tilde{N} / 2} u_{0} ‖}_{L^{r} (R^{n})} \end{matrix}$ by the Young inequality, where we have put $1 / q_{*} : = 1 + 1 / q - 1 / r$ and $\begin{matrix} I (q_{*}) : = {‖ F^{- 1} (| ξ |^{2 (N + 1 - \tilde{N} / 2)} e^{- t | ξ |^{2} - (1 - θ) (T - t) | ξ |^{2}}) ‖}_{L^{q_{*}} (R^{n})} . \end{matrix}$ We have $\begin{array}{rcl} I (\infty) & ⩽ & {‖ | ξ |^{2 (N + 1 - \tilde{N} / 2)} e^{- t | ξ |^{2} - (1 - θ) (T - t) | ξ |^{2}} ‖}_{L_{ξ}^{1} (R^{n})} \\ ⩽ & {‖ | ξ |^{2 (N + 1 - \tilde{N} / 2)} e^{- t | ξ |^{2}} ‖}_{L_{ξ}^{1} (R^{n})} \\ (6.4) & = & t^{- (N + 1) + \tilde{N} / 2 - n / 2} {‖ | η |^{2 (N + 1 - \tilde{N} / 2)} e^{- | η |^{2}} ‖}_{L_{η}^{1} (R^{n})}, \end{array}$ where we have put $η : = t^{1 / 2} ξ$ . Since we have $\begin{array}{l} F^{- 1} (| ξ |^{2 (N + 1 - \tilde{N} / 2)} e^{- t | ξ |^{2} - (1 - θ) (T - t) | ξ |^{2}}) \\ = {(2 π)}^{- n / 2} F^{- 1} (| ξ |^{2 (N + 1 - \tilde{N} / 2)} e^{- t | ξ |^{2}}) * F^{- 1} (e^{- (1 - θ) (T - t) | ξ |^{2}}) \\ = : {(2 π)}^{- n / 2} II * III, \end{array}$ we have $I (1) ⩽ {(2 π)}^{- n / 2} ‖ II ‖_{1} \cdot ‖ III ‖_{1}$ by the Young inequality. Since $II$ is rewritten as $\begin{matrix} II (x) = t^{- (N + 1) + \tilde{N} / 2 - n / 2} (F_{η}^{- 1} (| η |^{2 (N + 1 - \tilde{N} / 2)} e^{- | η |^{2}})) (t^{- 1 / 2} x), \end{matrix}$ we have $\begin{matrix} ‖ II ‖_{L^{1} (R^{n})} = t^{- (N + 1) + \tilde{N} / 2} {‖ F_{η}^{- 1} (| η |^{2 (N + 1 - \tilde{N} / 2)} e^{- | η |^{2}}) ‖}_{L^{1} (R^{n})}, \end{matrix}$ where we have put $η : = t^{1 / 2} ξ$ . Since $III$ is rewritten as $\begin{matrix} III (x) = {(1 - θ)}^{- n / 2} {(T - t)}^{- n / 2} (F_{η}^{- 1} e^{- | η |^{2}}) ({(1 - θ)}^{- 1 / 2} {(T - t)}^{- 1 / 2} x), \end{matrix}$ we have $‖ III ‖_{L^{1} (R^{n})} = ‖ F_{η}^{- 1} e^{- | η |^{2}} ‖_{L^{1} (R^{n})}$ . Thus, we have $\begin{matrix} (6.5) & I (1) ≲ ‖ II ‖_{L^{1} (R^{n})} \cdot ‖ III ‖_{L^{1} (R^{n})} ≲ t^{- (N + 1) + \tilde{N} / 2} . \end{matrix}$ Combining (6.4) and (6.5), we have $\begin{matrix} (6.6) & I (q_{*}) ⩽ I {(\infty)}^{1 - 1 / q_{*}} \cdot I {(1)}^{1 / q_{*}} ≲ t^{- (N + 1) + \tilde{N} / 2 - n (1 - 1 / q_{*}) / 2} . \end{matrix}$ We obtain the required result by (6.3) and (6.6). □

Now, let us prove the theorem. Let $v_{0}$ , $v_{1}$ and v be the functions defined in the theorem. Since $v_{0}$ satisfies $\begin{matrix} \partial_{x}^{α} v_{0} = \sum_{j = 0}^{N} \frac{{(T - t)}^{j}}{j!} e^{T Δ} {(- Δ)}^{j} \partial_{x}^{α} u_{0}, \end{matrix}$ we have $\begin{matrix} {‖ \partial_{x}^{α} v_{0} ‖}_{L^{r} (R^{n})} ⩽ \sum_{j = 0}^{N} \frac{{(T - t)}^{j}}{j!} {‖ {(- Δ)}^{j} \partial_{x}^{α} u_{0} ‖}_{L^{r} (R^{n})} ≲ ‖ u_{0} ‖_{W^{2 N + | α |, r} (R^{n})} \end{matrix}$ for any multi-index α by Lemma 2.1. Thus, we have $\begin{matrix} (6.7) & ‖ v_{0} ‖_{W^{ℓ, r} (R^{n})} = \sum_{| α | ⩽ ℓ} {‖ \partial_{x}^{α} v_{0} ‖}_{L^{r} (R^{n})} ≲ ‖ u_{0} ‖_{W^{2 N + ℓ, r} (R^{n})} \end{matrix}$ for $ℓ ⩾ 0$ . Replacing t and T with $t - s$ and $T - s$ in (6.2), respectively, we have $\begin{array}{rcl} e^{(t - s) Δ} & = & \sum_{j = 0}^{N} \frac{{((T - t) (- Δ))}^{j}}{j!} e^{(T - s) Δ} \\ + \frac{{((T - t) (- Δ))}^{N + 1}}{N!} \int_{0}^{1} {(1 - θ)}^{N} e^{(t - s) Δ + (1 - θ) (T - t) Δ} d θ . \end{array}$ Thus, we have $\begin{matrix} \int_{0}^{t} e^{(t - s) Δ} h (s) d s = v_{1} + v_{2}, \end{matrix}$ where we have put $h : = A f (u)$ and defined $v_{2}$ by $\begin{matrix} v_{2} (t, \cdot) : = \frac{{((T - t) (- Δ))}^{N + 1}}{N!} \int_{0}^{1} \int_{0}^{t} {(1 - θ)}^{N} e^{(t - s) Δ + (1 - θ) (T - t) Δ} h (s) d s d θ . \end{matrix}$ We have $\begin{array}{rcl} ‖ v_{2} ‖_{L^{r} (R^{n})} & ⩽ & \frac{{(T - t)}^{N + 1}}{N!} \int_{0}^{1} \int_{0}^{t} {(1 - θ)}^{N} {‖ {(- Δ)}^{N + 1} h (s) ‖}_{L^{r} (R^{n})} d s d θ \\ ⩽ & \frac{{(T - t)}^{N + 1}}{(N + 1)!} \int_{0}^{t} A (s) {‖ {(- Δ)}^{N + 1} f (u) (s) ‖}_{L^{r} (R^{n})} d s \end{array}$ by Lemma 2.1. Since we have $‖ {(- Δ)}^{N + 1} f (u) (s) ‖_{L^{r} (R^{n})} ≲ ‖ u ‖_{Y}^{p}$ by (5.1), and we also have $\int_{0}^{t} A (s) d s ⩽ 2 / n (p - 1) H$ by simple calculation, we obtain $\begin{matrix} (6.8) & ‖ v_{2} ‖_{L^{r} (R^{n})} ≲ \frac{{(T - t)}^{N + 1}}{H} ‖ u ‖_{Y}^{p} . \end{matrix}$ Analogously, we have $\begin{array}{rcl} ‖ v_{1} ‖_{W^{2, r} (R^{n})} & ≲ & \sum_{j = 0}^{N} \frac{{(T - t)}^{j}}{j!} \int_{0}^{t} {‖ {(- Δ)}^{j} h (s) ‖}_{W^{2, r} (R^{n})} d s \\ ≲ & \sum_{j = 0}^{N} \frac{{(T - t)}^{j}}{j!} \int_{0}^{t} A (s) d s \cdot ‖ u ‖_{Y}^{p} \\ (6.9) & ≲ & \sum_{j = 0}^{N} \frac{{(T - t)}^{j}}{j! H} ‖ u ‖_{Y}^{p} \end{array}$ by $\begin{matrix} {‖ {(- Δ)}^{j} h (s) ‖}_{W^{2, r} (R^{n})} ≲ A (s) {‖ f (u) ‖}_{W^{2 (N + 1), r} (R^{n})} ≲ A (s) ‖ u ‖_{Y}^{p} . \end{matrix}$

(1) Since v satisfies $v = v_{0} + v_{1}$ , we obtain $\begin{matrix} ‖ v ‖_{W^{2, r} (R^{n})} ≲ ‖ u_{0} ‖_{W^{2 (N + 1), r} (R^{n})} + \frac{‖ u ‖_{Y}^{p}}{H} \end{matrix}$ by (6.7) and (6.9). So that, $v \in L^{\infty} ((0, 1 / 2 H), W^{2, r} (R^{n}))$ as required.

(2) Since we have $u - v = e^{t Δ} u_{0} - v_{0} + v_{2}$ , we obtain $\begin{matrix} {‖ u (t) - v (t) ‖}_{L^{r} (R^{n})} ≲ \frac{{(T - t)}^{N + 1}}{(N + 1)! t^{N + 1}} ‖ u_{0} ‖_{L^{r} (R^{n})} + \frac{{(T - t)}^{N + 1}}{(p - 1) H} ‖ u ‖_{Y}^{p} \end{matrix}$ by Lemma 6.1 and (6.8), which is the required result.

Footnotes

Acknowledgements

This work was supported in part by JSPS KAKENHI Grant Number (B)16H03940 for the first author, and by JSPS KAKENHI Grant Number (C)19K03596 for the second author. The authors are thankful to the anonymous referee for several comments to revise the paper.

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