Bubble towers in the ancient solution of energy-critical heat equation

Abstract

We construct a radial smooth positive ancient solution for energy critical semi-linear heat equation in $\mathbb {R}^n$, $n\ge 7$. It blows up at the origin with the profile of multiple Talenti bubbles in the backward time infinity.

Appendices

Appendix A. Estimates for the data in the outer problem

We will prove Proposition 4.1 in this section. Throughout this section, we assume $\Vert \Psi \Vert _{\alpha ,\sigma }^{out,*} + \Vert \vec {\phi }\Vert _{a,\sigma }^{in,*}< \infty$, $\Vert \vec {\mu }_1\Vert _{\sigma }\le 1$.

The parameters are determined in the following order. First, we choose R as a large fixed positive constant. Second, we choose $\sigma >0$ small. Third, we choose $\delta >0$ small. Fourth, we choose $\epsilon >0$ small. Finally, we take $t_{0}$ very negative such that $\mu _{j} \approx \mu _{0j}$, for $j=1, \dots , k$, $\dot{\mu }_{j} \approx \dot{\mu }_{0j}$ for $j=2 , \dots , k$.

We introduce the notation $y_j=x/\mu _j$, $\bar{y}_{j}=x/\bar{\mu }_j$, $y_{0j}=x/\mu _{0j}$, $\bar{y}_{0j}=x/\bar{\mu }_{0j}$ for $j=1 , \dots , k$. One readily sees that $|y_j| \approx |y_{0j}|$, $|\bar{y}_j| \approx |\bar{y}_{0j}|$ for $j=1 , \dots , k$.

Lemma A.1

Consider the $U_j$ defined in (1.22). For $j=1,\cdots ,k-1$, one has

$$\begin{aligned} U_j< U_{j+1}\text { in }\{|x|<\bar{\mu }_{j+1}\} \text { and }U_j> U_{j+1}\text { in }\{|x|>\bar{\mu }_{j+1}\}. \end{aligned}$$

(A.1)

In $\{|x| \le \bar{\mu }_{0k} \}$,

$$\begin{aligned} U_{k} > rsim U_{k-1}> U_{k-2}> \dots > U_{1} . \end{aligned}$$

(A.2)

In $\{|x|\ge \bar{\mu }_{02} \}$,

$$\begin{aligned} U_{1} > rsim U_{2}> U_3>\dots > U_{k} . \end{aligned}$$

(A.3)

In $\{\bar{\mu }_{0,j+1}\le |x| \le \bar{\mu }_{0j} \}$, $j=2, \dots , k-1$,

$$\begin{aligned} U_{j} > rsim U_{j+1}> U_{j+2}> \dots> U_{k}, \quad U_{j} > rsim U_{j-1}> U_{j-2}> \dots > U_{1} . \end{aligned}$$

(A.4)

Moreover

$$\begin{aligned} \frac{U_{j+1}}{U_j}&\approx \lambda _{j+1}^{-\frac{n-2}{2}} \langle y_{j+1} \rangle ^{-(n-2)} \mathbf {1}_{\{ |x|\le \mu _{0j}\}} + \lambda _{j+1}^{\frac{n-2}{2}} \mathbf {1}_{\{ |x| > \mu _{0j}\}} \text{ for } j=1, \dots , k-1 , \end{aligned}$$

(A.5)

$$\begin{aligned} \frac{U_{j-1}}{U_j}&\approx \lambda _{j}^{\frac{n-2}{2}} \langle y_j \rangle ^{n-2} \mathbf {1}_{\{ |x|\le \mu _{0,j-1}\}} + \lambda _{j}^{-\frac{n-2}{2}} \mathbf {1}_{\{ |x| > \mu _{0,j-1}\}} \text{ for } j=2, \dots , k . \end{aligned}$$

(A.6)

Proof

(A.1)-(A.4) follow from that $\frac{U_{j+1}}{U_j}$ is strictly decreasing about |x| and $\frac{U_{j+1}}{U_j}(\bar{\mu }_{j+1})=1$, (see Fig. 1). Up to a multiplicity of the constant $\alpha _n$, $U_j=\mu _{j}^{\frac{2-n}{2}}(1+|y_j|^2)^{\frac{2-n}{2}}$ and

$$\begin{aligned} U_{j+1}=\frac{\mu _{j+1}^{\frac{n-2}{2}}}{(\mu _{j+1}^2+|x|^2)^{\frac{n-2}{2}}}=\frac{\mu _{j+1}^{\frac{n-2}{2}}\mu _j^{2-n}}{(\lambda _{j+1}^2+|y_j|^2)^{\frac{n-2}{2}}}, \end{aligned}$$

(A.7)

then

$$\begin{aligned} \frac{U_{j+1}}{U_j} =\lambda _{j+1}^{\frac{n-2}{2}}\frac{(1+|y_j|^2)^{\frac{n-2}{2}}}{(\lambda _{j+1}^2+|y_j|^2)^{\frac{n-2}{2}}} \approx \lambda _{j+1}^{-\frac{n-2}{2}} \langle y_{j+1} \rangle ^{-(n-2)} \mathbf {1}_{\{ |x|\le \mu _{0j}\}} + \lambda _{j+1}^{\frac{n-2}{2}} \mathbf {1}_{\{ |x| > \mu _{0j}\}} ,\qquad \end{aligned}$$

(A.8)

for $j=1, \dots , k-1$. This finishes the proof of (A.5). Similarly,

$$\begin{aligned} \frac{U_{j-1}}{U_j} = \lambda _j^{-\frac{n-2}{2}} \frac{(\lambda _j^2 + |y_{j-1}|^2)^{\frac{n-2}{2}}}{(1+|y_{j-1}|^2)^{\frac{n-2}{2}}} \approx \lambda _{j}^{\frac{n-2}{2}} \langle y_j \rangle ^{n-2} \mathbf {1}_{\{ |x|\le \mu _{0,j-1}\}} + \lambda _{j}^{-\frac{n-2}{2}} \mathbf {1}_{\{ |x| > \mu _{0,j-1}\}} \end{aligned}$$

(A.9)

for $j=2, \dots , k$. Then (A.6) holds. $\square$

Fig. 1

Relation for three bubbles

Lemma A.2

Consider $\varphi _0$ defined in (2.6). One has $|\varphi _0|\lesssim \sum \limits _{i=2}^k\lambda _i U_i\chi _i$.

Proof

By (2.6) and (2.22), we have

$$\begin{aligned} |\varphi _0|\lesssim \sum _{i=2}^k\mu _{i-1}^{-\frac{n-2}{2}}\langle y_i\rangle ^{-2}\chi _i. \end{aligned}$$

(A.10)

It follows from (2.4) that the support of $\chi _i$ are disjoint. More precisely, the support of $\chi _i$ is contained in $\{\lambda _{i+1}^{\frac{1}{2}}\le |y_i|\le \lambda _i^{-\frac{1}{2}}\}$. It is easy to verify that $\mu _{i-1}^{-\frac{n-2}{2}}\langle y_i\rangle ^{-2}\lesssim \lambda _i U_i$ in this set. $\square$

Lemma A.3

For $0<\alpha<a<1$, there exists R large enough and $t_0$ negative enough such that $B[\vec {\phi }]$ defined in (3.8) satisfies

$$\begin{aligned} \begin{aligned} \Vert B[\vec {\phi }\,]\Vert ^{out}_{\alpha ,\sigma }\lesssim R^{\alpha -a} \Vert \vec {\phi }\Vert ^{in,*}_{a,\sigma } \end{aligned} \end{aligned}$$

(A.11)

Proof

By the definition (6.1) and (4.4),

$$\begin{aligned} \langle y_j\rangle |\nabla _{y_j}\phi _j(y_j,t)|+|\phi _j(y_j,t)|\lesssim |t|^{\gamma _j}\mu _{0j}^{\frac{n-2}{2}}R^{n+1-a}\langle y_j\rangle ^{-n-1}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }. \end{aligned}$$

(A.12)

$\bullet$ First, using (3.7)

$$\begin{aligned} \begin{aligned} |\dot{\mu }_j\frac{\partial \varphi _j}{\partial \mu _j}\eta _j|=&\,\left| \dot{\mu }_j\mu _j^{-\frac{n}{2}}\left( \frac{n-2}{2} \phi _j(y_j,t)+y_j\cdot \nabla _{y_j} \phi _j(y_j,t)\right) \eta _{j}\right| \\ \lesssim&\, | \dot{\mu }_j| \mu _j^{-1}(-t)^{\gamma _j}R^{n+1-a}\langle y_j\rangle ^{-n-1}\eta _j\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma } \\ \lesssim&\, |t_0|^{-\epsilon } R^{n+1-a} w_{1j}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }. \end{aligned} \end{aligned}$$

(A.13)

Here we choose $\epsilon$ small such that $|\dot{\mu }_j \mu _j| \lesssim |t|^{-\epsilon }$ for $j=1, \dots , k$. We have used (4.20) in the last step.

$\bullet$ Second, (A.12) implies that

$$\begin{aligned} |\varphi _j(x,t)|\lesssim |t|^{\gamma _j}R^{-a}\Vert \phi _j\Vert _{j,a,\sigma }^{in,*} \text{ for } 2R\le |y_j| \le 4R . \end{aligned}$$

(A.14)

Using (3.2), we obtain

$$\begin{aligned} |\Delta \eta _j \varphi _j|\lesssim (R\mu _j)^{-2}|t|^{\gamma _j}R^{-a}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }\mathbf {1}_{\{2R\le |y_j|\le 4R\}}\lesssim R^{\alpha -a}w_{1j}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }. \end{aligned}$$

(A.15)

Similarly, we have

$$\begin{aligned} \begin{aligned} |\nabla _x\eta _j\cdot \nabla _x\varphi _j|\lesssim&\ R^{-1}\mu _j^{-1}|t|^{\gamma _j}\mu _j^{-1}R^{-1-a}\Vert \phi _j\Vert _{j,a,\sigma }^{in,*}\mathbf {1}_{\{2R\mu _j\le |x|\le 4R\mu _j\}} \lesssim R^{\alpha -a} w_{1j}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma } \end{aligned} \end{aligned}$$

(A.16)

and

$$\begin{aligned} \begin{aligned} |\partial _t\eta _j\varphi _j|&\lesssim R^{-1}\mu _j^{-2} |\dot{\mu }_j | |t|^{\gamma _j}R^{-a}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }\mathbf {1}_{\{2R\le |y_j|\le 4R\}} \lesssim R^{1+\alpha -a} |\dot{\mu }_j | w_{1j}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma } \\&\lesssim R^{1+\alpha -a} |t_0|^{-\epsilon } w_{1j}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma } , \end{aligned} \end{aligned}$$

(A.17)

when $|\dot{\mu }_j|\lesssim |t|^{-\epsilon }$ for $j=1, \dots , k$.

$\bullet$ Third, to estimate $\left| p(u_*^{p-1}-U_j^{p-1})\varphi _j\eta _j\right|$. We only give calculation details for $j=2, \dots , k-1$ since the case $j=1$ and $j=k$ can be dealt with similarly. Consider it in $\{\bar{\mu }_{0,j+1}\le |x|\le 4R\mu _{0j}\}$. By Lemma A.1 and Lemma A.2,

$$\begin{aligned} U_{j}^{-1}\left( \sum \limits _{i\ne j} U_{i} + \varphi _{0} \right) \ge 2^{-1}U_{j}^{-1}\left( \sum \limits _{i\ne j} U_{i} \right) -\lambda _{j}^{\frac{1}{2}} > -\frac{1}{2}, \end{aligned}$$

when $t_0$ is very negative. Consequently $u_*>\frac{1}{2}U_j$. Thus by the mean value theorem and (A.5) and (A.6),

$$\begin{aligned} \begin{aligned}&|u_*^{p-1}-U_j^{p-1}| \lesssim U_j^{p-1}\left( \frac{U_{j+1}}{U_j}+\frac{U_{j-1}}{U_j} + \lambda _j \right) \\ \lesssim&\mu _j^{-2} \langle y_j\rangle ^{-4} \left( \lambda _{j+1}^{-\frac{n-2}{2}} \langle y_{j+1} \rangle ^{2-n} \mathbf {1}_{\{\bar{\mu }_{0,j+1}\le |x|\le \mu _{0j}\}} + \lambda _{j+1}^{\frac{n-2}{2}} +\lambda _j^{\frac{n-2}{2}}\langle y_j\rangle ^{n-2}+\lambda _j\right) . \end{aligned}\nonumber \\ \end{aligned}$$

(A.18)

Therefore, using $|\varphi _j|\lesssim |t|^{\gamma _j}R^{n+1-a} \Vert \phi _j\Vert _{j,a,\sigma }^{in,*}$,

$$\begin{aligned} \begin{aligned}&\left| p(u_*^{p-1}-U_j^{p-1})\varphi _j\eta _j\right| \mathbf {1}_{\{\bar{\mu }_{0,j+1}\le |x|\le 4R\mu _j\}} \\ \lesssim&\ R^{n+1-a} \mu _j^{-2}|t|^{\gamma _j} \lambda _{j+1}^{-\frac{n-2}{2}} \langle y_{j+1}\rangle ^{2-n}\mathbf {1}_{\{\bar{\mu }_{0,j+1}\le |x|\le \mu _{0j} \}} \Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }\\&\ +R^{n+1-a}(\lambda _{j+1}^{\frac{n-2}{2}}+\lambda _j^{\frac{n-2}{2}} R^{n-2}+\lambda _j)w_{1j}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }\\ \lesssim&\ R^{n+1-a}|t_0|^{-\epsilon }(w_{2j}+w_{1j})\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }. \end{aligned} \end{aligned}$$

(A.19)

Here we have used the following fact.

$$\begin{aligned} \begin{aligned}&R^{n+1-a} \mu _j^{-2}|t|^{\gamma _j} \lambda _{j+1}^{-\frac{n-2}{2}} \langle y_{j+1}\rangle ^{2-n} \mathbf {1}_{\{\bar{\mu }_{0,j+1}\le |x|\le \mu _{0j} \}} \\ \approx&\, R^{n+1-a} \lambda _{j+1} \lambda _{j}^{\frac{n}{2} -1} |t|^{\sigma } |t|^{-2\sigma } \mu _{j+1}^{\frac{n}{2} -2} \mu _{j}^{-1} |x|^{2-n} \mathbf {1}_{\{\bar{\mu }_{0,j+1}\le |x|\le \mu _{0j} \}} \\ \lesssim&\, R^{n+1-a} |t_0|^{-\epsilon } w_{2j} . \end{aligned} \end{aligned}$$

In $\{\bar{\mu }_{0,m+1}\le |x|\le \bar{\mu }_{0m}\}$, $m=j+1,\cdots ,k$, one has $|u_*^{p-1}-U_j^{p-1}|\lesssim U_m^{p-1} \approx \mu _m^{-2} \langle y_m \rangle ^{-4}$ and $|\varphi _j|\lesssim (-t)^{\gamma _j}R^{n+1-a}\Vert \phi _j\Vert _{j,a,\sigma }^{in,*}$.

Then it is easy to see

$$\begin{aligned} \left| p(u_*^{p-1}-U_j^{p-1})\varphi _j\eta _j\right| \lesssim R^{n+1-a} |t|^{\gamma _j-\gamma _m} w_{1m}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }\lesssim R^{n+1-a}|t_0|^{-\epsilon }w_{1m}\Vert \phi _j\Vert ^{in,*}_{j,a,\sigma }. \end{aligned}$$

Taking $t_0$ very negative such that $|t_0|^{-\epsilon }<R^{\alpha -n-1}$, we obtain (A.11). $\square$

Recall $E^{out}$ defined in (3.10). We reorganize the terms as the following.

$$\begin{aligned} E^{out}=\bar{E}_{11}+\bar{E}_2+\bar{E}_3+\bar{E}_4+\bar{E}_5 \end{aligned}$$

(A.20)

where $\bar{E}_{11}$ is defined in (2.13), and

$$\begin{aligned} \bar{E}_2=&\mu _1^{-\frac{n+2}{2}}D_1[\vec {\mu }_1](1-\eta _1)+\sum _{j=2}^k\mu _j^{-\frac{n+2}{2}}D_j[\vec {\mu }_1] (\chi _j-\eta _j )+ \sum _{j=2}^k \mu _j^{-\frac{n+2}{2}}\Theta _j[\vec {\mu }_1]\chi _j \end{aligned}$$

(A.21)

$$\begin{aligned} \bar{E}_3=&\sum _{j=2}^kp(\bar{U}^{p-1}-U_j^{p-1})\varphi _{0j}\chi _j \end{aligned}$$

(A.22)

$$\begin{aligned} \bar{E}_4=&\sum _{j=2}^{k}\left( 2 \nabla _{x} \varphi _{0 j} \cdot \nabla _{x}\chi _{j}+ \Delta _{x}\chi _{j} \varphi _{0 j}\right) -\sum _{j=2}^{k} \partial _{t}\left( \varphi _{0 j} \chi _{j} \right) \end{aligned}$$

(A.23)

$$\begin{aligned} \bar{E}_5=&N_{\bar{U}}[\varphi _0]. \end{aligned}$$

(A.24)

Lemma A.4

There exist $\sigma ,\epsilon >0$ small and $t_0$ negative enough, such that

$$\begin{aligned}E^{out}\lesssim R^{-1} \left( \sum _{j=1}^kw_{1j}+\sum _{j=1}^{k-1} w_{2j}+w_3\right) . \end{aligned}$$

Proof

$\bullet$ Estimate of $\bar{E}_2$. Consider the first term in $\bar{E}_2$. The support of $1-\eta _1$ is $\{|y_{01}|\ge 2R\}$. Since we assume $\Vert \vec {\mu }\Vert _\sigma <1$, one has $|\dot{\mu }_{11}|\le \Vert \vec {\mu }_1\Vert _{\sigma } |t|^{-1-\sigma } \le |t|^{-1-\sigma }$. Then using (2.27)

$$\begin{aligned} |\mu _{1}^{-\frac{n-2}{2}} D_{1}[\vec {\mu }_1] (1-\eta _1)|\lesssim & {} |t|^{-1-\sigma } |x|^{2-n} \mathbf {1}_{\{ |x|\ge 2R \}}\\\lesssim & {} R^{4+\alpha -n}w_{11} + R^{-1} w_{3} \lesssim R^{-1} \left( w_{11} +w_{3} \right) . \end{aligned}$$

For $j\ge 2$, the support of $\chi _j-\eta _j$ is contained in $\{ |x|\le \bar{\mu }_{0,j+1}\}\cup \{ 2R \mu _{0j} \le |x|\le \bar{\mu }_{0j} \}$. In the first set (it is vacuum if $j=k$), one has $|\chi _j-\eta _j|\le \chi (2 \bar{y}_{0,j+1})$ and $\langle y_{0j}\rangle \approx 1$. It follows from (6.4) that

$$\begin{aligned} \left| \mu _j^{-\frac{n+2}{2}}D_j[\vec {\mu }_1](\chi _j-\eta _j)\right| \lesssim \mu _j^{-2} |t|^{\gamma _{j}} \chi (2 \bar{y}_{0,j+1}) \lesssim |t|^{-\epsilon } w_{1,j+1}. \end{aligned}$$

In the second set, since $|y_{0j}|\ge 2R$, then

$$\begin{aligned} \left| \mu _j^{-\frac{n+2}{2}}D_j[\vec {\mu }_1](\chi _j-\eta _j)\right| \lesssim \mu _{j}^{-2} |t|^{\gamma _{j}} |y_j|^{-4} \lesssim R^{-1 } w_{1j}. \end{aligned}$$

It is straightforward to have

$$\begin{aligned} |\mu _j^{-\frac{n+2}{2}}\Theta _j[\vec {\mu }_1]\chi _j| \lesssim |t|^{-\sigma } \mu _{j}^{-2} |t|^{\gamma _j} \langle y_j \rangle ^{-4} \chi _j \lesssim |t|^{-\sigma } w_{1j} . \end{aligned}$$

$\bullet$ Estimate of $\bar{E}_3$. In the support of $\chi _j$, by Lemma (A.1), we have

$$\begin{aligned} |\bar{U}^{p-1}-U_j^{p-1}|\lesssim U_j^{p-1}(\frac{U_{j+1}}{U_j}+\frac{U_{j-1}}{U_j}). \end{aligned}$$

It follows from (2.7) and (2.22) that $\varphi _{0j}\le |t|^{\gamma _j+\sigma }\langle y_i\rangle ^{-2}$. Using (A.18), similar to (A.19), we get

$$\begin{aligned}|\bar{E}_3|\lesssim \sum \limits _{j=2}^{k} \lambda _j^{\frac{n-2}{2}}|t|^{\sigma }w_{1j} + \sum \limits _{j=1}^{k-1} \lambda _{j+1} \lambda _j^{\frac{n-2}{2}}|t|^{2\sigma }w_{2j}\lesssim |t_0|^{-\epsilon }(\sum \limits _{j=2}^{k} w_{1j}+\sum \limits _{j=1}^{k-1}w_{2j}). \end{aligned}$$

$\bullet$ Estimate of $\bar{E}_4$. Notice

$$\begin{aligned} \left| \nabla _{x} \chi _{j}\right| \lesssim&\bar{\mu }_{0j}^{-1} \mathbf {1}_{\left\{ \frac{1}{2}\bar{\mu }_{0 j}<\left| x\right|< \bar{\mu }_{0 j}\right\} }+ \bar{\mu }_{0,j+1}^{-1} \mathbf {1}_{\left\{ \frac{1}{2} \bar{\mu }_{0,j+1}<\left| x\right|< \bar{\mu }_{0,j+1}\right\} }.\\ \left| \Delta _{x} \chi _{j}\right| \lesssim&\bar{\mu }_{0j}^{-2} \mathbf {1}_{\left\{ \frac{1}{2}\bar{\mu }_{0 j}<\left| x\right|< \bar{\mu }_{0 j}\right\} }+ \bar{\mu }_{0,j+1}^{-2} \mathbf {1}_{\left\{ \frac{1}{2} \bar{\mu }_{0,j+1}<\left| x\right| < \bar{\mu }_{0,j+1}\right\} }. \end{aligned}$$

By (2.22), one has $|\varphi _{0j}|\lesssim |t|^{\gamma _j+\sigma }\langle y_j\rangle ^{-2}$, $|\nabla _x\varphi _{0j}|\lesssim \lambda _{0j}^{\frac{n-2}{2}}\mu _j^{-\frac{n}{2}}\langle y_j\rangle ^{-3} \approx \mu _j^{-1}|t|^{\gamma _j+\sigma }\langle y_j\rangle ^{-3}$, then

$$\begin{aligned} |\nabla _x\varphi _{0j}\cdot \nabla _x \chi _j|\lesssim&|t|^\sigma \frac{\mu _j}{\bar{\mu }_j}w_{1j} +|t|^{\gamma _j-\gamma _{j+1}+\sigma } \left( \frac{\mu _{j+1}}{\mu _{j}} \right) ^{\frac{1-\alpha }{2}} w_{1,j+1}\lesssim |t_0|^{-\epsilon }(w_{1j}+w_{1,j+1}).\\ |\varphi _{0j} \Delta _x \chi _j|\lesssim&|t|^\sigma \frac{\mu _j}{\bar{\mu }_j}w_{1j} + |t|^{ \frac{n-2}{2}(\alpha _{j-1} -\alpha _{j}) + \frac{\alpha }{2}(\alpha _{j+1} - \alpha _{j}) + \sigma } w_{1,j+1}\lesssim |t_0|^{-\epsilon }(w_{1j}+w_{1,j+1}). \end{aligned}$$

where we have used that $\frac{n-2}{2}(\alpha _{j-1} -\alpha _{j}) + \frac{\alpha }{2}(\alpha _{j+1} - \alpha _{j}) = (\frac{\alpha }{n-6}-1) \left( \frac{n-2}{n-6}\right) ^{j-1} \le \frac{\alpha }{n-6}-1$.

For $\partial _t(\varphi _{0j}\chi _j)$, we have

$$\begin{aligned} \begin{aligned}&|\partial _t(\varphi _{0j})\chi _j|+|\varphi _{0j}\partial _t \chi _j | \lesssim |t|^{\gamma _j+\sigma -1}\langle y_j\rangle ^{-2}\left( \chi _j + | \nabla _{x} \chi _j | \right) \\&\quad \lesssim |t|^{\sigma -1} \mu _{j}^{2-\frac{\alpha }{2}} \mu _{j-1}^{\frac{\alpha }{2}} w_{1j} \lesssim |t_0|^{-\epsilon } w_{1j} . \end{aligned} \end{aligned}$$

$\bullet$ Estimate of $\bar{E}_5$. It follows from (2.10), (A.10) and $p\in (1,2)$ that

$$\begin{aligned} \begin{aligned}&|N_{\bar{U}}[\varphi _0]| \lesssim |\varphi _0|^p \lesssim \sum _{j=2}^k |t|^{(\gamma _j+\sigma )p}\langle y_j\rangle ^{-2p} \chi _j^p \\&\quad \approx \sum _{j=2}^k |t|^{2(\alpha _{j-1} - \alpha _j ) +\sigma } \langle y_j \rangle ^{2+\alpha -2p} \mu _{j}^{-2}|t|^{\gamma _j} \langle y_j \rangle ^{-2-\alpha } \chi _j^p . \end{aligned} \end{aligned}$$

If $2p \ge 2+\alpha$, it is easy to see $|N_{\bar{U}}[\varphi _0]| \lesssim |t_0|^{-\epsilon } \sum _{j=2}^kw_{1j}.$

If $2p < 2+\alpha$,

$$\begin{aligned} |N_{\bar{U}}[\varphi _0]| \lesssim \sum _{j=2}^k |t|^{\frac{2-\alpha +2p}{2}(\alpha _{j-1} - \alpha _j ) +\sigma } \mu _{j}^{-2} |t|^{\gamma _j} \langle y_j \rangle ^{-2-\alpha } \chi _j^p \lesssim |t_0|^{-\epsilon } \sum _{j=2}^kw_{1j} . \end{aligned}$$

$\bullet$ Estimate of $\bar{E}_{11}$. Regrouping the terms in (2.13), one obtain

$$\begin{aligned} \begin{aligned} \bar{E}_{11} =&\sum _{j=2}^{k}p U_j^{p-1}\left[ \sum _{l\ne j}U_l -U_{j-1}(0)\right] \chi _{j}+\sum _{j=2}^{k}\left[ {\bar{U}}^p-\sum _{i=1}^k U_i^p-p U_j^{p-1}\sum _{l\ne j}U_l\right] \chi _{j}\\&+ \left( -\sum _{j=2}^{k}(1-\chi _{j}) \partial _{t} U_{j} \right) +\left[ {\bar{U}}^p-\sum _{j=1}^{k} U_j^p\right] \left( 1-\sum _{i=2}^{k} \chi _{i}\right) \\ :=&J_1+J_2+J_3+J_4. \end{aligned} \end{aligned}$$

(A.25)

Claim:

$$\begin{aligned} \bar{E}_{11}(x,t)\lesssim |t_0|^{-\epsilon }\left( \sum _{j=1}^k w_{1j}+\sum _{j=1}^{k-1}w_{2j}+w_3\right) . \end{aligned}$$

• (1) Estimate of $J_1$.

  $$\begin{aligned} J_1=\sum _{j=2}^{k}pU_j^{p-1}\left( \sum _{l\ne j,j-1}U_l\right) \chi _{j}+\sum _{j=2}^kpU_j^{p-1}(U_{j-1}-U_{j-1}(0))\chi _j. \end{aligned}$$

  We will bound each term in the above equation. Fix $j\ge 2$. If $i\le j-2$,

  $$\begin{aligned}|pU_j^{p-1}U_i\chi _j|\lesssim \mu _j^{-2}\langle y_j \rangle ^{-4}\mu _{j-2}^{-\frac{n-2}{2}}\chi _j\lesssim |t|^{-\epsilon }w_{1j}.\end{aligned}$$

  If $i\ge j+1$, by Lemma A.1

  $$\begin{aligned} \begin{aligned}&|p U_j^{p-1}U_i\chi _j|\lesssim U_{j}^p\frac{U_{j+1}}{U_j}\chi _j\\&\quad \lesssim \mu _{j}^{-\frac{n+2}{2}}\langle y_j\rangle ^{-n-2} \left( \lambda _{j+1}^{-\frac{n-2}{2}} \langle y_{j+1}\rangle ^{2-n}\mathbf {1}_{\{\bar{\mu }_{0,j+1}\le |x|\le \mu _{0j} \}}+\lambda _{j+1}^{\frac{n-2}{2}} \right) \chi _j\\&\quad \lesssim \mu _{j+1}^{\frac{n-2}{2}}\mu _{j}^{-2} |x|^{2-n}\mathbf {1}_{\{\bar{\mu }_{0,j+1}\le |x|\le \mu _{0j} \}}+(\frac{\lambda _{j+1}}{\lambda _j})^{\frac{n-2}{2}} |t|^{\sigma }w_{1j}\\&\quad \lesssim |t|^{-\epsilon }(w_{2j}+w_{1j}) \end{aligned} \end{aligned}$$

  (A.26)

  when we choose $\sigma$ small first and then chose $\epsilon$ small enough. Using $|U_{j-1}-U_{j-1}(0)|\chi _j\lesssim \mu _{j-1}^{-\frac{n-2}{2}}\lambda _j$, we have

  $$\begin{aligned} |p U_j^{p-1}(U_{j-1}-U_{j-1}(0))\chi _j|\lesssim \mu _{j}^{-2}\langle y_j\rangle ^{-4}\mu _{j-1}^{-\frac{n-2}{2}}\lambda _j \chi _j \lesssim |t|^{-\epsilon }w_{1j}. \end{aligned}$$

  (A.27)

• (2) Estimate of $J_2$. By Lemma (A.1), we have

  $$\begin{aligned} \left| {\bar{U}}^p-\sum _{i=1}^k U_{i}^p-p U_j^{p-1}\sum _{l\ne j}U_l\right| \chi _j \lesssim \left( U_{j-1}^p+U_{j+1}^p\right) \chi _j.\end{aligned}$$

  Therein,

  $$\begin{aligned} \begin{aligned} U_{j-1}^p\chi _j&\approx |t|^{\frac{n+2}{2}\alpha _{j-1}} \chi _j \lesssim |t|^{\frac{n+2}{2}\alpha _{j-1}} \mu _{j}^2 |t|^{-\gamma _j} (\frac{\bar{\mu }_j}{\mu _j})^{2+\alpha } w_{1j} \chi _j \\&\approx (\frac{\mu _j}{\mu _{j-1}})^{\frac{2-\alpha }{2}} |t|^{\sigma } w_{1j} \chi _j \lesssim |t|^{-\epsilon }w_{1j}, \end{aligned} \end{aligned}$$

  and

  $$\begin{aligned}U_{j+1}^{p}\chi _j \approx \mu _{j+1}^{\frac{n+2}{2}}|x|^{-2-n}\chi _j\lesssim |t|^{2\sigma }\mu _{j+1}^3\mu _j|x|^{-4}w_{2j} \chi _{j}\lesssim |t|^{-\epsilon }w_{2j}, \end{aligned}$$

  when we take $\sigma$ small first and then take $\epsilon$ small enough.

• (3) Estimate of $J_3$. For $j = 2, \dots , k$, notice that

  $$\begin{aligned} |\partial _t U_j|=|\dot{\mu }_j\mu _j^{-\frac{n}{2}}Z_{n+1}(y_j)| \lesssim \mu _j^{-2} \mu _{j-1}^{-\frac{n-2}{2}} \langle y_j\rangle ^{2-n}. \end{aligned}$$

  (A.28)

  The support of $1-\chi _j$ is contained in $\{|x|\le \bar{\mu }_{0,j+1}\}\cup \{\frac{1}{2}\bar{\mu }_{0j}\le |x| < \bar{\mu }_{0j}\} \cup \{\bar{\mu }_{0j}\le |x|\}$. In the first set, it is easy to see $1-\chi _j= \chi (2|x|/\bar{\mu }_{0,j+1})$, then

  $$\begin{aligned} | (1-\chi _j)\partial _t U_j|\lesssim (\frac{\mu _{j+1}}{\mu _j})^{\frac{2-\alpha }{2}} (\frac{\mu _j}{\mu _{j-1}})^{\frac{n-2}{2}} |t|^{\sigma } w_{1,j+1} \chi (2|x|/\bar{\mu }_{0,j+1}) \lesssim |t|^{-\epsilon }w_{1,j+1} . \end{aligned}$$

  In the second set,

  $$\begin{aligned} | \partial _t U_j|\mathbf {1}_{\{\frac{1}{2}\bar{\mu }_{0j}\le |x|< \bar{\mu }_{0j}\}} \lesssim \left( \frac{\mu _{j}}{\mu _{j-1}} \right) ^{\frac{n-4- \alpha }{2}} |t|^{\sigma } w_{1j} \mathbf {1}_{\{\frac{1}{2}\bar{\mu }_{0j}\le |x| < \bar{\mu }_{0j}\}} \lesssim |t_0|^{-\epsilon }w_{1j} . \end{aligned}$$

  In the third set, we split it further to be $\{\bar{\mu }_{0 j}\le |x|\}=\cup _{m=2}^j\{\bar{\mu }_{0m}\le |x|\le \bar{\mu }_{0,m-1}\}\cup \{\bar{\mu }_{01}\le |x|\}$.

  Since $|y_j|$ is very large in the third set, (A.28) implies $|\partial _tU_j|\lesssim$ $\mu _j^{n-4}\mu _{j-1}^{-\frac{n-2}{2}}|x|^{2-n}$. Note that $\mu _j^{n-4}\mu _{j-1}^{-\frac{n-2}{2}}$ decreases about j up to some constant multiplicity. Then in $\{\bar{\mu }_{0 m}\le |x|\le \bar{\mu }_{0,m-1}\}$, $m = 2\dots , j$,

  $$|\partial _t U_j|\lesssim \mu _m^{n-4}\mu _{m-1}^{-\frac{n-2}{2}} |x|^{2-n} \lesssim |t|^{-\epsilon }w_{2,m-1}.$$

  In $\{\bar{\mu }_{01}\le |x|\}$, we have

  $$\begin{aligned}|\partial _t U_j|\lesssim \mu _2^{n-4} |x|^{2-n} \lesssim |t|^{-\epsilon }w_{3} .\end{aligned}$$

• (4) Estimate of $J_4$. Recall the definition of $\chi _i$ in (2.4), we have the support of $J_4$ is contained in the set $\cup _{m=3}^{k}\{\frac{1}{2}\bar{\mu }_{0m}\le |x|\le \bar{\mu }_{0m}\} \cup \{\frac{1}{2} \bar{\mu }_{02} \le |x| \}$.

  In $\{\frac{1}{2}\bar{\mu }_{0m}\le |x|\le \bar{\mu }_{0m}\}$, for $m=3, \dots , k$, by Lemma A.1, one has $U_m\approx U_{m-1}\approx \mu _{m-1}^{-\frac{n-2}{2}} \gg U_i$ for $i\ne m, m-1$. Therefore

  $$\begin{aligned}|J_4| \mathbf {1}_{\{\frac{1}{2}\bar{\mu }_{0m}\le |x|\le \bar{\mu }_{0m}\}} \lesssim \mu _{m-1}^{-\frac{n+2}{2}} \mathbf {1}_{\{\frac{1}{2}\bar{\mu }_{0m}\le |x|\le \bar{\mu }_{0m}\}} \lesssim (\frac{\mu _{m-1}}{\mu _m})^{\frac{\alpha -2}{2}} |t|^{\sigma } w_{1m} \lesssim |t_0|^{-\epsilon }w_{1m}.\end{aligned}$$

  In $\{ \frac{1}{2} \bar{\mu }_{02}\le |x|\}$, by Lemma A.1, when $\sigma <(n-6)^{-1}$,

  $$\begin{aligned} \begin{aligned}&|J_4| \lesssim U_1^{p-1}U_2\\&\quad \approx \mu _2^\frac{n-2}{2} \left( |x|^{2-n} \mathbf {1}_{\{ \frac{1}{2} \bar{\mu }_{02}\le |x| \le \bar{\mu }_{02} \}} + |x|^{2-n} \mathbf {1}_{\{ \bar{\mu }_{02}\le |x| \le 1 \}}\right. \\&\left. \quad + |x|^{-2-n} \mathbf {1}_{\{1 \le |x| \le \bar{\mu }_{01}\} } + |x|^{-2-n} \mathbf {1}_{ \{ \bar{\mu }_{01} \le |x| \}} \right) \\&\quad \lesssim |t_{0}|^{-\epsilon } \left( w_{12} + w_{21} + w_{11} +w_{3} \right) . \end{aligned} \end{aligned}$$

$\square$

Lemma A.5

There exist $\sigma >0$ small enough and $t_0$ negative enough, such that for $t<t_0$,

• (1) In $\{ |x|\le \bar{\mu }_{0i} \}$, $i = 1,\dots , k$, we have $w_{1j}^* \lesssim w_{1i}^*$ for $j = 1, \dots , i-1$ (it is vacuum if $i=1$)

• (2) In $\{ \bar{\mu }_{0i} \le |x| \}$, $i = 1,\dots , k$, we have $w_{1j}^* \lesssim w_{1i}^*$ for $j = i+1, \dots , k$ (it is vacuum if $i=k$).

• (3) In $\{|x|\le \bar{\mu }_{0j} \}$, $w_3^* \lesssim w_{1j}^*$ for $j = 1, \dots , k$. In $\{ |x|\ge |t|^{\frac{1}{2}} \}$, $w_3^* > rsim w_{1j}^*$ for $j = 1, \dots , k$ when $\delta \le (n-2-\alpha )^{-1}$.

Consequently,

$$\begin{aligned} \sum _{j=1}^kw_{1j}^*+w_{3}^*\lesssim {\left\{ \begin{array}{ll} w_{1k}^{*} &{} \text {if } |x| \le \bar{\mu }_{0k}, \\ w_{1i}^*+w_{1,i+1}^*&{} \text {if }\bar{\mu }_{0,i+1}\le |x|\le \bar{\mu }_{0i}, i=1,\cdots ,k-1,\\ w_{11}^{*} + w_{3}^*&{}\text {if }\bar{\mu }_{01}\le |x| \le |t|^{\frac{1}{2}} , \\ w_{3}^*&{}\text {if } |x| \ge |t|^{\frac{1}{2}} . \end{array}\right. } \end{aligned}$$

(A.29)

Proof

(1) For $j = 1, \dots , i-1$, in $\{|x|\le \bar{\mu }_{0i} \}$, we have $w^*_{1j}(x,t)= |t|^{\gamma _j}\le |t|^{\gamma _{i-1}}$ and $w^*_{1i}\ge |t|^{\gamma _i}\mu _i^\alpha \bar{\mu }_i^{-\alpha }$. It is easy to verify $|t|^{\gamma _{i-1}}\lesssim |t|^{\gamma _i}\mu _i^\alpha \bar{\mu }_i^{-\alpha }$ if $\alpha <n-6$.

(2) For $j = i+1, \dots , k$, in $\{ \bar{\mu }_{0i}\le |x|\le |t|^{\frac{1}{2}} \}$, $w_{1j}^*(x,t)= |t|^{\gamma _j}\mu _j^\alpha \bar{\mu }_j^{n-2-\alpha }|x|^{2-n}\approx |t|^{\gamma _j^*}|x|^{2-n}\le |t|^{\gamma _i^*}|x|^{2-n}\approx w_{1i}^*(x,t)$, because $\gamma _j^*$ is strictly decreasing on j, i.e.

$$\begin{aligned} \gamma _{1}^*> \gamma _{2}^*> \cdots > \gamma _k^*.\end{aligned}$$

In $\{|x|\ge |t|^{\frac{1}{2}} \}$, we have $w_{1j}^* \lesssim w_{1i}^*$ by the same reason.

(3) Due to (1), we only need to check $w_3^* \lesssim w_{11}^*$ in $\{|x|\le \bar{\mu }_{01} \}$. It is straightforward to have $R|t|^{-1-\sigma +\delta (4-n)} \lesssim |t|^{-1-\sigma }\langle x \rangle ^{-\alpha }$ in $\{|x|\le \bar{\mu }_{01} \}$. Due to (2), in $\{|x|\ge |t|^{\frac{1}{2}} \}$, we only need to check $w_3^* > rsim w_{11}^*$, which is easy to get when $\delta \le (n-2-\alpha )^{-1}$. $\square$

Lemma A.6

There exists $t_0$ negative enough such that

• (1) In $\{\bar{\mu }_{0,i+1}\le |x| \}$, we have $w^*_{2j}\lesssim w^*_{2i}$ for $j=i+1, \dots , k-1$ (it is vacuum if $i=k, k-1$). In $\{|x|\le \bar{\mu }_{0i} \}$, we have $w_{2,i-1}^* > rsim w_{2j}^*$ for $j= 1, \dots , i-2$ (it is vacuum if $i =1,2$).

• (2) In $\{|x|\ge \bar{\mu }_{01} \}$, $w_{2j}^*\lesssim w_3^*$ for $j = 1, \dots , k-1$.

Consequently

$$\begin{aligned} \sum _{j=1}^{k-1}w_{2j}^*\lesssim {\left\{ \begin{array}{ll} w_{2,k-1}^{*} &{}\text {if } |x|\le \bar{\mu }_{0k}, \\ w_{2i}^*+w_{2,i-1}^* &{}\text {if }\bar{\mu }_{0,i+1} \le |x|\le \bar{\mu }_{0i}, \text{ for } i=2, \dots , k-1, \\ w_{21}^{*} &{}\text {if }\bar{\mu }_{02}\le |x|\le \bar{\mu }_{01}, \\ w_3^*&{}\text {if }|x|\ge \bar{\mu }_{01}. \end{array}\right. } \end{aligned}$$

(A.30)

Proof

(1) In $\{\bar{\mu }_{0,i+1} \le |x|\le \bar{\mu }_{0i} \}$, it follows from (4.26) that $w_{2j}^*= |t|^{-2\sigma }\mu _{0,j+1}^{\frac{n}{2}-2}\mu _{0,j-1}|x|^{2-n}$ for $j=i+1, \dots , k-1$ and $w_{2i}^*=|t|^{-2\sigma }\mu _{0,i+1}^{\frac{n}{2}-2}\mu _{0i}^{-1}|x|^{4-n}$.

$$\begin{aligned} \frac{w_{2i}^{*}}{w_{2j}^{*}} > rsim (\frac{\mu _{0,i+1}}{\mu _{0,i+2}})^{\frac{n}{2} -2} \frac{\mu _{0,i+1}}{\mu _{0i}} \approx |t|^{(\frac{n}{2} -2)(\alpha _{i+2}- \alpha _{i+1}) - (\alpha _{i+1} -\alpha _{i})} > rsim 1 \end{aligned}$$

where we have used that $(\frac{n}{2} -2)(\alpha _{i+2}- \alpha _{i+1}) - (\alpha _{i+1} -\alpha _{i})= \frac{(n-4)^2 +4}{(n-6)^2} (\frac{n-2}{n-6})^{i-1}$. Thus $w_{2j}^*\lesssim w_{2i}^*$. It is easy to see that $w_{2j}^*\lesssim w_{2i}^*$ also holds in $\{|x|>\bar{\mu }_{0i} \}$. This deduces the first part. The second parts holds obviously by (4.26).

(2) In $\{\bar{\mu }_{01}\le |x| \le |t|^{\frac{1}{2}}\}$, by the definition (4.26) and (4.27), it is easy to see that $w_{2j}^{*}\lesssim w_{3}^{*}$ for $j=1,\dots , k-1$ since $w_{2j^*}$ have more time decay than $w_3^*$. In $\{|x| \ge |t|^{\frac{1}{2}} \}$, we have $w_{2j}^{*}\lesssim w_{3}^{*}$ by the similar reason. $\square$

Remark A.1

Lemma A.5 and A.6 help us consider much less terms in the topology of the outer problem in some special domains.

Lemma A.7

There exist $\sigma , \epsilon >0$ small and $t_0$ negative enough such that

$$\begin{aligned} \Vert {\mathcal T} ^{out}[V\Psi ]\Vert _{\alpha ,\sigma }^{out,*} \lesssim R^{-1} \Vert \Psi \Vert _{\alpha ,\sigma }^{out,*}. \end{aligned}$$

(A.31)

Proof

Without loss of generality, we assume $\Vert \Psi \Vert _{\alpha ,\sigma }^{out,*}=1$. By (3.9), we rewrite V as

$$\begin{aligned} V=pu_*^{p-1}(1-\sum _{j=1}^k\zeta _j)+\sum _{j=1}^k\zeta _jp(u_*^{p-1}-U_j^{p-1}) . \end{aligned}$$

(A.32)

We shall handle terms respectively.

Consider the first term in (A.32). Using (3.3), the support of $1-\sum _{j=1}^k\zeta _j$ is $\cup _{i=2}^{k}\{R\mu _{0i}\le |x|\le 2R^{-1}\mu _{0,i-1}\} \cup \{R\mu _{01}\le |x|\}$.

$\bullet$ In $\{R\mu _{01}\le |x|\}$, we have $p u_*^{p-1}\lesssim \mu _1^2|x|^{-4}\le R^{-1}|x|^{-3}$ by Lemma A.1 and A.2. Split the region into $\{R\mu _{01}\le |x|\le \bar{\mu }_{01}\}\cup \{\bar{\mu }_{01}\le |x| \le |t|^{\frac{1}{2}}\} \cup \{|x|\ge |t|^{\frac{1}{2}}\}$. In the first set, one has $|\Psi |\lesssim w_{11}^*+w_{12}^*+w_{21}^*$ by (A.29) and (A.30). Notice $w_{12}^* + w_{21}^* \lesssim w_{11}^*$ in $\{R\mu _{01}\le |x| \le \bar{\mu }_{01}\}$. Therefore

$$\begin{aligned} \big | p u_{*}^{p-1} (1-\sum _{j=1}^{k} \zeta _{j}) \Psi \big | \lesssim R^{-1} |x|^{-3} w_{11}^* \lesssim R^{-1} w_{11} . \end{aligned}$$

(A.33)

In the second set, by (A.29) and (A.30), one has $|\Psi |\lesssim w_{11}^{*} + w_3^*$. Then

$$\begin{aligned} \big | pu_{*}^{p-1} (1-\sum _{j=1}^{k} \zeta _{j}) \Psi \big | \lesssim \ R^{-1} |x|^{-3} (w_{11}^{*} + w_3^* ) \lesssim \ R^{-1} (w_{11} + w_3 ) . \end{aligned}$$

(A.34)

In the third set, similarly we have

$$\begin{aligned} \big | pu_{*}^{p-1} (1-\sum _{j=1}^{k} \zeta _{j}) \Psi \big | \lesssim R^{-1} |x|^{-3} w_3^* \lesssim \ R^{-1} w_3 . \end{aligned}$$

(A.35)

$\bullet$ Consider the region $\{R\mu _{0i}\le |x| \le 2R^{-1}\mu _{0,i-1}\}$, $i = 2, \dots , k$. We divide it further into two parts, $\{R\mu _{0i}\le |x|\le \bar{\mu }_{0i}\}\cup \{\bar{\mu }_{0i}\le |x|\le 2R^{-1}\mu _{0,i-1}\}$. In $\{R\mu _{0i}\le |x| \le \bar{\mu }_{0i}\}$, we have $p u_*^{p-1}\lesssim \mu _i^{2}|x|^{-4}$ by Lemma A.1 and A.2. Moreover, one has $|\Psi |\le w_{1i}^*+w_{1,i+1}^*+w_{2,i-1}^*+w_{2i}^*$ by Lemma A.5 and A.6. One readily has $w_{1,i+1}^* \lesssim w_{1i}^*$ in $\{ R\mu _{0i}\le |x| \le \bar{\mu }_{0i} \}$. Thus

$$\begin{aligned} \begin{aligned} \big | p u_{*}^{p-1} (1-\sum _{j=1}^{k} \zeta _{j}) \Psi \big | \lesssim&\, \mu _{i}^{2} |x|^{-4 } \left( w_{1i}^* + w_{1,i+1}^* + w_{2,i-1}^* + w_{2i}^* \right) \\ \lesssim&\, \mu _{i}^{2} |x|^{-4 } \left( w_{1i}^* + w_{2,i-1}^* + w_{2i}^* \right) \\ \lesssim&\, R^{-1} \left( w_{1i} + w_{2i} \right) , \end{aligned} \end{aligned}$$

(A.36)

where we have used the fact that in $\{R\mu _{0i}\le |x|\le \bar{\mu }_{0i} \}$,

$$\begin{aligned}&\mu _i^{2}|x|^{-4}w_{1i}^*\lesssim |t|^{\gamma _i}\mu _i^{2+\alpha }|x|^{-4-\alpha } \lesssim R^{-2}w_{1i}, \\&\mu _i^{2}|x|^{-4}w_{2,i-1}^* \lesssim \mu _i^{2}|x|^{-4}|t|^{-2\sigma }\mu _{i-1}^{1-\frac{n}{2}} \lesssim R^{\alpha -2} |t|^{-\sigma } w_{1i}, \\&\mu _i^{2}|x|^{-4}w_{2,i}^* \lesssim \mu _i^{2}|x|^{-4}|t|^{-2\sigma }\mu _{i+1}^{\frac{n}{2}-2}\mu _{i}^{-1} |x|^{4-n} \lesssim R^{-2}w_{2i}. \end{aligned}$$

In the other part $\{\bar{\mu }_{0i} \le |x| \le 2R^{-1}\mu _{0,i-1}\}$, we have $p u_*^{p-1}\lesssim U_{i-1}^{p-1}\lesssim \mu _{i-1}^{-2}$ and $w_{2,i-2}^*\lesssim w_{1,i-1}^*$ (which is vacuum if $i=2$). Then

$$\begin{aligned} \begin{aligned} \big | p u_*^{p-1} (1-\sum _{j=1}^{k} \zeta _{j}) \Psi \big | \lesssim&\, \mu _{i-1}^{-2} \left( w_{1,i-1}^* + w_{1i}^* + w_{2,i-2}^* + w_{2,i-1}^* \right) \\ \lesssim&\, \mu _{i-1}^{-2} \left( w_{1,i-1}^* + w_{1i}^* + w_{2,i-1}^* \right) \\ \lesssim&\, \left( \mu _{i-1}^{-2} |t|^{\gamma _{i-1}} \mathbf {1}_{\{ |x| \le 2R^{-1}\mu _{i-1} \}} + R^{-2}w_{2,i-1} \right) , \end{aligned} \end{aligned}$$

(A.37)

where we have used the fact that in $\{\bar{\mu }_{0i}\le |x|\le 2R^{-1}\mu _{0,i-1}\}$,

$$\begin{aligned}&\mu _{i-1}^{-2}w_{2,i-1}^*\lesssim \mu _{i-1}^{-2}|t|^{-2\sigma }\mu _i^{\frac{n}{2}-2}\mu _{i-1}^{-1}|x|^{4-n}\lesssim R^{-2} |t|^{-2\sigma }\mu _i^{\frac{n}{2}-2}\mu _{i-1}^{-1}|x|^{2-n}\lesssim R^{-2}w_{2,i-1} , \\&\mu _{i-1}^{-2}w_{1i}^*\le \mu _{i-1}^{-2}|t|^{\gamma _i}\mu _i^{\alpha }\bar{\mu }_i^{n-2-\alpha }|x|^{2-n} \lesssim |t|^{\gamma _i+2\sigma }\mu _i^{\frac{\alpha }{2}+1}\mu _{i-1}^{\frac{n-4-\alpha }{2}}w_{2,i-1}\lesssim |t|^{-\epsilon }w_{2,i-1} . \end{aligned}$$

By Lemma B.2,

$$\begin{aligned} \begin{aligned}&{\mathcal T} ^{out} [ \mu _{i-1}^{-2} |t|^{\gamma _{i-1} } \mathbf {1}_{\{|x|\le 2R^{-1}\mu _{0,i-1}\}} ]\\ \lesssim&\, {\left\{ \begin{array}{ll} |t|^{\gamma _{i-1}} R^{-2} &{} \text{ if } |x|\le R^{-1} \mu _{0,i-1}, \\ \mu _{i-1}^{-2} |t|^{\gamma _{i-1}} (R^{-1} \mu _{i-1})^n |x|^{2-n} &{} \text{ if } R^{-1} \mu _{0,i-1} \le |x| \le |t|^{\frac{1}{2}}, \\ R^{-n} (|x|^2)^{(2-n)\alpha _{i-1} +\gamma _{i-1}} |x|^{2-n} &{} \text{ if } |x| \ge |t|^{\frac{1}{2}} . \end{array}\right. }\\ \lesssim&\, R^{-1} w_{1,i-1}^*. \end{aligned} \end{aligned}$$

(A.38)

Next we consider the second term in (A.32). Recall the support of $\zeta _j$ (3.4) is contained in $\{ R^{-1} \mu _{0j} \le |x| \le 2R \mu _{0j}\}$, which are mutually disjoint.

$\bullet$ For $j=1$, we have $|\zeta _1(u_*^{p-1}-U_1^{p-1})|\lesssim U_2^{p-1}\zeta _{1} \lesssim \mu _2^{2}|x|^{-4}\mathbf {1}_{\{ R^{-1}\mu _{01} \le |x| \le 2 R\mu _{01}\}}$ since $\varphi _{0} =0$ in the support of $\zeta _{1}$. Then

$$\begin{aligned} \left| \zeta _{1} ( u_*^{p-1} - U_j^{p-1} ) \Psi \right| \lesssim \mu _2^2 |x|^{-4} \mathbf {1}_{\{ R^{-1}\mu _{01} \le |x| \le 2 R\mu _{01} \}} \left( w_{11}^* + w_{12}^* + w_{21}^* \right) \lesssim |t|^{-\epsilon } w_{11} , \end{aligned}$$

(A.39)

where we have used the fact that in $\{R^{-1}\mu _{01}\le |x|\le 2R\mu _{01}\}$,

$$\begin{aligned}&\mu _2^2|x|^{-4}w_{11}^*\lesssim \mu _2^2|x|^{-4}|t|^{\gamma _1}(1+|x|)^{-\alpha }\le R^{4}\mu _2^2 w_{11}\le R^4|t|^{-2\epsilon } w_{11},\\&\mu _2^2|x|^{-4}w_{21}^*\lesssim \mu _2^2|x|^{-4} |t|^{-2\sigma }\mu _2^{\frac{n}{2}-2}|x|^{2-n}\lesssim R^{n+2}|t|^{-2\epsilon }w_{11},\\&\mu _2^2|x|^{-4}w_{12}^*\lesssim \mu _2^2|x|^{-4}|t|^{\gamma _2}\mu _2^\alpha \bar{\mu }_2^{n-2-\alpha }|x|^{2-n}\lesssim R^{n+2}|t|^{-2\epsilon }w_{11}. \end{aligned}$$

$\bullet$ For $j =2,\dots , k$, we have $|\zeta _j(u_*^{p-1}-U_j^{p-1})|\lesssim (U_{j-1}^{p-1}+U_{j+1}^{p-1} + \varphi _{0j}^{p-1})|\zeta _j|\lesssim \mu _{j-1}^{-2} |\zeta _j|$ where we have used the fact that $\varphi _{0}= \varphi _{0j} \chi _{j}$ in $\{ R^{-1} \mu _{0j} \le |x| \le 2R \mu _{0j}\}$ and $U_{j+1}$ is vacuum if $j=k$. Then

$$\begin{aligned} \left| \zeta _{j} ( u_{*}^{p-1} - U_j^{p-1} ) \Psi \right| \lesssim \mu _{j-1}^{-2} \left( w_{1j}^* + w_{1,j+1}^* + w_{2,j-1}^* + w_{2j}^* \right) |\zeta _j| \lesssim |t|^{-\epsilon } \left( w_{1j} + w_{2j} \right) , \end{aligned}$$

(A.40)

where we have used the fact that in $\{R^{-1}\mu _{0j}\le |x|\le 2R\mu _{0j}\}$, $w_{2,j-1}^* \lesssim w_{1j}^*$ and

$$\begin{aligned}&\mu _{j-1}^{-2}w_{1j}^*\lesssim \mu _{j-1}^{-2}|t|^{\gamma _j}\lesssim \lambda _j^{2}\langle y_j\rangle ^{2+\alpha }w_{1j}\lesssim R^{2+\alpha }|t|^{-2 \epsilon }w_{1j},\\&\mu _{j-1}^{-2}w_{1,j+1}^*\lesssim \mu _{j-1}^{-2}|t|^{\gamma _{j+1}}\mu _{j+1}^\alpha \bar{\mu }_{j+1}^{n-2-\alpha }|x|^{2-n}\lesssim \mu _{j-1}^{-2}\mu _j^{1-\frac{\alpha }{2}}\mu _{j+1}^{1+\frac{\alpha }{2}}|t|^{\sigma }w_{2j}\lesssim |t|^{-\epsilon }w_{2j},\\&\mu _{j-1}^{-2}w_{2j}^*\lesssim \mu _{j-1}^{-2}|t|^{-2\sigma }\mu _{j+1}^{\frac{n}{2}-2}\mu _j^{-1}|x|^{4-n}\lesssim \mu _{j-1}^{-2}(R\mu _i)^2w_{2j}\lesssim R^2|t|^{-2\epsilon }w_{2j}. \end{aligned}$$

Combining the above calculations of the two terms in (A.32), we get the conclusion. $\square$

Lemma A.8

There exist $\sigma ,\epsilon >0$ small and $t_0$ negative enough such that

$$\begin{aligned}\Vert \mathcal {N}[\vec {\phi }, \Psi , \vec {\mu }_{1}]\Vert ^{out}_{\alpha ,\sigma }\lesssim |t_0|^{-\epsilon }\left( \Vert \vec {\phi }\Vert ^{in,*}_{a,\sigma }+\Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma }\right) ^p.\end{aligned}$$

Proof

By (3.9) and some elementary inequality

$$\begin{aligned} |\mathcal {N}[\vec {\phi },\Psi ,\vec {\mu }_{1}]|\lesssim \sum _{j=1}^k \mu _j^{-\frac{n+2}{2}}|\phi _j|^p\eta _j+|\Psi |^p. \end{aligned}$$

(A.41)

For the first part on the RHS, recalling (A.12), we obtain

$$\begin{aligned} \begin{aligned} \mu _j^{-\frac{n+2}{2}}|\phi _j|^p\eta _j \lesssim&\, |t|^{\gamma _jp} \frac{R^{(n+1-a)p}}{\langle y_j\rangle ^{(n+1)p}} \mathbf {1}_{\{|x|\le 4R\mu _{0j} \}}\left( \Vert \phi _j\Vert ^{in,*}_{j,a,\sigma } \right) ^p\\&\quad \lesssim R^{(n+1-a)p}|t|^{-2\epsilon } w_{1j} \left( \Vert \phi _j\Vert ^{in,*}_{j,a,\sigma } \right) ^p . \end{aligned} \end{aligned}$$

(A.42)

For the second part on the RHS of (A.41),

$\bullet$ In $\{|x|\ge \bar{\mu }_{01} \}$, by (A.29), we have $|\Psi |\lesssim (w_{11}^{*} + w_3^{*}) \Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma }$ in $\{ \bar{\mu }_{01} \le |x| \le |t|^{\frac{1}{2}} \}$ and $|\Psi |\lesssim w_3^{*} \Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma }$ in $\{ |x| > |t|^{\frac{1}{2}}\}$. Notice

$$\begin{aligned} \begin{aligned} (w_3^*)^p =&\, R^{p-1}|t|^{1+(1-p)\sigma } |x|^{-4} w_3\lesssim |t|^{-1}w_3 \text{ if } |x| > |t|^{\frac{1}{2}} , \\ (w_3^*)^p =&\, R^{p-1} |t|^{-(1+\sigma )(p-1)} |x|^{-\frac{2n-12}{n-2}} w_{3} \lesssim |t|^{-\epsilon } w_{3} \text{ if } \bar{\mu }_{01} \le |x| \le |t|^{\frac{1}{2}} , \\ (w_{11}^*)^p =&\, R^{-1} |t|^{-\frac{4}{n-2}(1+\sigma ) +\delta (n-2-p\alpha )} w_{3} \lesssim |t|^{-\epsilon } w_{3} \text{ if } \bar{\mu }_{01} \le |x| \le |t|^{\frac{1}{2}}, \end{aligned} \end{aligned}$$

where we take $\delta \le 2(n-2)^{-2}$ in the last formula. Thus

$$\begin{aligned} |\Psi |^p \lesssim \ |t|^{-\epsilon } w_3 \left( \Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma } \right) ^p . \end{aligned}$$

(A.43)

$\bullet$ In $\{1\le |x|\le \bar{\mu }_{01}\}$, we have $|\Psi |\lesssim (w_{11}^*+w_{12}^*+w_{21}^*)\Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma }\lesssim w_{11}^*\Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma }$ since $\left( w_{12}^* + w_{21}^* \right) \mathbf {1}_{\{1\le |x|\le \bar{\mu }_{01}\}} \lesssim w_{11}^*$. Therefore

$$\begin{aligned} |\Psi |^p \lesssim \ \left( w_{11}^*\right) ^p \left( \Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma } \right) ^p \lesssim |t|^{-(1+\sigma )\frac{4}{n-2} +2\delta } w_{11} \left( \Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma } \right) ^p \lesssim |t|^{-\epsilon } w_{11} \left( \Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma } \right) ^p , \end{aligned}$$

(A.44)

for $\delta \le (n-2)^{-1}$.

$\bullet$ In $\{\bar{\mu }_{02}\le |x|\le 1\}$, we have

$$\begin{aligned} \begin{aligned}&(w_{11}^*)^p\lesssim |t|^{-(1+\sigma )p}\lesssim |t|^{-\epsilon }w_{11}, \\&(w_{12}^*)^p\lesssim |t|^{\gamma _2p}\bar{\mu }_2^{n+2}|x|^{-n-2}\lesssim |t|^{\gamma _2p}\bar{\mu }_2^{n-2}|x|^{2-n}\lesssim |t|^{-\epsilon }w_{21}, \\&(w_{21}^*)^p\lesssim |t|^{-2\sigma p}\mu _2^{(\frac{n}{2}-2)p}|x|^{(4-n)p}\lesssim |t|^{-2\sigma p}\mu _2^{\frac{n}{2}-1}|x|^{2-n}\lesssim |t|^{-\epsilon } w_{21}, \end{aligned} \end{aligned}$$

where we have used $n-2\le (n-4)p$ when $n\ge 6$ in the last inequality. Then

$$\begin{aligned} |\Psi |^p\lesssim [(w_{11}^*)^p + (w_{12}^*)^p+(w_{21}^*)^p](\Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma })^p\lesssim |t|^{-\epsilon }(w_{11}+w_{21})(\Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma })^p. \end{aligned}$$

(A.45)

$\bullet$ In $\{\bar{\mu }_{0,i+1}\le |x|\le \bar{\mu }_{0i}\}$, $i=2,\dots , k$, we have

$$\begin{aligned} (w_{1i}^*)^p\lesssim&\, |t|^{\gamma _ip}\langle y_i\rangle ^{-\alpha p} \lesssim \mu _{i}^{2} |t|^{\gamma _{i}(p-1)} \langle y_i \rangle ^{2+\alpha -\alpha p} w_{1i} \lesssim |t|^{-\sigma (p-1)} \mu _{i}\mu _{i-1}^{-1} w_{1i} \lesssim |t|^{-\epsilon }w_{1i} \end{aligned}$$

provided $\epsilon <\frac{2}{n-6}$.

$$\begin{aligned} (w_{1,i+1}^*)^p\lesssim&\ |t|^{p \gamma _{i+1}}\bar{\mu }_{i+1}^{n+2}|x|^{-n-2}\lesssim \mu _{i}^{-\frac{n+2}{2}}|t|^{-\sigma p}\bar{\mu }_{i+1}^{n-2}|x|^{2-n}\\ \approx&\ \lambda _{i+1}\mu _{i+1}^{\frac{n}{2}-2}\mu ^{-1}_i|t|^{-\sigma p}|x|^{2-n}\lesssim |t|^{-\epsilon }w_{2i} \end{aligned}$$

provided $\epsilon <-2\sigma +\sigma p+\frac{2}{n-6}$.

$$\begin{aligned} (w_{2,i-1}^*)^p\lesssim&\ |t|^{-2\sigma p}\mu _{i-1}^{-\frac{n+2}{2}}\lesssim |t|^{-2\sigma p}\mu _{i-1}^{-\frac{n+2}{2}}(\frac{\bar{\mu }_i}{\mu _i})^{2+\alpha }\langle y_i\rangle ^{-2-\alpha }\end{aligned}$$

(A.46)

$$\begin{aligned} \lesssim&\ |t|^{-2\sigma p}(\frac{\mu _i}{\mu _{i-1}})^{1-\frac{\alpha }{2}}\mu _{i}^{-2}\mu _{i-1}^{1-\frac{n}{2}}\langle y_i\rangle ^{-2-\alpha }\lesssim |t|^{-\epsilon }w_{1i} \end{aligned}$$

(A.47)

provided $\epsilon <\frac{1}{n-6}$.

$$\begin{aligned} (w_{2i}^*)^p\lesssim&\ |t|^{-2\sigma p}\mu _{i+1}^{(\frac{n}{2}-2)p}\mu _i^{-p}|x|^{(4-n)p}\lesssim |t|^{-2\sigma p}\mu _{i+1}^{(\frac{n}{2}-2)p}\mu _i^{-p}|x|^{2-n}\bar{\mu }_{i+1}^{(4-n)p-(2-n)}\end{aligned}$$

(A.48)

$$\begin{aligned} \approx&\ \lambda _{j+1} |t|^{-2\sigma p}\mu _{i+1}^{\frac{n}{2}-2}\mu _i^{-1}|x|^{2-n}\lesssim |t|^{-\epsilon }w_{2i} \end{aligned}$$

(A.49)

provide $\epsilon <\frac{2}{n-6}$. Here we have used $(4-n)p-(2-n)\le 0$ when $n\ge 6$.

Therefore, for $i =2,\dots , k$,

$$\begin{aligned} |\Psi |^p \lesssim&\left( w_{1i}^* + w_{1,i+1}^* + w_{2,i-1}^* + w_{2i}^* \right) ^p \left( \Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma } \right) ^p \lesssim |t|^{-\epsilon } \left( w_{1i} + w_{2i} \right) \left( \Vert \Psi \Vert ^{out,*}_{\alpha ,\sigma } \right) ^p, \end{aligned}$$

where $w_{1,i+1}^*, w_{2i}^*$ are vacuum if $i=k$ and $w_{1,i+1}^*$ are vacuum if $i=k,k-1$. $\square$

Proof of Proposition 4.1

This is a combination of results in Lemma A.3, A.4, A.7, A.8 and Lemma 4.4, 4.54.6. $\square$

Appendix B. Some estimates for the outer problem

1.1 Basic estimates

Let G(x, t) denote the standard heat kernel on $\mathbb {R}^n$, that is

$$\begin{aligned} G(x,t)=\frac{1}{(4\pi t)^{n/2}}e^{-\frac{|x|^2}{4t}} . \end{aligned}$$

(B.1)

Recall the ${\mathcal T} ^{out}[g](x,t)$ defined by (4.19).

Lemma B.1

Suppose $n >2$, $a\ge 0$, $d_1\le d_2\le \frac{1}{2}$ and b satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{n}{2}-b+d_2(a-n)>1 &{} \text{ if } a<n, \\ \frac{n}{2}-b+d_1(a-n)>1 &{} \text{ if } a\ge n, \end{array}\right. } \end{aligned}$$

(B.2)

$0\le c_1,c_2\le c_{**}$. Then there exists C depending on $n,a,b,d_1,d_2,c_{**}$ such that for $t<-1$

$$\begin{aligned}{\mathcal T} ^{out}\left[ \frac{ |t|^{b} }{|x|^{a} } \mathbf {1}_{\{c_1|t|^{d_1}\le |x|\le c_2|t|^{d_2}\}}\right] (0,t)\le C{\left\{ \begin{array}{ll}|t|^{b}(c_1|t|^{d_1})^{2-a}&{}\text {if }a\in (2,\infty ), \\ |t|^{b}\ln (c_2|t|^{d_2}/(c_1|t|^{d_1})) &{}\text {if }a=2, \\ |t|^{b}(c_2|t|^{d_2})^{2-a}&{}\text {if }a\in [0,2).\end{array}\right. }\end{aligned}$$

Proof

Using (B.1), we obtain

$$\begin{aligned}&\int _{-\infty }^t\int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|y|^2}{4(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{c_1|s|^{d_1}\le |y|\le c_2|s|^{d_2}\}}dyds\nonumber \\&\quad \approx \int _{-\infty }^t\int _{c_1|s|^{d_1}}^{c_2|s|^{d_2}}\frac{|s|^{b}}{(t-s)^{n/2}}e^{-\frac{r^2}{4(t-s)}}r^{n-1-a}drds \approx \int _{-\infty }^t\int _{\frac{c_1^2|s|^{2d_1}}{4(t-s)}}^{\frac{c_2^2|s|^{2d_2}}{4(t-s)}}e^{-z}z^{\frac{n-a-2}{2}}\frac{|s|^{b}}{(t-s)^{a/2}}dzds\nonumber \\&\quad =\int _{-\infty }^t\frac{|s|^{b}}{(t-s)^{a/2}}F\left( \frac{c_1^2|s|^{2d_1}}{4(t-s)},\frac{c_2^2|s|^{2d_2}}{4(t-s)}\right) ds, \end{aligned}$$

(B.3)

where

$$\begin{aligned}F(A,B):=\int _A^Be^{-z}z^{\frac{n-a-2}{2}}dz. \end{aligned}$$

We shall split (B.3) into four integrals $J_1,J_2,J_3,J_4$ according to the regions of s. First, in the region $s\in [t-c_1^2|t|^{2d_1},t]$, one has $|s-t|\le c_1^2 |t|$ when $t<-1$. Therefore

$$\begin{aligned} \begin{aligned} J_1=&\int _{t-c_1^2|t|^{2d_1}}^t\frac{|s|^{b}}{(t-s)^{a/2}}F\left( \frac{c_1^2|s|^{2d_1}}{4(t-s)},\frac{c_2^2|s|^{2d_2}}{4(t-s)}\right) ds \\ \lesssim&\ |t|^{b}\int _{t-c_1^2|t|^{2d_1}}^t\frac{1}{(t-s)^{a/2}} e^{-\frac{c_1^2|t|^{2d_1}}{4(t-s)} (1+c_{**}^2)^{2(d_1)^{-}} }ds \\ \approx&\ c_1^{2-a}|t|^{b+d_1(2-a)}\int _{\frac{(1+c_{**}^2)^{2(d_1)^{-}}}{4}}^{\infty }e^{-\tilde{s}}\tilde{s}^{\frac{a}{2} -2}d\tilde{s}\lesssim c_1^{2-a}|t|^{b+d_1(2-a)}, \end{aligned} \end{aligned}$$

where $(d_1)^{-}=\min \{0,d_1\}$.

Second, in the region $s\in [t-c_2^2|t|^{2d_2},t-c_1^2|t|^{2d_1}]$,

$$\begin{aligned} \begin{aligned} J_2=&\int _{t-c_2^2|t|^{2d_2}}^{t-c_1^2|t|^{2d_1}}\frac{|s|^{b}}{(t-s)^{a/2}}F\left( \frac{c_1^2|s|^{2d_1}}{4(t-s)},\frac{c_2^2|s|^{2d_2}}{4(t-s)}\right) ds \\ \lesssim&\int _{t-c_2^2|t|^{2d_2}}^{t-c_1^2|t|^{2d_1}}\frac{|s|^{b}}{(t-s)^{a/2}} F\left( \frac{c_1^2|t|^{2d_1}(1+c_{**}^2)^{2(d_1)^{-}}}{4(t-s)},\frac{c_2^2|t|^{2d_2}(1+c_{**}^2)^{2(d_{2})^{+}}}{4(t-s)}\right) ds \\ \lesssim&\int _{t-c_2^2|t|^{2d_2}}^{t-c_1^2|t|^{2d_1}}\frac{|t|^{b}}{(t-s)^{a/2}} {\left\{ \begin{array}{ll} 1 &{} \text{ if } a<n, \\ \left| \ln \left( \frac{c_1^2|t|^{2d_1}(1+c_{**}^2)^{2(d_1)^{-}}}{4(t-s)} \right) \right| &{} \text{ if } a=n, \\ \left( \frac{c_1^2 |t|^{2d_1}}{t-s} \right) ^{\frac{n-a}{2}} &{} \text{ if } a>n, \end{array}\right. } ds\\ \lesssim&{\left\{ \begin{array}{ll}c_1^{2-a}|t|^{b+d_1(2-a)}&{}\text {if }2<a,\\ |t|^{b}\log (c_2/c_1|t|^{2(d_2-d_1)}) &{}\text {if }a=2,\\ c_2^{2-a}|t|^{b+d_2(2-a)}&{}\text {if }0\le a<2, \end{array}\right. } \end{aligned} \end{aligned}$$

(B.4)

where $(d_2)^{+} = \max \{0,d_2\}$.

Third, when $s\in [2t-c_2^2|t|^{2d_2},t-c_2^2|t|^{2d_2}]$,

$$\begin{aligned} \begin{aligned} J_3=&\int _{2t-c_2^2|t|^{2d_2}}^{t-c_2^2|t|^{2d_2}}\frac{|s|^{b}}{(t-s)^{a/2}}F\left( \frac{c_1^2|s|^{2d_1}}{4(t-s)},\frac{c_2^2|s|^{2d_2}}{4(t-s)}\right) ds \\ \lesssim&\int _{2t-c_2^2|t|^{2d_2}}^{t-c_2^2|t|^{2d_2}}\frac{|t|^{b}}{(t-s)^{a/2}}F\left( \frac{c_1^2|t|^{2d_1}(2+c_{**}^2)^{2(d_2)^{-}}}{4(t-s)},\frac{c_2^2|t|^{2d_2} (2+c_{**}^2)^{2(d_2)^{+}}}{4(t-s)}\right) ds \\ \lesssim&\int _{2t-c_2^2|t|^{2d_2}}^{t-c_2^2|t|^{2d_2}}\frac{|t|^{b}}{(t-s)^{a/2}} {\left\{ \begin{array}{ll} \left( \frac{c_2^2|t|^{2d_2}}{t-s}\right) ^{\frac{n-a}{2}} &{} \text{ if } a<n, \\ \ln \left( \frac{c_2^2|t|^{2d_2} (2+c_{**}^2)^{2(d_2)^{+}}}{c_1^2|t|^{2d_1}(2+c_{**}^2)^{2(d_2)^{-}}}\right) &{} \text{ if } a=n, \\ \left( \frac{c_1^2|t|^{2d_1}}{t-s}\right) ^{\frac{n-a}{2}} &{} \text{ if } a>n, \end{array}\right. } ds \\ \lesssim&{\left\{ \begin{array}{ll} c_2^{2-a}|t|^{b+d_2(2-a)} &{} \text{ if } a<n, \\ c_{2}^{2-n} |t|^{b+d_2(2-n) } \ln \left( \frac{c_2^2|t|^{2d_2} (2+c_{**}^2)^{2(d_2)^{+}}}{c_1^2|t|^{2d_1}(2+c_{**}^2)^{2(d_2)^{-}}}\right) &{} \text{ if } a=n, \\ |t|^b c_{1}^{n-a} |t|^{d_{1}(n-a)} c_{2}^{2-n} |t|^{d_{2}(2-n)} &{} \text{ if } a>n. \end{array}\right. } \end{aligned} \end{aligned}$$

(B.5)

Fourth, when $s\in (-\infty ,2t-c_2^2|t|^{2d_2}]$, we have $\frac{-s}{2} \le t-s \le -s$. For $a<n$,

$$\begin{aligned} \begin{aligned} J_4 =&\int _{-\infty }^{2t-c_2^2|t|^{2d_2}}\frac{|s|^{b}}{(t-s)^{a/2}}F\left( \frac{c_1^2|s|^{2d_1}}{4(t-s)},\frac{c_2^2|s|^{2d_2}}{4(t-s)}\right) ds\\ \lesssim&\int _{-\infty }^{2t-c_2^2|t|^{2d_2}} |s|^{b-\frac{a}{2}} F\left( \frac{c_1^2|s|^{2d_1}}{4|s|},\frac{c_2^2|s|^{2d_2}}{2|s|}\right) ds \\ \lesssim&\int _{-\infty }^{2t-c_2^2|t|^{2d_2}}|s|^{b-\frac{a}{2}} c_2^{n-a} |s|^{(2d_2-1)\frac{n-a}{2}} ds \\ \lesssim&c_2^{n-a}|t|^{b+d_2(n-a)-\frac{n}{2}+1} \lesssim c_2^{2-a}|t|^{b+d_2(2-a)}, \end{aligned} \end{aligned}$$

where (B.2) is needed to guarantee the integrability and the last step is using $d_2\le 1/2$, $c_2\le c_{**}$ and $n>2$. For $a\ge n$, similarly we have

$$\begin{aligned} \begin{aligned} J_4 \lesssim&{\left\{ \begin{array}{ll} c_{1}^{n-a} |t|^{b+d_{1}(n-a) +1-\frac{n}{2}} &{} \text{ if } a>n, \\ |t|^{b+1-\frac{a}{2}}\ln \left( \frac{2c_2|t|^{d_2}}{c_1|t|^{d_1}} \right) &{} \text{ if } a=n, \end{array}\right. } \\ \lesssim&c_1^{2-a}|t|^{b+d_1(2-a)} . \end{aligned} \end{aligned}$$

Collecting the estimates of $J_1$ to $J_4$, we get the conclusion. $\square$

Remark A.2

After close examination of the proof, only (B.4) needs the comparison between a and 2. In fact, if $a<2$, one can let $c_1=0$ to get

$$\begin{aligned} {\mathcal T} ^{out}\left[ \frac{|t|^{b}}{|x|^a}\mathbf {1}_{\{ |x|\le c_2|t|^{d_2}\}}\right] (0,t)\le C c_2^{2-a}|t|^{b+d_2(2-a)}.\end{aligned}$$

Lemma B.2

Suppose $n >2$, $a\ge 0$, $d_1\le d_2\le \frac{1}{2}$ and b satisfies (B.2), $0\le c_1,c_2\le c_{**}$. Denote

$$\begin{aligned} u(x,t)=\mathcal {T}^{out}\left[ \frac{|t|^{b}}{|x|^a}\mathbf {1}_{\{ c_1|t|^{d_1}\le |x|\le c_2|t|^{d_2}\}}\right] (x,t). \end{aligned}$$

Then there exists C depending on $n,a,b,d_1,d_2,c_{**}$ such that for $t<-1$

$$\begin{aligned} u(x,t)\le C {\left\{ \begin{array}{ll}c_1^{2-a}|t|^{b+d_1(2-a)}&{}\text {if }a\in (2,\infty ),\\ c_2^{2-a}|t|^{b+d_2(2-a)}&{}\text {if }a\in [0,2).\end{array}\right. } \end{aligned}$$

(B.6)

Moreover, when $|x|>2c_2|2t|^{d_2}$, $a<n$,

$$\begin{aligned} u(x,t) \le C c_2^{n-a}{\left\{ \begin{array}{ll} |t|^{b+d_2(n-a)}|x|^{2-n}&{}\text {if }2c_2|2t|^{d_2}\le |x|\le |t|^{\frac{1}{2}},\\ |x|^{2b+2d_2(n-a)+2-n}&{}\text {if }|x|\ge |t|^{\frac{1}{2}}. \end{array}\right. } \end{aligned}$$

(B.7)

When $|x|\ge 2c_1 |2t|^{d_1}$, $a>n$,

$$\begin{aligned} u(x,t) \le C c_1^{n-a}{\left\{ \begin{array}{ll} |t|^{b+d_{1}(n-a)} |x|^{2-n} &{} \text{ if } 2c_1 |2t|^{d_1} \le |x|\le |t|^{\frac{1}{2}}, \\ |x|^{2b+2d_1(n-a)} |x|^{2-n} &{} \text{ if } |x|\ge |t|^{\frac{1}{2}} . \end{array}\right. } \end{aligned}$$

(B.8)

Proof

$\bullet$ Since

$$\begin{aligned}\frac{|t|^{b}}{|x|^a} \mathbf {1}_{\{ c_1|t|^{d_1}\le |x|\le c_2|t|^{d_2}\}}\le |t|^b\min \left\{ \frac{1}{c_1^a|t|^{d_1a}},\frac{1}{|x|^a}\right\} \mathbf {1}_{\{|x|\le c_2|t|^{d_2}\}}=f(x,t) , \end{aligned}$$

then

$$\begin{aligned} u(x,t)\lesssim \int _{-\infty }^t\int _{\mathbb {R}^n}G(x-y,t-s)f(y,s)dyds . \end{aligned}$$

Since G and f are both decreasing functions for each time slice, using Hardy-Littlewood rearrangement inequality, then

$$\begin{aligned}u(x,t)\le u(0,t)=J_1+J_2,\end{aligned}$$

where

$$\begin{aligned} J_1=&\int _{-\infty }^t\int _{\mathbb {R}^n}G(y,t-s)|s|^b\min \left\{ \frac{1}{c_1^a|s|^{d_1a}},\frac{1}{|y|^a}\right\} \mathbf {1}_{\{|y|\le c_1|s|^{d_1}\}}dyds, \end{aligned}$$

(B.9)

$$\begin{aligned} J_2=&\int _{-\infty }^t\int _{\mathbb {R}^n}G(y,t-s)|s|^b\min \left\{ \frac{1}{c_1^a|s|^{d_1a}},\frac{1}{|y|^a}\right\} \mathbf {1}_{\{c_1|s|^{d_1}\le |y|\le c_2|s|^{d_2}\}}dyds. \end{aligned}$$

(B.10)

Applying Lemma B.1 and Remark A.2, we obtain

$$\begin{aligned} J_1\lesssim c_1^{2-a}|t|^{b+d_1(2-a)},\quad J_2\lesssim {\left\{ \begin{array}{ll}c_1^{2-a}|t|^{b+d_1(2-a)}&{}\text {if }a\in (2,\infty ),\\ c_2^{2-a}|t|^{b+d_2(2-a)}&{}\text {if }a\in (0,2).\end{array}\right. } \end{aligned}$$

Therefore (B.6) is established.

$\bullet$ Next we will establish (B.7) when $d_2\le 0$. In this case, for $x>2c_2|t|^{d_2}$, one has

$$\begin{aligned} \frac{1}{2}|x|\le |x-y|\le 2|x| \text { for } y \text { with } |y|\le |s|^{d_2} \text { and }s\le t. \end{aligned}$$

(B.11)

Then

$$\begin{aligned} u(x,t)\lesssim&\int _{-\infty }^t\int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{c_1|s|^{d_1}\le |y|\le c_2|s|^{d_2}\}}dyds\\ \approx&\int ^t_{-\infty }\int _{c_1|s|^{d_1}}^{c_2|s|^{d_2}}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}|s|^{b}r^{n-1-a}drds\\ \lesssim&c_2^{n-a}\int ^t_{-\infty }\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}|s|^{b+d_2(n-a)}ds\\ \le&c_2^{n-a}\left( \max \{|t|,|x|^2\}\right) ^{b+d_2(n-a)}|x|^{2-n}. \end{aligned}$$

The last step follows from the following two facts

$$\begin{aligned}&\int _{2t}^t\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}|s|^{b+d_2(n-a)}ds \approx |t|^{b+d_2(n-a)}\int _{2t}^t\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}ds\\&\quad \approx |t|^{b+d_2(n-a)}|x|^{2-n}\int _{\frac{|x|^2}{16|t|}}^\infty e^{-z}z^{\frac{n}{2}-2}dz \lesssim {\left\{ \begin{array}{ll} |t|^{b+d_2(n-a)}|x|^{2-n}&{}\text {if }|x|<|t|^{\frac{1}{2}},\\ |t|^{b+d_2(n-a)}\int _{\frac{|x|^2}{16|t|}}^\infty e^{-z}z^{\frac{n}{2}-2}dz&{}\text {if }|x|\ge |t|^{\frac{1}{2}}. \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned}&\int _{-\infty }^{2t}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}|s|^{b+d_2(n-a)}ds\approx \int _{-\infty }^{2t}e^{-\frac{|x|^2}{16|s|}}|s|^{b+d_2(n-a)-\frac{n}{2}}ds\\&\quad \approx |x|^{2b+2d_2(n-a)-n+2}\int _{0}^{\frac{|x|^2}{32|t|}}e^{-z}z^{-b-d_2(n-a)+\frac{n}{2}-2}dz \approx {\left\{ \begin{array}{ll} |t|^{b+d_2(n-a)-\frac{n}{2}+1}&{}\text {if }|x|<|t|^{\frac{1}{2}},\\ |x|^{2b+2d_2(n-a)-n+2}&{}\text {if }|x|\ge |t|^{\frac{1}{2}}, \end{array}\right. } \end{aligned}$$

where $-b-d_2(n-a)+\frac{n}{2}-2>-1$ is needed to guarantee the integrability. Thus (B.7) is established.

$\bullet$ Next we will establish (B.7) when $d_2>0$. We do not have (B.11) anymore. In this case, $|x|\ge 2c_2|2t|^{d_2}$ is equivalent to $-(\frac{|x|}{2c_2})^{1/d_2}\le 2t$. We write

$$\begin{aligned}(-\infty ,t)=(-(\frac{|x|}{2c_2})^{1/d_2},t)\cup (-(\frac{2|x|}{c_2})^{1/d_2},-(\frac{|x|}{2c_2})^{1/d_2})\cup (-\infty ,-(\frac{2|x|}{c_2})^{1/d_2})\end{aligned}$$

and thus

$$\begin{aligned} u\lesssim \int _{-\infty }^t\int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{|y|\le c_2|s|^{d_2}\}}dyds=u_1+u_2+u_3 \end{aligned}$$

(B.12)

where $u_1,u_2,u_3$ are the integrations according to the three intervals respectively. We shall verify that $u_1,u_2,u_3$ all satisfy (B.7). For $u_1$, one has $c_2|s|^{d_2}\le |x|/2$ for such s. Then

$$\begin{aligned} u_1=&\int ^t_{-(|x|/2c_2)^{1/d_2}}\int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{|y|\le c_2|s|^{d_2}\}}dyds\\ \lesssim&\int ^t_{-(|x|/2c_2)^{1/d_2}}\int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{|y|\le c_2|s|^{d_2}\}}dyds\\ \lesssim&\left( \int _{2t}^t+\int ^{2t}_{-(|x|/2c_2)^{1/d_2}}\right) \int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{|y|\le c_2|s|^{d_2}\}}dyds\\ \lesssim&c_2^{n-a} \left( \max \{|t|,|x|^2\}\right) ^{b+d_2(n-a)}|x|^{2-n}. \end{aligned}$$

The last step follows from some easy integration which has been done many times in this section. For $u_2$,

$$\begin{aligned} u_2\lesssim&\int _{-(2|x|/c_2)^{1/d_2}}^{-(|x|/2c_2)^{1/d_2}}\int _{\mathbb {R}^n} \frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{|y|\le c_2|s|^{d_2}\}}dyds\\ \lesssim&(|x|/c_2)^{b/d_2}\int _{-(2|x|/c_2)^{1/d_2}}^{-(|x|/2c_2)^{1/d_2}}\int _{\mathbb {R}^n} \frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}\frac{1}{|y|^a}\mathbf {1}_{\{|y|\le 2|x|\}}dyds\\ \lesssim&c_2^{-b/d_2}|x|^{b/d_2+n-a}\int _{-(2|x|/c_2)^{1/d_2}}^{-(|x|/2c_2)^{1/d_2}} \frac{1}{(t-s)^{n/2}}ds\lesssim c_2^{(-b+\frac{n}{2}-1)/d_2}(|x|^{1/d_2})^{b+d_2(n-a)+1-\frac{n}{2}}\\ \lesssim&c_2^{n-a}{\left\{ \begin{array}{ll} |t|^{b+d_2(n-a)}|x|^{2-n}&{}\text {if }2c_2|2t|^{d_2}\le |x|<|t|^{\frac{1}{2}},\\ |x|^{2b+2d_2(n-a)+2-n}&{}\text {if }|x|\ge |t|^{\frac{1}{2}}. \end{array}\right. } \end{aligned}$$

The last step follows from $0<d_2\le \frac{1}{2}$ and $b+d_2(n-a)+1-\frac{n}{2}<0$. Similarly, for $u_3$,

$$\begin{aligned} u_3\lesssim&\int ^{-(2|x|/c_2)^{1/d_2}}_{-\infty }\int _{\mathbb {R}^n} \frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{|y|\le c_2|s|^{d_2}\}}dyds\\ \lesssim&(|x|/c_2)^{(b+1-\frac{n}{2})/d_2}|x|^{n-a}+\int ^{-(2|x|)^{1/d_2}}_{-\infty }\frac{1}{|s|^{\frac{n}{2}}}e^{-\frac{|y|^2}{16(t-s)}}\frac{|s|^a}{|y|^b}\mathbf {1}_{\{2|x|\le |y|\le c_2|s|^{d_2}\}}dyds\\ \lesssim&c_2^{(-b+\frac{n}{2}-1)/d_2}(|x|^{1/d_2})^{b+d_2(n-a)+1-\frac{n}{2}}\\ \lesssim&c_2^{n-a}{\left\{ \begin{array}{ll} |t|^{b+d_2(n-a)}|x|^{2-n}&{}\text {if }2c_2|2t|^{d_2}\le |x|<|t|^{\frac{1}{2}},\\ |x|^{2b+2d_2(n-a)+2-n}&{}\text {if }|x|\ge |t|^{\frac{1}{2}}. \end{array}\right. } \end{aligned}$$

Collecting the results of $u_1,u_2,u_3$, we can get (B.7).

$\bullet$ By the similar calculation like (B.7), we will get (B.8). $\square$

Lemma B.3

Suppose $2<a<n$, $0\le d_2\le \frac{1}{2}$, $\frac{n}{2} -b>1$ and $0<c_2\le c_{**}$. Then there exists C depending on $n,a,b,d_2,c_{**}$ such that for $t<-1$,

$$\begin{aligned} \mathcal {T}^{out}\left[ \frac{|t|^{b}}{|x|^a}\mathbf {1}_{\{|x|\le c_2|t|^{d_2}\}}\right] \le C |t|^{b}|x|^{2-a} \quad \text {for }|x|<c_2|t|^{d_2}. \end{aligned}$$

(B.13)

Proof

We divide u into three parts

$$\begin{aligned} u(x,t)= \int _{-\infty }^t\int _{\mathbb {R}^n}G(x-y,t-s)\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{|y|\le c_2|s|^{d_2}\}}dyds=u_1+u_2+u_3, \end{aligned}$$

where $u_1$ is the term with $\mathbf {1}_{\{|y|\le \frac{1}{2}|x|\}}$ inside the integrand, $u_2$ is the one with $\mathbf {1}_{\{\frac{1}{2}|x|\le |y|\le 2|x|\}}$ and $u_3$ is the one with $\{ 2|x|\le |y|\le c_2|s|^{d_2}\}$. Since most of the calculation are similar to the proof of the previous lemma. We omit some details here. For $u_1$, we proceed as

$$\begin{aligned} u_1\lesssim&\int _{-\infty }^t\int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{|y|\le \frac{1}{2}|x|\}}dyds\\ \approx&\,|x|^{n-a}\int _{-\infty }^t\frac{1}{(t-s)^{n/2}}e^{-\frac{|x|^2}{16(t-s)}}|s|^{b}ds\\ \lesssim&\, |t|^{b}|x|^{2-a}\int _{\frac{|x|^2}{16(-t)}}^\infty e^{-z}z^{\frac{n}{2}-2}dz+|x|^{2b+2-a}\int _0^{\frac{|x|^2}{32(-t)}}e^{-z}z^{-b+\frac{n}{2}-2}dz\\ \lesssim&|t|^{b}|x|^{2-a}+|x|^{n-a}|t|^{1-\frac{n}{2}+b}, \end{aligned}$$

where we have used the fact that $\frac{n}{2} -b>1$. For $u_2$, we have

$$\begin{aligned} u_2\lesssim&\int _{-\infty }^t\int _{\mathbb {R}^n} \frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{\frac{1}{2}|x|\le |y|\le 2|x|\}}dyds\\ \lesssim&|x|^{-a}\int _{-\infty }^t\int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}|s|^{b}\mathbf {1}_{\{|x-y|\le 3|x|\}}dydx\\ \approx&|x|^{-a}\int _{-\infty }^t\int _0^{3|x|}\frac{1}{(t-s)^{n/2}}e^{-\frac{r^2}{4(t-s)}}|s|^{b}r^{n-1}dyds \approx |x|^{-a}\int _{-\infty }^t\int _0^{\frac{9|x|^2}{4(t-s)}}|s|^{b}e^{-z}z^{\frac{n}{2}-1} d z d s\\ \approx&|x|^{-a}\int _{-\infty }^t|s|^{b}\min \left\{ 1,\left( \frac{|x|^2}{t-s}\right) ^{\frac{n}{2}}\right\} d s \approx |t|^{b}|x|^{2-a}+|t|^{1+b-\frac{n}{2}}|x|^{n-a}. \end{aligned}$$

For $u_3$, we have

$$\begin{aligned} u_3\lesssim&\int _{-\infty }^t\int _{\mathbb {R}^n} \frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{2|x|\le |y|\le c_2|s|^{d_2}\}}dyds\\ \lesssim&\int _{-\infty }^t\int _{\mathbb {R}^n} \frac{1}{(t-s)^{n/2}}e^{-\frac{|y|^2}{16(t-s)}}\frac{|s|^{b}}{|y|^a}\mathbf {1}_{\{2|x|\le |y|\le c_2 |s|^{d_2}\}}dyds\\ \approx&\int _{-\infty }^t\int _{2|x|}^{c_2|s|^{d_2}}\frac{1}{(t-s)^{n/2}}e^{-\frac{r^2}{16(t-s)}}|s|^{b}r^{n-1-a}drds\\ \approx&\int _{-\infty }^t\int ^{\frac{c_2|s|^{d_2}}{16(t-s)}}_{\frac{|x|^2}{4(t-s)}}(t-s)^{-\frac{a}{2}}|s|^{b}e^{-z}z^{\frac{n-a}{2}-1}dzds\lesssim |t|^{b} |x|^{2-a} . \end{aligned}$$

Combining the estimate of $u_1,u_2,u_3$ and using the fact that $|x|^{n-a}|t|^{1-\frac{n}{2}+b}\le |t|^{b}|x|^{2-a}$ because $|x|\le c_2|t|^{d_2}$, we get (B.13). $\square$

Corollary B.4

Suppose that $n>2$, $d_1\le d_2\le \frac{1}{2}$, b satisfies (B.2), $0\le c_1,c_2\le c_{**}$. Denote

$$\begin{aligned} u(x,t) = \mathcal {T}^{out}\left[ |t|^{b} |x|^{-a}\mathbf {1}_{\{ c_1|t|^{d_1}\le |x|\le c_2|t|^{d_2}\}}\right] . \end{aligned}$$

Then there exists C depending on $n,a,b,d_1,d_2,c_{**}$ such that for $t<-1$, if $0\le a <2$,

$$\begin{aligned} u(x,t) \le C {\left\{ \begin{array}{ll} c_2^{2-a}|t|^{b+d_2(2-a)} &{}\text {if }|x|\le c_2|t|^{d_2}, \\ c_2^{n-a}|t|^{b+d_2(n-a)}|x|^{2-n}&{}\text {if }c_2|t|^{d_2}\le |x|\le |t|^{\frac{1}{2}}, \\ c_2^{n-a} |x|^{2b+2d_2(n-a)+2-n}&{}\text {if }|x|\ge |t|^{\frac{1}{2}}. \end{array}\right. } \end{aligned}$$

(B.14)

If $2<a<n$,

$$\begin{aligned} u(x,t) \le C {\left\{ \begin{array}{ll} c_1^{2-a}|t|^{b+d_1(2-a)}&{}\text {if }|x|\le c_1|t|^{d_1}, \\ |t|^{b}|x|^{2-a}&{}\text {if }c_1|t|^{d_1}\le |x|\le c_2|t|^{d_2}, \\ c_2^{n-a}|t|^{b+d_2(n-a)}|x|^{2-n}&{}\text {if }c_2|t|^{d_2}\le |x|\le |t|^{\frac{1}{2}}, \\ c_2^{n-a} |x|^{2b+2d_2(n-a)+2-n}&{}\text {if }|x|\ge |t|^{\frac{1}{2}}. \end{array}\right. } \end{aligned}$$

(B.15)

If $a>n$,

$$\begin{aligned} u(x,t) \le C {\left\{ \begin{array}{ll} c_1^{2-a}|t|^{b+d_1(2-a)}&{}\text {if }|x|\le c_1|t|^{d_1}, \\ c_1^{n-a}|t|^{b+d_1(n-a)}|x|^{2-n} &{}\text {if }c_1|t|^{d_1}\le |x|\le |t|^{\frac{1}{2}}, \\ c_1^{n-a} |x|^{2b+2d_1(n-a)+2-n}&{}\text {if }|x|\ge |t|^{\frac{1}{2}}. \end{array}\right. } \end{aligned}$$

(B.16)

Proof

(B.14) and (B.16) follow Lemma B.2 directly. (B.15) follows from (B.7) and (B.13). $\square$

Lemma B.5

Suppose that $\frac{a}{2} -b >1$. Then

$$\begin{aligned} \begin{aligned}&{\mathcal T} ^{out} \left[ |t|^{b} |x|^{-a} \mathbf {1}_{\{|x|\ge |t|^{\frac{1}{2}}\}} \right] \\ \lesssim&\, |t|^{1+b-\frac{a}{2}} \mathbf {1}_{\{|x|\le |t|^{\frac{1}{2}}\} } + \mathbf {1}_{\{|x|\ge |t|^{\frac{1}{2}}\}} |x|^{-a} {\left\{ \begin{array}{ll} |t|^{1+b}, &{} \text{ if } b<-1, \\ 1+ \ln \left( \frac{|x|^2}{|t|} \right) &{} \text{ if } b= - 1, \\ |x|^{2+2b} &{} \text{ if } b> - 1 . \end{array}\right. } \end{aligned} \end{aligned}$$

(B.17)

Proof

Denote $u(x,t) = {\mathcal T} ^{out} \left[ |t|^{b} |x|^{-a} \mathbf {1}_{[|x|\ge |t|^{\frac{1}{2}}]} \right]$.

$\bullet$ For $|x|\le \frac{1}{2} |t|^{\frac{1}{2}}$, we have $|x-y|\ge \frac{1}{2}|y|$ for $|y|\ge |s|^{\frac{1}{2}}\ge |t|^{\frac{1}{2}}$. Then

$$\begin{aligned} \begin{aligned} u(x,t) \lesssim&\, \int _{-\infty }^t \int _{{{\mathbb R}}^n} \frac{1}{ (t-s)^{n/2}} e^{-\frac{|y|^2}{16(t-s)}} |s|^{b} |y|^{-a} \mathbf {1}_{\{|y|\ge |s|^{\frac{1}{2}}\}} \,\mathrm {d}\xi \,\mathrm {d}s \\ \lesssim&\, \left( \int _{-\infty }^{2t} + \int _{2t}^t \right) |s|^{b} (t-s)^{-a/2} e^{-\frac{(-s)}{32(t-s)}} \,\mathrm {d}s \lesssim \ |t|^{1+b-\frac{a}{2}} , \end{aligned} \end{aligned}$$

(B.18)

where $\frac{a}{2} -b >1$ is used to guarantee the integrability in the last step.

$\bullet$ Consider $|x|\ge 4|t|^{\frac{1}{2}}$. We make the following decomposition.

$$\begin{aligned} u(x,t) =&\, \left( \int _{-\frac{1}{4}|x|^2}^t+\int _{-4|x|^2}^{-\frac{1}{4}|x|^2}+\int ^{-4|x|^2}_{-\infty }\right) \int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}|s|^{b}|y|^{-a}\mathbf {1}_{\{|y|\ge |s|^{\frac{1}{2}}\}}\,\mathrm {d}y\,\mathrm {d}s \\ :=&\, P_1 + P_2 +P_3 . \end{aligned}$$

For $P_1$, we divide it further to be

$$\begin{aligned} P_1=\int _{-\frac{1}{4}|x|^2}^t\int _{\mathbb {R}^n}\frac{1}{(t-s)^{n/2}}e^{-\frac{|x-y|^2}{4(t-s)}}|s|^{b}|y|^{-a}\mathbf {1}_{\{|y|\ge |s|^{\frac{1}{2}}\}}dyds=P_{11}+P_{12}+P_{13} \end{aligned}$$

where $P_{11}$ is the term with $\mathbf {1}_{\{|s|^{\frac{1}{2}}\le |y|\le \frac{1}{2}|x|\}}$ in the integrand, $P_{12}$ is the one with $\mathbf {1}_{\{\frac{1}{2}|x|\le |y|\le 2|x|\}}$ and $P_{13}$ is the one with $\mathbf {1}_{\{2|x|\le |y|\}}$. For $P_{11}$, when $a<n$,

$$\begin{aligned} \begin{aligned} P_{11} \lesssim&\, \int _{ -\frac{1}{4}|x|^2 }^t \int _{{{\mathbb R}}^n} \frac{1}{(t-s)^{n/2}} e^{-\frac{|x|^2}{16(t-s)}} |s|^{b} |y|^{-a} \mathbf {1}_{\{ |s|^{\frac{1}{2}} \le |y|\le \frac{|x|}{2}\}} \,\mathrm {d}y \,\mathrm {d}s \\ \lesssim&\, \int _{ -\frac{1}{4}|x|^2 }^t \frac{1}{(t-s)^{n/2}} e^{-\frac{|x|^2}{16(t-s)}} |s|^{b} |x|^{n-a} \,\mathrm {d}s \\ =&\, |x|^{n-a} \int _{ -\frac{1}{4}|x|^2 }^{t} \frac{1}{(t-s)^{n/2}} e^{-\frac{|x|^2}{16(t-s)}} |s|^{b} \,\mathrm {d}s \lesssim \ |x|^{2-a+2b} . \end{aligned} \end{aligned}$$

(B.19)

When $a\ge n$, by similar calculation, $P_{11} \lesssim |x|^{2-a+2b}$ still holds.

For $P_{12}$,

$$\begin{aligned} \begin{aligned} P_{12} \lesssim&\, |x|^{-a} \int _{ -\frac{1}{4}|x|^2 }^t \int _{{{\mathbb R}}^n} \frac{1}{(t-s)^{n/2}} e^{-\frac{|x-y|^2}{4(t-s)}} |s|^{b} \mathbf {1}_{\{ \frac{|x|}{2} \le |y| \le 2|x|\}} \,\mathrm {d}y \,\mathrm {d}s \\ \lesssim&\, |x|^{-a} \left( \int _{2t }^t + \int _{ -\frac{1}{4}|x|^2 }^{2t} \right) \int _{0}^{3|x|} \frac{1}{(t-s)^{n/2}} e^{-\frac{r^2}{4(t-s)}} |s|^{b} r^{n-1} \,\mathrm {d}r \,\mathrm {d}s \\ \lesssim&\, |x|^{-a} |t|^{1+b} + |x|^{-a} {\left\{ \begin{array}{ll} |t|^{1+b} &{} \text{ if } b< - 1, \\ 1+ \ln \left( \frac{|x|^2}{-t} \right) &{} \text{ if } b=-1, \\ (|x|^2)^{1+b} &{} \text{ if } b>- 1 . \end{array}\right. } \end{aligned} \end{aligned}$$

(B.20)

For $P_{13}$,

$$\begin{aligned} \begin{aligned} P_{13} \lesssim&\, \int _{ -\frac{1}{4}|x|^2 }^t \int _{{{\mathbb R}}^n} \frac{1}{(t-s)^{n/2}} e^{-\frac{|y|^2}{ 16(t-s)}} |s|^{b} |y|^{-a} \mathbf {1}_{\{ 2|x| \le |y|\}} \,\mathrm {d}y \,\mathrm {d}s \\ \approx&\, \int _{ -\frac{1}{4}|x|^2 }^t \int _{2|x|}^\infty \frac{(-s)^{b}}{(t-s)^{n/2}} e^{-\frac{ r^2}{ 16(t-s)}} r^{n-1-a} \,\mathrm {d}r \,\mathrm {d}s \\ =&\, \left( \int _{2t }^{t} + \int _{ -\frac{1}{4}|x|^2 }^{2t} \right) \int _{2|x|}^\infty \frac{(-s)^{b}}{(t-s)^{n/2}} e^{-\frac{ r^2}{ 16(t-s)}} r^{n-1-a} \,\mathrm {d}r \,\mathrm {d}s \\ \lesssim&\, (-t)^{b} |x|^{2-a} e^{-\frac{|x|^2}{16(-t)}} + |x|^{2+2b- a} . \end{aligned} \end{aligned}$$

(B.21)

For $P_2$, since $-\frac{1}{4}|x|^2\le 4t$ in this case,

$$\begin{aligned} \begin{aligned} P_2 \lesssim&\, \int _{-4|x|^2}^{ -\frac{1}{4}|x|^2 } \int _{{{\mathbb R}}^n} \frac{1}{ |s|^{n/2}} e^{-\frac{|x-y|^2}{4(-s)}} |s|^{b} |y|^{-a} \mathbf {1}_{\{|y|\ge \frac{ |x|}{2} \}} \,\mathrm {d}y \,\mathrm {d}s \\ \approx&\, |x|^{-n+2b} \int _{-4|x|^2}^{ -\frac{1}{4}|x|^2 } \int _{{{\mathbb R}}^n} e^{-\frac{|x-y|^2}{4(-s)}} |y|^{-a} \left( \mathbf {1}_{\{\frac{ |x|}{2} \le |y|\le 2|x|\}} + \mathbf {1}_{\{2|x|\le |y| \}} \right) \,\mathrm {d}y \,\mathrm {d}s \\ \lesssim&\, |x|^{2+2b-a} . \end{aligned} \end{aligned}$$

(B.22)

For $P_3$, in this case, $|x|\le \frac{1}{2} |s|^{\frac{1}{2}}$,

$$\begin{aligned} \begin{aligned} P_3 \lesssim&\, \int _{-\infty }^{-4|x|^2} \int _{{{\mathbb R}}^n} \frac{1}{(t-s )^{n/2}} e^{-\frac{|y|^2}{ 16(t-s)}} |s|^{b} |y|^{-a} \mathbf {1}_{\{|y|\ge |s|^{\frac{1}{2}}\}} \,\mathrm {d}y \,\mathrm {d}s \\ \lesssim&\, \int _{-\infty }^{-4|x|^2} \int _{{{\mathbb R}}^n} (-s)^{b-\frac{n}{2}} e^{-\frac{|y|^2}{ 16(-s)}} |y|^{-a} \mathbf {1}_{\{|y|\ge |s|^{\frac{1}{2}}\}} \,\mathrm {d}y \,\mathrm {d}s \lesssim \ |x|^{2+2b-a} , \end{aligned} \end{aligned}$$

(B.23)

where $\frac{a}{2} -b>1$ is required to guarantee the integrability. Combining the above estimates of $P_1$, $P_2$ and $P_3$, we get, when $|x|\ge 4|t|^{\frac{1}{2}}$

$$\begin{aligned} u(x,t) \lesssim |x|^{-a} {\left\{ \begin{array}{ll} |t|^{1+b} &{} \text{ if } b<-1, \\ 1+ \ln \left( \frac{|x|^2}{-t} \right) &{} \text{ if } b=-1, \\ |x|^{2+2b} &{} \text{ if } b>-1 . \end{array}\right. } \end{aligned}$$

(B.24)

$\bullet$ Consider the case $\frac{1}{2} |t|^{\frac{1}{2}} \le |x|\le 4|t|^{\frac{1}{2}}$,

$$\begin{aligned} \begin{aligned} u(x,t) =&\, \int _{-\infty }^t \int _{{{\mathbb R}}^n} \frac{1}{(4\pi (t-s))^{n/2}} e^{-\frac{|x-y|^2}{4(t-s)}} |s|^{b} |y|^{-a} \mathbf {1}_{[|y|\ge |s|^{\frac{1}{2}}]} \,\mathrm {d}y \,\mathrm {d}s \\ \lesssim&\, \left( \int _{64t}^t + \int _{-\infty }^{64t} \right) \int _{{{\mathbb R}}^n} \frac{1}{(t-s)^{n/2}} e^{-\frac{|x-y|^2}{4(t-s)}} |s|^{b} |y|^{-a} \mathbf {1}_{\{|y|\ge |s|^{\frac{1}{2}}\}} \,\mathrm {d}y \,\mathrm {d}s \\ \lesssim&\, |t|^{1+b -\frac{a}{2}} . \end{aligned} \end{aligned}$$

(B.25)

$\square$

In order to get the gradient estimate of $\bar{\varphi }$, we need the following lemma.

Lemma B.6

For $d_{1}\le d_{2} \le \frac{1}{2}$, $\frac{n}{2} -b-d_2 n>0$, $c_1, c_2 \approx 1$, we have

$$\begin{aligned} {\mathcal T} ^{d}[|t|^{b} \mathbf {1}_{[c_1 |t|^{d_1} \le |x| \le c_2 |t|^{d_2}]}] \lesssim {\left\{ \begin{array}{ll} |t|^{b+d_2} &{} \text{ if } |x| \le |t|^{d_2}, \\ |t|^{b+d_2 n} |x|^{1-n} &{} \text{ if } |t|^{d_2} \le |x| \le |t|^{\frac{1}{2}}, \\ (|x|^2)^{b+d_2n} |x|^{1-n} &{} \text{ if } |x| \ge |t|^{\frac{1}{2}}. \end{array}\right. } \end{aligned}$$

(B.26)

For $d_{1}\le d_{2} \le \frac{1}{2}$, $a>n$, $\frac{n}{2} -b -d_1(n-a)>0$, $c_1, c_2 \approx 1$, we have

$$\begin{aligned} {\mathcal T} ^{d}[\frac{|t|^{b}}{|x|^a} \mathbf {1}_{[c_1 |t|^{d_1} \le |x| \le c_2 |t|^{d_2}]}] \lesssim {\left\{ \begin{array}{ll} |t|^{b+d_1(1-a)} &{} \text{ if } |x| \le |t|^{d_1} , \\ |t|^{b+d_1(n-a)} |x|^{1-n} &{} \text{ if } |t|^{d_1} \le |x| \le |t|^{\frac{1}{2}}, \\ (|x|^2)^{b+d_1(n-a)} |x|^{1-n} &{} \text{ if } |x| \ge |t|^{\frac{1}{2}} . \end{array}\right. } \end{aligned}$$

(B.27)

We omit the proof since it relies splitting integral domain like above.

1.2 Proofs of three lemmas in the outer problem

Proof of Lemma 4.4

For $j=2,\dots ,k$, by Corollary B.4,

$$\begin{aligned}&{\mathcal T} ^{out}[\mu _{0j}^{-2}(t)|t|^{\gamma _j} \mathbf {1}_{\{ |x|\le \mu _{0j} \}}] \lesssim {\left\{ \begin{array}{ll} |t|^{\gamma _{j}} &{} \text{ if } |x|\le \mu _{0j}, \\ |t|^{\gamma _{j}} \mu _{0j}^{n-2} |x|^{2-n} &{} \text{ if } \mu _{0j} \le |x|\le |t|^{\frac{1}{2}}, \\ |x|^{2\gamma _{j} +(4-2n)\alpha _j +2-n} &{} \text{ if } |x|\ge |t|^{\frac{1}{2}}, \end{array}\right. }\\&{\mathcal T} ^{out} [\mu _{0j}^{\alpha }(t)|t|^{\gamma _j} |x|^{-2-\alpha } \mathbf {1}_{ \{\mu _{0j} \le |x|\le \bar{\mu }_{0j} \} }] \lesssim {\left\{ \begin{array}{ll} |t|^{\gamma _{j}} &{} \text{ if } |x|\le \mu _{0j}, \\ |t|^{\gamma _{j}} \mu _{0j}^{\alpha } |x|^{-\alpha } &{} \text{ if } \mu _{0j} \le |x|\le \bar{\mu }_{0j} , \\ |t|^{\gamma _{j}} \mu _{0j}^{\alpha } \bar{\mu }_{0j}^{n-2-\alpha } |x|^{2-n} &{} \text{ if } \bar{\mu }_{0j} \le |x| \le |t|^{\frac{1}{2}} , \\ |x|^{2\gamma _j^*+2-n} &{} \text{ if } |x| \ge |t|^{\frac{1}{2}} . \end{array}\right. } \end{aligned}$$

Thus ${\mathcal T} ^{out}[w_{1j}] \lesssim w_{1j}^{*}$.

For $j=1$, the first part of $w_{11}$ can be dealt with by the same method above. For the rest part, by Corollary B.4,

$$\begin{aligned} \begin{aligned} {\mathcal T} ^{out}[|t|^{\gamma _{1}} \bar{\mu }_{01}^{n-2-\alpha } |x|^{-1-n} \mathbf {1}_{\{ \bar{\mu }_{01} \le |x| \le |t|^{\frac{1}{2}} \}}] \lesssim&\, {\left\{ \begin{array}{ll} |t|^{\gamma _{1} } \bar{\mu }_{01}^{-1-\alpha } &{} \text{ if } |x|\le \bar{\mu }_{01} , \\ |t|^{\gamma _1} \bar{\mu }_{01}^{n-3-\alpha } |x|^{2-n} &{} \text{ if } \bar{\mu }_{01} \le |x|\le |t|^{\frac{1}{2}} , \\ |x|^{2(\gamma _1 +\delta (n-3-\alpha )) +2-n} &{} \text{ if } |x| \ge |t|^{\frac{1}{2}} . \end{array}\right. } \\ \lesssim&\, w_{11}^{*} . \end{aligned} \end{aligned}$$

$\square$

Proof of Lemma 4.5

This just follows from (B.15). $b=-2\sigma -(\frac{n}{2}-2)\alpha _{j+1}+\alpha _j$, $d_2=-\frac{1}{2}(\alpha _j+\alpha _{j-1})$, $a=n-2$. $\square$

Proof of Lemma 4.6

By Corollary B.4, we have

$$\begin{aligned} {\mathcal T} ^{out}[|t|^{-1-\sigma } |x|^{2-n} \mathbf {1}_{\{ \bar{\mu }_{01} \le |x| \le |t|^{\frac{1}{2}}\}}] \lesssim {\left\{ \begin{array}{ll} |t|^{-1-\sigma } \bar{\mu }_{01}^{4-n} &{} \text{ if } |x|\le \bar{\mu }_{01}, \\ |t|^{-1-\sigma } |x|^{4-n} &{} \text{ if } \bar{\mu }_{01} \le |x|\le |t|^{\frac{1}{2}}, \\ |x|^{2-2\sigma -n} &{} \text{ if } |x| \ge |t|^{\frac{1}{2} }. \end{array}\right. } \end{aligned}$$

By Lemma B.5, we have

$$\begin{aligned} {\mathcal T} ^{out}[|t|^{-1-\sigma } |x|^{2-n} \mathbf {1}_{\{ |x| \ge |t|^{\frac{1}{2}}\}}] \lesssim {\left\{ \begin{array}{ll} |t|^{1-\sigma -\frac{n}{2}} &{} \text{ if } |x|\le |t|^{\frac{1}{2}}, \\ |t|^{-\sigma }|x|^{2-n} &{} \text{ if } |x| \ge |t|^{\frac{1}{2} }. \end{array}\right. } \end{aligned}$$

Then (4.27) follows when $\delta \le \frac{1}{2}$. $\square$