$\displaystyle\int_{-81/100}^{t}\int_{B_{L\lambda(s)}^{c}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}$
$\displaystyle\quad\quad\quad\quad\times\lambda(s)^{\frac{n-4}{2}}|\lambda^{\prime}(s)||y|^{2-n}dyds$
$\displaystyle=$
$\displaystyle\int_{-81/100}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\lambda(s)^{\frac{n-4}{2}}|\lambda^{\prime}(s)|$
$\displaystyle\quad\times\left[\int_{B_{L\lambda(s)}^{c}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}|y|^{2-n}dy\right]ds.$

We still divide this integral into three parts, $\mathrm{I}$ being on the
interval $(-81/100,t-\lambda(t)^{2\mu})$, $\mathrm{II}$ being on the interval
$(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})$, and $\mathrm{III}$ on
$(t-K^{2\beta}\lambda(t)^{2},t)$.

Estimate of $\mathrm{I}$. Direct calculation gives

$\displaystyle\mathcal{P}$ $\displaystyle:=$
$\displaystyle\int_{B_{L\lambda(s)}^{c}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}|y|^{2-n}dy$
$\displaystyle\lesssim$
$\displaystyle\left(t-s\right)^{-\frac{n-2}{2}}\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}\int_{\frac{L\lambda(s)}{\sqrt{t-s}}}^{+\infty}e^{-c\left(\frac{|x|}{\sqrt{t-s}}-r\right)^{2}}rdr.$

Recall that $K\lambda(t)/2\leq|x|\leq 4K\lambda(t)$, $t-s\geq
K^{2\beta}\lambda(t)^{2}$. There are two cases.

  * •

If $L\lambda(s)\geq 6K\lambda(t)$,

$\mathcal{P}\lesssim\left(t-s\right)^{-\frac{n-2}{2}}\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}.$

  * •

If $L\lambda(s)\leq 6K\lambda(t)$,

$\displaystyle\mathcal{P}$ $\displaystyle\lesssim$
$\displaystyle\left(t-s\right)^{-\frac{n-2}{2}}\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}$
$\displaystyle\lesssim$
$\displaystyle\left(t-s\right)^{-\frac{n-2-\gamma}{2}}L^{-\gamma}\lambda(s)^{-\gamma}.$

Hence

$\displaystyle\mathrm{I}$ $\displaystyle\lesssim$
$\displaystyle|x|^{-\gamma}\int_{(-81/100,t-\lambda(t)^{2\mu})\cap\\{L\lambda(s)\geq
6K\lambda(t)\\}}(t-s)^{-\frac{n-2-\gamma}{2}}\lambda(s)^{\frac{n-4}{2}}|\lambda^{\prime}(s)|$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\times\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}$
$\displaystyle+$ $\displaystyle
L^{-\gamma}|x|^{-\gamma}\int_{(-81/100,t-\lambda(t)^{2\mu})\cap\\{L\lambda(s)\leq
6K\lambda(t)\\}}(t-s)^{-\frac{n-2}{2}+\gamma}\lambda(s)^{\frac{n-4}{2}-\gamma}|\lambda^{\prime}(s)|.$

Because

$\displaystyle(t-s)^{-\frac{n-2-\gamma}{2}}\lambda(s)^{\frac{n-6}{2}}\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}$
$\displaystyle=$
$\displaystyle(t-s)^{-\frac{n+2-2\gamma}{4}}\left[\frac{\lambda(s)}{\sqrt{t-s}}\right]^{\frac{n-6}{2}}\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}$
$\displaystyle\lesssim$ $\displaystyle
L^{-\frac{n-6}{2}-\gamma}\lambda(s)^{-\gamma}(t-s)^{-\frac{n+2}{4}+\gamma},$

by noting that we have assumed $L\lambda(s)\geq 6K\lambda(t)$, the first
integral is controlled by

$\displaystyle
L^{-\frac{n-6}{2}}\left[K\lambda(t)\right]^{-2\gamma}\int_{-81/100}^{t-\lambda(t)^{2\mu}}(t-s)^{-\frac{n+2}{4}+\gamma}|\lambda(s)\lambda^{\prime}(s)|$
$\displaystyle\lesssim$ $\displaystyle
L^{-\frac{n-6}{2}}K^{-2\gamma}\lambda(t)^{-\frac{n-2}{2}\mu+2(\mu-1)\gamma}.$

In the last step we used only the estimate $|\lambda\lambda^{\prime}|\leq C$.

For the second integral, by the estimate $|\lambda\lambda^{\prime}|\leq C$
again, we obtain

$\displaystyle
L^{-\gamma}|x|^{-\gamma}\int_{(-81/100,t-\lambda(t)^{2\mu})\cap\\{L\lambda(s)\leq
6K\lambda(t)\\}}(t-s)^{-\frac{n-2}{2}+\gamma}\lambda(s)^{\frac{n-4}{2}-\gamma}|\lambda^{\prime}(s)|$
$\displaystyle\lesssim$ $\displaystyle
L^{-\frac{n-6}{2}}K^{\frac{n-6}{2}-2\gamma}\lambda(t)^{\frac{n-6}{2}-2\gamma}\int_{-81/100}^{t-\lambda(t)^{2\mu}}(t-s)^{-\frac{n-2}{2}+\gamma}$
$\displaystyle\lesssim$ $\displaystyle
L^{-\frac{n-6}{2}}K^{\frac{n-6}{2}-2\gamma}\lambda(t)^{-(n-4)\mu+\frac{n}{2}-3+2(\mu-1)\gamma}.$

Because

$-(n-4)\mu+\frac{n}{2}-3+2(\mu-1)\gamma>-\frac{n-2}{2}\mu+2(\mu-1)\gamma,$

combining these two estimates, we get

$\mathrm{I}\lesssim
L^{-\frac{n-6}{2}}K^{-2\gamma}\lambda(t)^{-\frac{n-2}{2}\mu+2(\mu-1)\gamma}.$

Estimate of $\mathrm{II}$. As in the previous case, we have

$\displaystyle\mathrm{II}$ $\displaystyle\lesssim$
$\displaystyle|x|^{-\gamma}\int_{(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})\cap\\{L\lambda(s)\geq
6K\lambda(t)\\}}(t-s)^{-\frac{n-2-\gamma}{2}}\lambda(s)^{\frac{n-4}{2}}|\lambda^{\prime}(s)|$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\times\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}$
$\displaystyle+$ $\displaystyle
L^{-\gamma}|x|^{-\gamma}\int_{(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})\cap\\{L\lambda(s)\leq
6K\lambda(t)\\}}(t-s)^{-\frac{n-2}{2}+\gamma}\lambda(s)^{\frac{n-4}{2}-\gamma}|\lambda^{\prime}(s)|$
$\displaystyle\lesssim$ $\displaystyle
L^{-\frac{n-6}{2}}\left[K\lambda(t)\right]^{-2\gamma}\int_{t-\lambda(t)^{2\mu}}^{t-K^{2\beta}\lambda(t)^{2}}(t-s)^{-\frac{n+2}{4}+\gamma}|\lambda(s)\lambda^{\prime}(s)|$
$\displaystyle+$ $\displaystyle
L^{-\frac{n-6}{2}}K^{\frac{n-6}{2}-2\gamma}\lambda(t)^{\frac{n-6}{2}-2\gamma}\int_{t-\lambda(t)^{2\mu}}^{t-K^{2\beta}\lambda(t)^{2}}(t-s)^{-\frac{n-2}{2}+\gamma}|\lambda(s)\lambda^{\prime}(s)|$
$\displaystyle\lesssim$ $\displaystyle
L^{-\frac{n-6}{2}}\left[K^{-\frac{n-2}{2}\beta+2(\beta-1)\gamma}+K^{-\frac{n-2}{2}\beta-\left(\frac{n-6}{2}-2\gamma\right)\left(\beta-1\right)}\right]$
$\displaystyle\quad\times\lambda(t)^{-\frac{n-2}{2}}\|\lambda\lambda^{\prime}\|_{L^{\infty}(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})}$
$\displaystyle\lesssim$ $\displaystyle
K^{-\frac{n-2}{2}\beta+2(\beta-1)\gamma}\lambda(t)^{-\frac{n-2}{2}}\|\lambda\lambda^{\prime}\|_{L^{\infty}(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})}.$

Estimate of $\mathrm{III}$. By (14.21),

$\displaystyle\mathrm{III}$ $\displaystyle\lesssim$
$\displaystyle\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\int_{B_{1}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}$
$\displaystyle\quad\quad\quad\quad\quad\times\lambda(s)^{\frac{n-4}{2}}|\lambda^{\prime}(s)||y|^{2-n}$
$\displaystyle=$
$\displaystyle\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\lambda(s)^{\frac{n-4}{2}}|\lambda^{\prime}(s)|$
$\displaystyle\quad\quad\quad\quad\times\left[\underbrace{\int_{B_{1}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}|y|^{2-n}dy}_{\mbox{Lemma
\ref{lem heat kernel integral 1}}}\right]ds$ $\displaystyle\lesssim$
$\displaystyle\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\left(|x|+\sqrt{t-s}\right)^{2-n}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{2\gamma}\lambda(s)^{\frac{n-4}{2}}|\lambda^{\prime}(s)|ds$
$\displaystyle\lesssim$
$\displaystyle\left[K\lambda(t)\right]^{2-n}K^{2(\beta-1)\gamma}\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\lambda(s)^{\frac{n-4}{2}}|\lambda^{\prime}(s)|ds$
$\displaystyle\lesssim$ $\displaystyle
K^{2-n+2\beta+2(\beta-1)\gamma}\lambda(t)^{-\frac{n-2}{2}}\|\lambda\lambda^{\prime}\|_{L^{\infty}(t-K^{2\beta}\lambda(t)^{2},t)}.$

Because $n>6$ and $\beta$ is sufficiently close to $1$, we have

$2-n+2\beta+2(\beta-1)\gamma<-\frac{n-2}{2}\beta+2(\beta-1)\gamma.$

Adding up estimates for $\mathrm{I}$, $\mathrm{II}$ and $\mathrm{III}$ we
finish the proof. ∎

######  Lemma 14.4 (Estimates on $\overline{\phi}_{4}$).

For $K\lambda(t)/2\leq|x|\leq 4K\lambda(t)$,

$\overline{\phi}_{4}(x,t)\lesssim\lambda(t)^{-\frac{n-2}{2}\mu+2(\mu-1)\gamma}+K^{-\frac{n-2}{2}(\beta-1)+2(\beta-1)\gamma}\left[K\lambda(t)\right]^{-\frac{n-2}{2}}\|\lambda\xi^{\prime}\|_{L^{\infty}(t-\lambda(t)^{2\mu},t)}.$

###### Proof.

We still have the heat kernel representation

$\displaystyle\overline{\phi}_{4}(x,t)$ $\displaystyle\lesssim$
$\displaystyle\int_{-81/100}^{t}\int_{B_{L\lambda(s)}^{c}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}$
$\displaystyle\quad\quad\quad\quad\times\lambda(s)^{\frac{n-2}{2}}|\xi^{\prime}(s)||y|^{1-n}dyds$
$\displaystyle=$
$\displaystyle\int_{-81/100}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\lambda(s)^{\frac{n-2}{2}}|\xi^{\prime}(s)|$
$\displaystyle\quad\times\left[\int_{B_{L\lambda(s)}^{c}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}|y|^{1-n}dy\right]ds.$

We still divide this integral into three parts, $\mathrm{I}$ being on the
interval $(-81/100,t-\lambda(t)^{2\mu})$, $\mathrm{II}$ being on the interval
$(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})$, and $\mathrm{III}$ on the
interval $(t-K^{2\beta}\lambda(t)^{2},t)$.

Estimate of $\mathrm{I}$. Direct calculation gives

$\displaystyle\mathcal{P}$ $\displaystyle:=$
$\displaystyle\int_{B_{L\lambda(s)}^{c}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}|y|^{1-n}dy$
$\displaystyle\lesssim$
$\displaystyle\left(t-s\right)^{-\frac{n-1}{2}}\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}\int_{\frac{L\lambda(s)}{\sqrt{t-s}}}^{+\infty}e^{-c\left(\frac{|x|}{\sqrt{t-s}}-r\right)^{2}}dr.$

Recall that $K\lambda(t)/2\leq|x|\leq 4K\lambda(t)$ and $t-s\geq
K^{2\beta}\lambda(t)^{2}$. There are two cases.

  * •

If $L\lambda(s)\geq 6K\lambda(t)$,

$\mathcal{P}\lesssim\left(t-s\right)^{-\frac{n-1}{2}}\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}.$

  * •

If $L\lambda(s)\leq 6K\lambda(t)$,

$\mathcal{P}\lesssim\left(t-s\right)^{-\frac{n-1-\gamma}{2}}L^{-\gamma}\lambda(s)^{-\gamma}.$

Plugging these estimates into the formula for $\mathrm{I}$, we obtain

$\displaystyle\mathrm{I}$ $\displaystyle\lesssim$
$\displaystyle|x|^{-\gamma}\int_{(-81/100,t-\lambda(t)^{2\mu})\cap\\{L\lambda(s)\geq
6K\lambda(t)\\}}(t-s)^{-\frac{n-1-\gamma}{2}}\lambda(s)^{\frac{n-2}{2}}|\xi^{\prime}(s)|$
$\displaystyle\quad\quad\quad\quad\quad\quad\times\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}$
$\displaystyle+$ $\displaystyle
L^{-\gamma}|x|^{-\gamma}\int_{(-81/100,t-\lambda(t)^{2\mu})\cap\\{L\lambda(s)\leq
6K\lambda(t)\\}}(t-s)^{-\frac{n-1}{2}+\gamma}\lambda(s)^{\frac{n-2}{2}-\gamma}|\xi^{\prime}(s)|.$

Because

$\displaystyle(t-s)^{-\frac{n-1-\gamma}{2}}\lambda(s)^{\frac{n-4}{2}}\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}$
$\displaystyle=$
$\displaystyle(t-s)^{-\frac{n+2}{4}+\frac{\gamma}{2}}\left[\frac{\lambda(s)}{\sqrt{t-s}}\right]^{\frac{n-4}{2}}\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}$
$\displaystyle\lesssim$ $\displaystyle
L^{-\gamma}\lambda(s)^{-\gamma}(t-s)^{-\frac{n+2}{4}+\gamma},$

the first integral is controlled by

$\displaystyle\left[K\lambda(t)\right]^{-2\gamma}\int_{-81/100}^{t-\lambda(t)^{2\mu}}(t-s)^{-\frac{n+2}{4}+\gamma}|\lambda(s)\xi^{\prime}(s)|$
$\displaystyle\lesssim$ $\displaystyle
K^{-2\gamma}\lambda(t)^{-\frac{n-2}{2}\mu+2(\mu-1)\gamma}.$

In the last step we used only the estimate $|\lambda\xi^{\prime}|\leq C$.

Similarly, for the second integral, we have

$\displaystyle
L^{-\gamma}|x|^{-\gamma}\int_{(-81/100,t-\lambda(t)^{2\mu})\cap\\{L\lambda(s)\leq
4K\lambda(t)\\}}(t-s)^{-\frac{n-1}{2}+\gamma}\lambda(s)^{\frac{n-2}{2}-\gamma}|\xi^{\prime}(s)|$
$\displaystyle\lesssim$ $\displaystyle
L^{-\frac{n-4}{2}}K^{\frac{n-4}{2}-2\gamma}\lambda(t)^{\frac{n-4}{2}-2\gamma}\int_{-81/100}^{t-\lambda(t)^{2\mu}}(t-s)^{-\frac{n-1}{2}+\gamma}$
$\displaystyle\lesssim$ $\displaystyle
L^{-\frac{n-4}{2}}K^{\frac{n-4}{2}-2\gamma}\lambda(t)^{\frac{n-4}{2}-(n-3)\mu+2(\mu-1)\gamma}.$

Because

$-\frac{n-2}{2}\mu+2(\mu-1)\gamma<\frac{n-4}{2}-(n-3)\mu+2(\mu-1)\gamma,$

we get

$\mathrm{I}\lesssim\lambda(t)^{-\frac{n-2}{2}\mu+2(\mu-1)\gamma}.$

Estimate of $\mathrm{II}$. We have

$\displaystyle\mathrm{II}$ $\displaystyle\lesssim$
$\displaystyle|x|^{-\gamma}\int_{(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})\cap\\{L\lambda(s)\geq
6K\lambda(t)\\}}(t-s)^{-\frac{n-1-\gamma}{2}}\lambda(s)^{\frac{n-2}{2}}|\xi^{\prime}(s)|$
$\displaystyle\quad\quad\quad\quad\quad\quad\times\left[1+\frac{\sqrt{t-s}}{L\lambda(s)}\right]^{\gamma}e^{-c\frac{L^{2}\lambda(s)^{2}}{t-s}}$
$\displaystyle+$ $\displaystyle
L^{-\gamma}|x|^{-\gamma}\int_{(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})\cap\\{L\lambda(s)\leq
6K\lambda(t)\\}}(t-s)^{-\frac{n-1}{2}+\gamma}\lambda(s)^{\frac{n-2}{2}-\gamma}|\xi^{\prime}(s)|$
$\displaystyle\lesssim$
$\displaystyle\left[K\lambda(t)\right]^{-2\gamma}\int_{t-\lambda(t)^{2\mu}}^{t-K^{2\beta}\lambda(t)^{2}}(t-s)^{-\frac{n+2}{4}+\gamma}|\lambda(s)\xi^{\prime}(s)|$
$\displaystyle+$ $\displaystyle
L^{-\frac{n-4}{2}}K^{\frac{n-4}{2}-2\gamma}\lambda(t)^{\frac{n-4}{2}-2\gamma}\int_{t-\lambda(t)^{2\mu}}^{t-K^{2\beta}\lambda(t)^{2}}(t-s)^{-\frac{n-1}{2}+\gamma}|\lambda(s)\xi^{\prime}(s)|$
$\displaystyle\lesssim$
$\displaystyle\left[K^{-\frac{n-2}{2}\beta+2(\beta-1)\gamma}+L^{-\frac{n-4}{2}}K^{-\beta(n-3-2\gamma)+\frac{n-4}{2}-2\gamma}\right]$
$\displaystyle\quad\quad\times\lambda(t)^{-\frac{n-2}{2}}\|\lambda\xi^{\prime}\|_{L^{\infty}(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})}$
$\displaystyle\lesssim$ $\displaystyle
K^{-\frac{n-2}{2}\beta+2(\beta-1)\gamma}\lambda(t)^{-\frac{n-2}{2}}\|\lambda\xi^{\prime}\|_{L^{\infty}(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})}.$

Estimate of $\mathrm{III}$. As in the $\overline{\phi}_{3}$ case, we have

$\displaystyle\mathrm{II}$ $\displaystyle=$
$\displaystyle\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\lambda(s)^{\frac{n-2}{2}}|\xi^{\prime}(s)|$
$\displaystyle\quad\quad\quad\quad\times\left[\int_{B_{1}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}|y|^{1-n}dy\right]ds$
$\displaystyle\lesssim$
$\displaystyle|x|^{1-n}\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{2\gamma}\lambda(s)^{\frac{n-2}{2}}|\xi^{\prime}(s)|ds$
$\displaystyle\lesssim$ $\displaystyle
K^{1-n+2\beta+2(\beta-1)\gamma}\lambda(t)^{-\frac{n-2}{2}}\|\lambda\xi^{\prime}\|_{L^{\infty}(t-K^{2\beta}\lambda(t)^{2},t)}.$

Because

$1-n+2\beta+2(\beta-1)\gamma<-\frac{n-2}{2}\beta+2(\beta-1)\gamma,$

adding up estimates for $\mathrm{I}$, $\mathrm{II}$ and $\mathrm{III}$ we
finish the proof. ∎

######  Lemma 14.5 (Estimates on $\overline{\phi}_{5}$).

For $K\lambda(t)/2\leq|x|\leq 4K\lambda(t)$,

$\displaystyle\overline{\phi}_{5}(x,t)$ $\displaystyle\lesssim$
$\displaystyle\left[K^{-\left(\frac{n-2}{2}-\gamma\right)(\beta-1)}+K^{\frac{n-2}{2}+2\beta+2(\beta-1)\gamma}e^{-cK}+\frac{K^{\frac{n-2}{2}+2\beta+(\beta-1)\gamma}}{e^{cL}}\right]$
$\displaystyle\quad\quad\quad\quad\times\left[K\lambda(t)\right]^{-\frac{n-2}{2}}\left\|\left(\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\lambda\lambda^{\prime},\lambda\xi^{\prime}\right)\right\|_{L^{\infty}(t-\lambda(t)^{2\mu},t)}$
$\displaystyle+$ $\displaystyle\lambda(t)^{-\frac{n-2}{2}\mu+(\mu-1)\gamma}.$

###### Proof.

By the heat kernel representation, we have

$\displaystyle\overline{\phi}_{5}(x,t)$ $\displaystyle\leq$
$\displaystyle\int_{-81/100}^{t}\int_{B_{L\lambda(s)}^{c}}G(x,y;t,s)\lambda(s)^{-\frac{n+2}{2}}e^{-c\frac{|y|}{\lambda(s)}}$
$\displaystyle\times\left[\frac{|a(s)|}{\lambda(s)}+\left|\lambda(s)a^{\prime}(s)-\mu_{0}\frac{a(s)}{\lambda(s)}\right|+\frac{|a(s)|}{\lambda(s)}\left(|\lambda(s)\lambda^{\prime}(s)|+|\lambda(s)\xi^{\prime}(s)|\right)\right].$

We still divide this integral into three parts, $\mathrm{I}$ being on the
interval $(-81/100,t-\lambda(t)^{2\mu})$, $\mathrm{II}$ being on the interval
$(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})$, and $\mathrm{III}$ on
$(t-K^{2\beta}\lambda(t)^{2},t)$.

Estimate of $\mathrm{I}$. For $\mathrm{I}$, just using the estimate

$\left\|\left(\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\frac{a}{\lambda}\left(|\lambda\lambda^{\prime}|+|\lambda\xi^{\prime}|\right)\right)\right\|_{L^{\infty}(-1,t)}\leq
C,$

by the operator bound on $G$ (see (14.19)) and Hölder inequality, we get

$\displaystyle\mathrm{I}$ $\displaystyle\lesssim$
$\displaystyle\int_{-81/100}^{t-\lambda(t)^{2\mu}}\int_{B_{1}}G(x,t;y,s)\lambda(s)^{-\frac{n+2}{2}}e^{-c\frac{|y|}{\lambda(s)}}dyds$
$\displaystyle\lesssim$
$\displaystyle\int_{-81/100}^{t-\lambda(t)^{2\mu}}(t-s)^{-\frac{n+2}{4}}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}ds$
$\displaystyle\lesssim$
$\displaystyle\lambda(t)^{-\frac{n-2}{2}\mu+(\mu-1)\gamma}.$

Estimate of $\mathrm{II}$. For $\mathrm{II}$, in the same way we obtain

$\displaystyle\mathrm{II}$ $\displaystyle\lesssim$
$\displaystyle\left\|\left(\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\frac{a}{\lambda}\left(|\lambda\lambda^{\prime}|+|\lambda\xi^{\prime}|\right)\right)\right\|_{L^{\infty}(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})}$
$\displaystyle\quad\times\int_{t-\lambda(t)^{2\mu}}^{t-K^{2\beta}\lambda(t)^{2}}\int_{B_{1}}G(x,y;t,s)\lambda(s)^{-\frac{n+2}{2}}e^{-c\frac{|y|}{\lambda(s)}}dyds$
$\displaystyle\lesssim$ $\displaystyle\left\|\left(\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\lambda\lambda^{\prime},\lambda\xi^{\prime}\right)\right\|_{L^{\infty}(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})}$
$\displaystyle\quad\times\int_{-1}^{t-K^{2\beta}\lambda(t)^{2}}(t-s)^{-\frac{n+2}{4}}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}ds$
$\displaystyle\lesssim$ $\displaystyle
K^{-\frac{n-2}{2}\beta+(\beta-1)\gamma}\lambda(t)^{-\frac{n-2}{2}}\left\|\left(\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\lambda\lambda^{\prime},\lambda\xi^{\prime}\right)\right\|_{L^{\infty}(t-\lambda(t)^{2\mu},t-K^{2\beta}\lambda(t)^{2})}.$

Estimate of $\mathrm{III}$. In this case, first note that

$\displaystyle\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\int_{B_{L\lambda(s)}^{c}}G(x,t;y,s)\lambda(s)^{-\frac{n+2}{2}}e^{-c\frac{|y|}{\lambda(s)}}dyds$
$\displaystyle\lesssim$
$\displaystyle\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\int_{B_{L\lambda(s)}^{c}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\times\lambda(s)^{-\frac{n+2}{2}}e^{-c\frac{|y|}{\lambda(s)}}dyds$
$\displaystyle\lesssim$
$\displaystyle\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\lambda(s)^{-\frac{n+2}{2}}$
$\displaystyle\quad\times\left[\underbrace{\int_{B_{L\lambda(s)}^{c}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}e^{-c\frac{|y|}{\lambda(s)}}dy}_{\mbox{
Lemma \ref{lem heat kernel integral 2}}}\right]ds$ $\displaystyle\lesssim$
$\displaystyle\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\lambda(s)^{-\frac{n+2}{2}}$
$\displaystyle\quad\times\left[e^{-c\frac{|x|}{\lambda(s)}}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}+\left(1+\frac{\lambda(s)}{\sqrt{t-s}}\right)^{n+\gamma}e^{-cL-c\frac{|x|^{2}}{t-s}}\right]ds$
$\displaystyle\lesssim$
$\displaystyle\left[K^{2\beta+2(\beta-1)\gamma}e^{-cK}+K^{2\beta+(\beta-1)\gamma}e^{-cL}\right]\lambda(t)^{-\frac{n-2}{2}},$

where we have used (14.21) to deduce that last inequality.

Then similar to the estimate of $\mathrm{II}$, we get

$\displaystyle\mathrm{III}$ $\displaystyle\lesssim$
$\displaystyle\left[K^{2\beta+2(\beta-1)\gamma}e^{-cK}+K^{2\beta+(\beta-1)\gamma}e^{-cL}\right]\lambda(t)^{-\frac{n-2}{2}}$
$\displaystyle\quad\quad\quad\quad\times\left\|\left(\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\lambda\lambda^{\prime},\lambda\xi^{\prime}\right)\right\|_{L^{\infty}(t-K^{2\beta}\lambda(t)^{2},t)}.$

Adding up estimates for $\mathrm{I}$, $\mathrm{II}$ and $\mathrm{III}$ we
finish the proof. ∎

####  14.3. Estimate of $\mathcal{O}(t)$

As an application of the pointwise estimates on
$\overline{\phi}_{1},\cdots,\overline{\phi}_{5}$ obtained in the previous
subsection, we give an estimate on $\mathcal{O}(t)$.

######  Proposition 14.6.

There exist two constants $C(K,L)$ and $\sigma(K,L)\ll 1$ (depending on $K$
and $L$) such that for any $t\in(-81/100,81/100)$,

$\mathcal{O}(t)\leq
C(K,L)\lambda(t)^{2+2\kappa}+\sigma(K,L)\left\|\left(\mathcal{I},\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\lambda\lambda^{\prime},\lambda\xi^{\prime}\right)\right\|_{L^{\infty}(t-\lambda(t)^{2\mu},t)}.$
(14.24)

###### Proof.

First by the five lemmas in the previous subsection, we obtain

$\displaystyle\left[K\lambda(t)\right]^{\frac{n-2}{2}}\sup_{B_{2K\lambda(t)}(\xi(t))\setminus
B_{K\lambda(t)}(\xi(t))}|\phi(x,t)|$ $\displaystyle\leq$
$\displaystyle\left[K\lambda(t)\right]^{\frac{n-2}{2}}\sup_{B_{2K\lambda(t)}\setminus
B_{K\lambda(t)}}\left[\overline{\phi}_{1}(x,t)+\overline{\phi}_{2}(x,t)+\overline{\phi}_{3}(x,t)+\overline{\phi}_{4}(x,t)+\overline{\phi}_{5}(x,t)\right]$
$\displaystyle\leq$ $\displaystyle
C(K,L)\lambda(t)^{2+2\kappa}+\sigma(K,L)\left\|\left(\mathcal{I},\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\lambda\lambda^{\prime},\lambda\xi^{\prime}\right)\right\|_{L^{\infty}(t-\lambda(t)^{2\mu},t)}.$

Next, by a rescaling and standard interior gradient estimates for heat
equation, we get

$\displaystyle\left[K\lambda(t)\right]^{\frac{n}{2}}\sup_{B_{2K\lambda(t)}(\xi(t))\setminus
B_{K\lambda(t)}(\xi(t))}|\nabla\phi_{1}(x,t)|$ $\displaystyle\lesssim$
$\displaystyle\left[K\lambda(t)\right]^{\frac{n-2}{2}}\sup_{\left(B_{4K\lambda(t)(\xi(t))}\setminus
B_{K\lambda(t)/2(\xi(t))}\right)\times(t-K^{2}\lambda(t)^{2},t)}|\phi_{1}(y,s)|$
$\displaystyle\lesssim$ $\displaystyle
K^{\frac{n-2}{2}-\gamma}\lambda(t)^{\frac{n-2}{2}-\gamma}.$

The same estimate holds for $\phi_{2}$, if we note that $F_{2}\equiv 0$ in
$B_{4K\lambda(t)(\xi(t))}\setminus B_{K\lambda(t)/2(\xi(t))}$.

In the same way, for $\phi_{3}$ we have

$\displaystyle\left[K\lambda(t)\right]^{\frac{n}{2}}\sup_{B_{2K\lambda(t)}(\xi(t))\setminus
B_{K\lambda(t)}(\xi(t))}|\nabla\phi_{3}(x,t)|$ $\displaystyle\lesssim$
$\displaystyle\left[K\lambda(t)\right]^{\frac{n-2}{2}}\sup_{\left(B_{4K\lambda(t)(\xi(t))}\setminus
B_{K\lambda(t)/2(\xi(t))}\right)\times(t-K^{2}\lambda(t)^{2},t)}\left[|\phi_{3}(y,s)|+\left(K\lambda(t)\right)^{2}|F_{3}(y,s)|\right]$
$\displaystyle\lesssim$ $\displaystyle
K^{-\frac{n-2}{2}(\beta-1)+2(\beta-1)\gamma}\|\lambda\lambda^{\prime}\|_{L^{\infty}(t-\lambda(t)^{2\mu},t)}+K^{\frac{n-2}{2}}\lambda(t)^{2+2\kappa}$
$\displaystyle+$ $\displaystyle
K^{-\frac{n-6}{2}}\left\|\lambda\lambda^{\prime}\right\|_{L^{\infty}(t-K^{2}\lambda(t)^{2},t)}.$

The same estimates hold for $\phi_{4}$ and $\phi_{5}$.

Putting these gradient estimates together and arguing as in (14.3), we get the
estimate on the remaining terms in $\mathcal{O}(t)$. ∎

###  15\. A Harnack inequality for $\lambda$

In this section, we combine the estimates in the previous two sections to
establish an Harnack inequality for $\lambda$.

For any $t\in(-81/100,81/100)$, denote

$\mathcal{D}(t):=\|\phi_{in}(t)\|_{\alpha;1+\theta}+\left|\lambda(t)\xi^{\prime}(t)\right|+\left|\lambda(t)\lambda^{\prime}(t)\right|+\left|\lambda(t)a^{\prime}(t)-\mu_{0}\frac{a(t)}{\lambda(t)}\right|+\left|\frac{a(t)}{\lambda(t)}\right|.$

For $r\in(0,81/100)$, let

$g(r):=\sup_{-r<t<r}\mathcal{D}(t)$

and

$M(r):=\sup_{-r<t<r}\lambda(t)^{2},\quad m(r):=\inf_{-r<t<r}\lambda(t)^{2}.$

By Proposition 13.4 and Lemma 13.6, there exist three universal constants $C$,
$T\gg 1$ and $\sigma\ll 1$ such that

$\mathcal{D}(t)\leq\sigma\sup_{[t-T\lambda(t)^{2},t+T\lambda(t)^{2}]}\mathcal{D}(s)+C\sup_{[t-T\lambda(t)^{2},t+T\lambda(t)^{2}]}\mathcal{O}(s).$
(15.1)

By Proposition 14.6,

$\mathcal{O}(t)\leq
C(K,L)\lambda(t)^{2+2\kappa}+\sigma(K,L)\sup_{[t-\lambda(t)^{2\mu},t]}\mathcal{D}(s).$
(15.2)

Combining these two estimates we get

$\mathcal{D}(t)\leq\frac{1}{2}\sup_{[t-2\lambda(t)^{2\mu},t+2K\lambda(t)^{2}]}\mathcal{D}(s)+C\sup_{[t-2\lambda(t)^{2\mu},t+2K\lambda(t)^{2}]}\lambda(s)^{2+2\kappa}.$
(15.3)

For any $r\in(0,1)$, taking supremum over
$t\in\left(-r+2M(r)^{\mu},r-2M(r)^{\mu}\right)$ in (15.3), we obtain

$g\left(r-2M(r)^{\mu}\right)\leq\frac{1}{2}g(r)+CM(r)^{1+\kappa}.$ (15.4)

For any $r_{1}<r_{2}$, an iteration of this inequality from $r_{2}$ to $r_{1}$
in $\lfloor(r_{2}-r_{1})/M(r_{2})^{\mu}\rfloor$ steps leads to

$g(r_{1})\leq
g(r_{2})e^{-c(r_{2}-r_{1})M(r_{2})^{-\mu}}+CM(r_{2})^{1+\kappa}.$

For any $r\in(0,1)$, $M(r)\ll 1$, so

$e^{-cM(r)^{-\mu/2}}\lesssim M(r)^{1+\kappa}.$

Hence by choosing $r_{2}=r$ and $r_{1}=r-M(r)^{\mu/2}$, we get

$g\left(r-M(r)^{\mu/2}\right)\lesssim M(r)^{1+\kappa}.$

By this estimate, integrating $\lambda\lambda^{\prime}$ on
$[-r+M(r)^{\mu/2},r-M(r)^{\mu/2}]$, we find a constant $C_{H}$ such that

$M\left(r-M(r)^{\mu/2}\right)\leq
m\left(r-M(r)^{\mu/2}\right)+C_{H}M(r)^{1+\kappa}.$ (15.5)

######  Lemma 15.1.

There exists an $r_{H}\in[1/2,3/4]$ such that

$M\left(r_{H}-M(r_{H})^{\mu/2}\right)\geq 2C_{H}M(r_{H})^{1+\kappa}.$

###### Proof.

Assume that for any $r\in[1/2,3/4]$,

$M\left(r-M(r)^{\mu/2}\right)\leq 2C_{H}M(r)^{1+\kappa}.$ (15.6)

Set

$r_{0}:=3/4,\quad a_{0}:=M(r_{0})$

and for $k\in\mathbb{N}$,

$r_{k+1}:=r_{k}-a_{k}^{\mu/2},\quad a_{k+1}:=M(r_{k+1}).$

Then (15.6) says

$a_{k+1}\leq 2C_{H}a_{k}^{1+\kappa}.$

By our assumption, $a_{0}\ll 1$. An induction gives

$a_{k}\leq a_{0}^{k+1},\quad r_{k}\geq
r_{0}-\sum_{k=0}^{+\infty}a_{0}^{(k+1)\frac{\mu}{2}}\geq 1/2.$

As a consequence, $M(1/2)=0$. This is impossible. ∎

For this $r_{H}$, (15.5) can be written as a Harnack inequality

$m(r_{H})\geq\left[1-C_{H}M(r_{H})^{\kappa}\right]M(r_{H})\geq\frac{1}{2}M(r_{H}).$
(15.7)

After a scaling $u(x,t)\mapsto r_{H}^{(n-2)/2}u(r_{H}x,r_{H}^{2}t)$, from here
to Section 18 we work in the following setting:

  * •

we denote

$\varepsilon:=\lambda(0),$ (15.8)

and after a translation in the spatial direction, we assume $\xi(0)=0$;

  * •

by the above Harnack inequality (15.7), for any $t\in[-1,1]$,

$\lambda(t)=\varepsilon+O\left(\varepsilon^{1+\kappa}\right)$ (15.9)

and

$\|\phi_{in}(t)\|+\left|\lambda(t)\lambda^{\prime}(t)\right|+\left|\lambda(t)\xi^{\prime}(t)\right|+\left|\lambda(t)a^{\prime}(t)-\mu_{0}\frac{a(t)}{\lambda(t)}\right|+\left|\frac{a(t)}{\lambda(t)}\right|\lesssim\varepsilon^{1+\kappa}.$
(15.10)

  * •

integrating $\xi^{\prime}$ and using the above estimate, we obtain

$|\xi(t)|\lesssim\varepsilon^{\kappa},\quad\mbox{for any}~{}~{}t\in[-1,1].$
(15.11)

###  16\. Inner problem and outer problem again

In this section, under the assumptions (15.8)-(15.10), we prove

######  Proposition 16.1.

For any $\gamma>0$, there exists a constant $C(\gamma)$ such that

$g\left(\frac{8}{9}\right)\leq C(\gamma)\varepsilon^{\frac{n-2}{2}-\gamma}.$
(16.1)

This gives an improvement on the estimate of scaling parameters etc.. A more
precise estimate on $\phi$ also follows:

######  Proposition 16.2.

For any $\gamma>0$, there exists a constant $C(\gamma)>0$ such that

$|\phi(x,t)|\leq
C(\gamma)\left(\varepsilon+|x-\xi(t)|\right)^{-\gamma}\quad\mbox{in}\quad
Q_{7/8}.$ (16.2)

To prove these propositions, estimates on $\overline{\phi}_{2}$,
$\overline{\phi}_{3}$, $\overline{\phi}_{4}$ and $\overline{\phi}_{5}$ in
Section 14 are not sufficient. We need to upgrade them.

######  Lemma 16.3 (Estimate on $\overline{\phi}_{2}$).

If $|x|\geq K\lambda(t)/2$, then

$\overline{\phi}_{2}(x,t)\lesssim\left[K^{-\left(\frac{n-2}{2}-\gamma\right)\beta}\frac{\varepsilon^{\gamma}}{|x|^{\gamma}}+L^{\frac{n-2}{2}-\gamma}K^{\beta\gamma}\frac{\varepsilon^{n-2}}{|x|^{n-2}}\right]\varepsilon^{-\frac{n-2}{2}}\|\mathcal{I}\|_{L^{\infty}(-1,t)}.$
(16.3)

###### Proof.

As in the proof of Lemma 14.2, we have the heat kernel representation for
$\overline{\phi}_{2}$. But now we just divide the integral into two parts,
$\mathrm{I}$ being on $(-1,t-K^{2\beta}\lambda(t)^{2})$ and $\mathrm{II}$ on
$(t-K^{2\beta}\lambda(t)^{2},t)$.

For $\mathrm{I}$, similar to (14.2), we have

$\displaystyle\mathrm{I}$ $\displaystyle\lesssim$
$\displaystyle|x|^{-\gamma}\int_{-1}^{t-K^{2\beta}\lambda(t)^{2}}(t-s)^{-\frac{n+2}{4}+\frac{\gamma}{2}}\mathcal{I}(s)ds$
$\displaystyle\lesssim$ $\displaystyle
K^{-\left(\frac{n-2}{2}-\gamma\right)\beta}\lambda(t)^{-\frac{n-2}{2}+\gamma}|x|^{-\gamma}\|\mathcal{I}\|_{L^{\infty}(-1,t-K^{2\beta}\lambda(t)^{2})}$
$\displaystyle\lesssim$ $\displaystyle
K^{-\left(\frac{n-2}{2}-\gamma\right)\beta}\varepsilon^{-\frac{n-2}{2}+\gamma}|x|^{-\gamma}\|\mathcal{I}\|_{L^{\infty}(-1,t-K^{2\beta}\lambda(t)^{2})}$

The estimate for $\mathrm{II}$ is almost the same as the one for
$\mathrm{III}$ in the proof of Lemma 14.2, that is,

$\displaystyle\mathrm{II}$ $\displaystyle\lesssim$
$\displaystyle\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\frac{\mathcal{I}(s)}{L^{\alpha}\lambda(s)^{\alpha}}\left(\frac{|\lambda^{\prime}(s)|}{\lambda(s)}+\frac{|\xi^{\prime}(s)|}{L\lambda(s)}+\frac{1}{L^{2}\lambda(s)^{2}}\right)\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}$
$\displaystyle\times\left[\int_{B_{2L\lambda(s)}\setminus
B_{L\lambda(s)}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}dy\right]$
$\displaystyle\lesssim$
$\displaystyle\|\mathcal{I}\|_{L^{\infty}(t-K^{2\beta}\lambda(t)^{2},t)}L^{\frac{n-2}{2}}\lambda(t)^{\frac{n-2}{2}}\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{L\lambda(t)}\right)^{\gamma}$
$\displaystyle\lesssim$
$\displaystyle\|\mathcal{I}\|_{L^{\infty}(t-K^{2\beta}\lambda(t)^{2},t)}L^{\frac{n-2}{2}}\lambda(t)^{\frac{n-2}{2}}\left(1+\frac{K^{\beta}}{L}\right)^{\gamma}\int_{t-K^{2\beta}\lambda(t)^{2}}^{t}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x|^{2}}{t-s}}$
$\displaystyle\lesssim$
$\displaystyle\|\mathcal{I}\|_{L^{\infty}(t-K^{2\beta}\lambda(t)^{2},t)}L^{\frac{n-2}{2}}\lambda(t)^{\frac{n-2}{2}}\left(1+\frac{K^{\beta}}{L}\right)^{\gamma}|x|^{2-n}$
$\displaystyle\lesssim$ $\displaystyle
L^{\frac{n-2}{2}-\gamma}K^{\beta\gamma}\varepsilon^{\frac{n-2}{2}}|x|^{2-n}\|\mathcal{I}\|_{L^{\infty}(t-K^{2}\lambda(t)^{2},t)}.$

Putting these two estimates together, we obtain (16.4). ∎

######  Corollary 16.4.

If $|x|\geq K\lambda(t)/2$, then

$\overline{\phi}_{2}(x,t)\lesssim\left[K^{-\frac{n-2}{2}(\beta-1)+\beta\gamma}+\frac{L^{\frac{n-2}{2}-\gamma}}{K^{\frac{n-2}{2}-\beta\gamma}}\right]\left(K\varepsilon\right)^{-\frac{n-2}{2}}\|\mathcal{I}\|_{L^{\infty}(-1,t)}.$
(16.4)

######  Lemma 16.5 (Estimate on $\overline{\phi}_{3}$).

If $|x|\geq K\lambda(t)/2$, then

$\overline{\phi}_{3}(x,t)\lesssim\varepsilon^{\frac{n-6}{2}}|x|^{4-n}\|\lambda\lambda^{\prime}\|_{L^{\infty}(-1,t)}.$
(16.5)

###### Proof.

As in the proof of Lemma 14.3, we still have the heat kernel representation

$\displaystyle\overline{\phi}_{3}(x,t)$ $\displaystyle\lesssim$
$\displaystyle\int_{-81/100}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\lambda(s)^{\frac{n-4}{2}}\left|\lambda^{\prime}(s)\right|$
$\displaystyle\times\left[\int_{B_{1}}(t-s)^{-\frac{n}{2}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}e^{-c\frac{|x-y|^{2}}{t-s}}|y|^{2-n}dy\right]ds$
$\displaystyle\lesssim$
$\displaystyle\varepsilon^{\frac{n-6}{2}}\|\lambda\lambda^{\prime}\|_{L^{\infty}(-1,t)}$
$\displaystyle\times\int_{-81/100}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{2\gamma}\left(\sqrt{t-s}+|x|\right)^{2-n}ds$
$\displaystyle\lesssim$
$\displaystyle\varepsilon^{\frac{n-6}{2}}\|\lambda\lambda^{\prime}\|_{L^{\infty}(-1,t)}|x|^{4-n}.$

Here we have used Lemma B.1 to deduce the second inequality. ∎

######  Corollary 16.6.

If $|x|\geq K\lambda(t)/2$, then

$\overline{\phi}_{3}(x,t)\lesssim
K^{-\frac{n-6}{2}}\left(K\varepsilon\right)^{-\frac{n-2}{2}}\|\lambda\lambda^{\prime}\|_{L^{\infty}(-1,t)}.$
(16.6)

######  Lemma 16.7 (Estimate on $\overline{\phi}_{4}$).

If $|x|\geq K\lambda(t)/2$, then

$\overline{\phi}_{4}(x,t)\lesssim\varepsilon^{\frac{n-4}{2}}|x|^{3-n}\|\lambda\xi^{\prime}\|_{L^{\infty}(-1,t)}.$
(16.7)

###### Proof.

As in the proof of Lemma 14.4, we still have the heat kernel representation

$\displaystyle\overline{\phi}_{4}(x,t)$ $\displaystyle\lesssim$
$\displaystyle\int_{-81/100}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}\lambda(s)^{\frac{n-2}{2}}\left|\xi^{\prime}(s)\right|$
$\displaystyle\times\left[\int_{B_{1}}(t-s)^{-\frac{n}{2}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}e^{-c\frac{|x-y|^{2}}{t-s}}|y|^{1-n}dy\right]ds$
$\displaystyle\lesssim$
$\displaystyle\varepsilon^{\frac{n-4}{2}}\|\lambda\xi^{\prime}\|_{L^{\infty}(-1,t)}$
$\displaystyle\times\int_{-81/100}^{t}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{2\gamma}\left(\sqrt{t-s}+|x|\right)^{1-n}ds$
$\displaystyle\lesssim$
$\displaystyle\varepsilon^{\frac{n-4}{2}}\|\lambda\xi^{\prime}\|_{L^{\infty}(-1,t)}|x|^{3-n}.$

Here we have used Lemma B.1 to deduce the second inequality. ∎

######  Corollary 16.8.

If $|x|\geq K\lambda(t)/2$, then

$\overline{\phi}_{4}(x,t)\lesssim
K^{-\frac{n-4}{2}}\left(K\varepsilon\right)^{-\frac{n-2}{2}}\|\lambda\xi^{\prime}\|_{L^{\infty}(-1,t)}.$
(16.8)

######  Lemma 16.9 (Estimate on $\overline{\phi}_{5}$).

If $|x|\geq K\lambda(t)/2$, then

$\overline{\phi}_{5}(x,t)\lesssim\varepsilon^{\frac{n-4}{2}}|x|^{3-n}\left\|\left(\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\lambda\lambda^{\prime},\lambda\xi^{\prime}\right)\right\|_{L^{\infty}(-1,t)}.$
(16.9)

###### Proof.

For $|y|\geq L\lambda(s)$,

$\lambda(s)^{-\frac{n+2}{2}}e^{-c|y|/\lambda(s)}\lesssim\lambda(s)^{\frac{n-4}{2}}|y|^{1-n}.$

Hence

$\displaystyle\partial_{t}\overline{\phi}_{5}-\Delta\overline{\phi}_{5}-\left[C+\frac{\delta}{|x|^{2}}\right]\overline{\phi}_{5}+\xi^{\prime}(t)\cdot\nabla\overline{\phi}_{5}$
$\displaystyle\lesssim$
$\displaystyle\left[\frac{|a(t)|}{\lambda(t)}+\Big{|}\lambda(t)a^{\prime}(t)-\mu_{0}\frac{a(t)}{\lambda(t)}\Big{|}+|\lambda(t)\lambda^{\prime}(t)|+|\lambda(t)\xi^{\prime}(t)|\right]\lambda(t)^{\frac{n-4}{2}}|x|^{1-n}\chi_{B_{L\lambda(t)}^{c}}.$

Then we can proceed as in the previous lemma to conclude. ∎

######  Corollary 16.10.

If $|x|\geq K\lambda(t)/2$, then

$\overline{\phi}_{5}(x,t)\lesssim
K^{-\frac{n-4}{2}}\left(K\varepsilon\right)^{-\frac{n-2}{2}}\left\|\left(\frac{a}{\lambda},\lambda
a^{\prime}-\mu_{0}\frac{a}{\lambda},\lambda\lambda^{\prime},\lambda\xi^{\prime}\right)\right\|_{L^{\infty}(-1,t)}.$
(16.10)

Combining Corollary 16.4-Corollary 16.10 with Lemma 14.1, proceeding as in the
proof of Proposition 14.6 (to estimate $|\nabla\phi_{1}|$-$|\nabla\phi_{5}|$),
we find two constants $C(K,L)$ and $\sigma(K,L)\ll 1$ such that, for any
$t\in(-81/100,81/100)$,

$\mathcal{O}(t)\leq
C(K,L)\varepsilon^{\frac{n-2}{2}-\gamma}+\sigma(K,L)\sup_{-1<s<t}\mathcal{D}(s).$
(16.11)

Combining this inequality with Proposition 13.4, we obtain (compare this with
(15.4))

$\mathcal{D}(t)\leq\frac{1}{2}\sup_{[t-C\varepsilon^{2},t+C\varepsilon^{2}]}\mathcal{D}(s)+C\varepsilon^{\frac{n-2}{2}-\gamma}.$
(16.12)

Here we have used the fact that, by (15.9) and the definition of inner time
variable $\tau$ in Section 13, we have

$\frac{d\tau}{dt}\sim\varepsilon^{-2}.$ (16.13)

Similar to Section 15, an iteration of (16.12) from $r=1$ to $r=8/9$ gives
(16.1). This finishes the proof of Proposition 16.1. Plugging (16.1) into
Lemma 16.3-Lemma 16.9, we get Proposition 16.2.

###  17\. Improved estimates on $\phi$

From now on we need to write down the dependence on $i$ explicitly. For
$u_{i}$, the decomposition in Section 12 reads as

$\phi_{i}(x,t):=u_{i}(x,t)-W_{\xi_{i}(t),\lambda_{i}(t)}(x)-a_{i}(t)Z_{0,\xi_{i}(t),\lambda_{i}(t)}(x).$
(17.1)

The parameters satisfy the assumptions (15.8)-(15.10). In particular,

$\varepsilon_{i}:=\lambda_{i}(0),\quad\xi_{i}(0)=0.$

In this section we prove some uniform (in $\varepsilon_{i}$) estimates on
$\phi_{i}$.

For simplicity of notation, we will denote

$W_{i}(x,t):=W_{\xi_{i}(t),\lambda_{i}(t)}(x),\quad
Z_{\ast,i}(x,t):=\left(Z_{j,\xi_{i}(t),\lambda_{i}(t)}(x)\right)_{j=0,\cdots,n+1}.$

####  17.1. Uniform $L^{\infty}$ bound

The first one is a uniform $L^{\infty}$ bound on $\phi_{i}$.

######  Proposition 17.1.

There exists a universal constant $C$ (independent of $i$) such that

$|\phi_{i}(x,t)|\leq C\quad\mbox{in}\quad Q_{6/7}.$ (17.2)

First by (16.2), we get

$u_{i}(x,t)\lesssim\left(\frac{\varepsilon_{i}}{\varepsilon_{i}^{2}+|x-\xi_{i}(t)|^{2}}\right)^{\frac{n-2}{2}}+|x-\xi_{i}(t)|^{-\gamma}.$
(17.3)

Hence $\overline{\phi}_{i,1}$ satisfies, for some constant $C_{0}>0$,

$\partial_{t}\overline{\phi}_{i,1}-\Delta\overline{\phi}_{i,1}+\xi_{i}^{\prime}(t)\cdot\nabla\overline{\phi}_{i,1}\leq
C_{0}\left[\varepsilon_{i}^{2}|x|^{-4}\chi_{B_{L\varepsilon_{i}}^{c}}+|x|^{-(p-1)\gamma}\right]\overline{\phi}_{i,1}.$
(17.4)

As in the proof of Lemma 14.1, for each $\varepsilon>0$, consider the problem

$\left\\{\begin{aligned} &-\Delta
w_{\varepsilon}=C_{0}\varepsilon^{2}|x|^{-4}\chi_{B_{L\varepsilon}^{c}}w_{\varepsilon}\quad&\mbox{in
}~{}~{}B_{1},\\\ &w_{\varepsilon}=1\quad&\mbox{on }~{}~{}\partial
B_{1}.\end{aligned}\right.$ (17.5)

######  Lemma 17.2.

There exists a radially symmetric solution $w_{\varepsilon}$ of (17.5).
Moreover, if $L$ is universally large, for any $\varepsilon>0$,

$1\leq w_{\varepsilon}\leq 2\quad\mbox{in }B_{1}.$

###### Proof.

Consider the initial value problem

$\left\\{\begin{aligned}
&w^{\prime\prime}(s)+(n-2)w^{\prime}(s)+C_{0}e^{-2s}w(s)=0\quad\mbox{in
}~{}~{}[\log L,+\infty),\\\ &w(\log L)=1,\quad w^{\prime}(\log
L)=0.\end{aligned}\right.$

Global existence and uniqueness of the solution follows from standard ordinary
differential equation theory.

If $w>0$ in $[\log L,s_{0}]$ for some $s_{0}>\log L$, then $w^{\prime}<0$, and
consequently, $0<w<1$ in $(\log L,s_{0})$. Integrating the equation of $w$ we
get

$\displaystyle w^{\prime}(s)$ $\displaystyle=$ $\displaystyle-
C_{0}e^{-(n-2)s}\int_{\log L}^{s}e^{(n-4)\tau}w(\tau)d\tau$
$\displaystyle\geq$ $\displaystyle-\frac{C_{0}}{n-4}e^{-2s}.$

Hence, if $L$ is large enough,

$w(s_{0})\geq 1-\frac{C_{0}}{n-4}L^{-2}\geq\frac{1}{2}.$

This holds for any $s_{0}$ provided that $w>0$ in $[\log L,s_{0}]$, so
$w(s)\geq 1/2$ for any $s\geq\log L$.

For each $\varepsilon>0$, the function

$w_{\varepsilon}(x):=\frac{w\left(\log(\varepsilon^{-1}|x|)\right)}{w\left(\log\varepsilon^{-1}\right)}$

satisfies all of the requirements. ∎

######  Corollary 17.3.

There exists a constant $C$ independent of $\varepsilon$ such that

$|\nabla w_{\varepsilon}|\leq C\quad\mbox{in }~{}~{}B_{1}.$

###### Proof.

By definition, we have

$|\nabla
w_{\varepsilon}(x)|=\frac{w^{\prime}\left(\log(\varepsilon^{-1}|x|)\right)}{w\left(\log\varepsilon^{-1}\right)}\frac{\varepsilon}{|x|}.$

If $|x|\leq L\varepsilon$, $w^{\prime}=0$ and there is nothing to prove. For
$|x|\geq L\varepsilon$, by (17.1),

$\left|w^{\prime}\left(\log(\varepsilon^{-1}|x|)\right)\right|\leq\frac{C_{0}}{n-4}.$

Because $w\left(\log\varepsilon^{-1}\right)\geq 1/2$, we get

$|\nabla w_{\varepsilon}(x)|\leq\frac{2C_{0}}{n-4}.\qed$

The function $\varphi_{i}:=\overline{\phi}_{i,1}/w_{\varepsilon_{i}}$
satisfies

$\partial_{t}\varphi_{i}-w_{\varepsilon_{i}}^{-2}\mbox{div}\left(w_{\varepsilon_{i}}^{2}\nabla\varphi_{i}\right)+\xi_{i}^{\prime}\cdot\nabla\varphi_{i}\leq\left[C_{0}|x|^{-(p-1)\gamma}-\xi_{i}^{\prime}\cdot\nabla
w_{\varepsilon_{i}}\right]\varphi_{i}.$

By Lemma 17.2, this is a uniformly parabolic equation. Because $\gamma$ is
very small, $|x|^{-(p-1)\gamma}\in L^{2n}(B_{1})$. By (10.1) and Corollary
17.3, $\xi_{i}^{\prime}\cdot\nabla w_{\varepsilon_{i}}$ are uniformly bounded
in $L^{\infty}(Q_{6/7})$. Then standard Moser iteration gives

######  Lemma 17.4.

There exists a universal constant $C$ independent of $i$ such that

$\sup_{Q_{6/7}}\varphi_{i}\lesssim\int_{Q_{1}}\varphi_{i}.$ (17.7)

Combining this lemma with Lemma 17.2 and the definition of
$\overline{\phi}_{i,1}$ (see (14.13)), we get

$\|\phi_{i,1}\|_{L^{\infty}(Q_{6/7})}\leq C.$

Starting from this estimate, following the iteration argument in Section 16,
we deduce that

$g(5/6)\lesssim\varepsilon_{i}^{\frac{n-2}{2}}.$ (17.8)

Substituting this estimate into Lemma 16.3-Lemma 16.9, and noting that now the
heat kernel for $\partial_{t}-\Delta-\xi_{i}^{\prime}\cdot\nabla-V_{i}$ enjoys
the standard Gaussian bound

$G_{i}(x,t;y,s)\lesssim(t-s)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}},$

i.e. we can take $\gamma=0$, we obtain

######  Lemma 17.5.

For any $x\in B_{K\varepsilon}^{c}$ and $t\in(-5/6,5/6)$, we have

$|\phi_{i,2}(x,t)|+|\phi_{i,3}(x,t)|+|\phi_{i,4}(x,t)|+|\phi_{i,5}(x,t)|\lesssim_{K}\frac{\varepsilon^{n-4}}{|x|^{n-4}}.$

Combining (17.7), (17.8) with this lemma, we finish the proof of Proposition
17.1.

Once we know that $\phi_{i}$ are uniformly bounded in $Q_{1}$ (after a scaling
of the domain), the above estimates are improved to

######  Proposition 17.6.

$\displaystyle\|\phi_{i,1}\|_{L^{\infty}(Q_{5/6})}\lesssim\|\phi_{i}\|_{L^{\infty}(Q_{1})},$
$\displaystyle
g(5/6)\lesssim\varepsilon_{i}^{\frac{n-2}{2}}\left(\|\phi_{i}\|_{L^{\infty}(Q_{1})}+e^{-c\varepsilon_{i}^{-2}}\right),$

and for any $(x,t)\in Q_{5/6}$,

$|\phi_{i,2}(x,t)|+|\phi_{i,3}(x,t)|+|\phi_{i,4}(x,t)|+|\phi_{i,5}(x,t)|\lesssim_{K}\frac{\varepsilon^{n-4}}{|x|^{n-4}}\left(\|\phi_{i}\|_{L^{\infty}(Q_{1})}+e^{-c\varepsilon_{i}^{-2}}\right).$

Here the term $e^{-c\varepsilon_{i}^{-2}}$ appears because when we apply Lemma
13.6 to estimate $a/\lambda$, there is a boundary term

$\frac{a_{i}(1)}{\lambda_{i}(1)}e^{-c\left[\tau(1)-\tau(t)\right]},$

where, because $d\tau/dt\sim\varepsilon_{i}^{-2}$,

$\tau(1)-\tau(t)\geq c\varepsilon_{i}^{-2},\quad\mbox{for any }~{}~{}t<25/36.$

Because $\xi_{i}(0)=0$, by the estimate of $g(5/6)$ in Proposition 17.6, now
(15.11) is upgraded to

$|\xi_{i}(t)|\lesssim\varepsilon_{i}^{\frac{n-4}{2}},\quad\mbox{for
any}~{}~{}t\in[-25/36,25/36].$ (17.9)

####  17.2. Gradient estimates

In this subsection we establish a uniform $C^{1+\theta,(1+\theta)/2}$ estimate
on $\phi_{i}$.

With the estimates in Proposition 17.6, now we have

######  Lemma 17.7.

In $Q_{5/6}$,

$\left|\partial_{t}\phi_{i}-\Delta\phi_{i}\right|\lesssim\left(\frac{\varepsilon_{i}^{2}}{\varepsilon_{i}^{4}+|x|^{4}}+1\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$
(17.10)

###### Proof.

By (17.8), now (17.9) can be upgraded to

$|\xi_{i}(t)|\lesssim\varepsilon_{i}^{\frac{n-4}{2}},\quad\mbox{for any
}~{}~{}t\in[-25/36,25/36].$ (17.11)

Then by the definition of $\phi_{i}$, we have

$\displaystyle u_{i}(x,t)$ $\displaystyle\lesssim$ $\displaystyle
W_{i}(x,t)+|a_{i}(t)|Z_{0,i}(x,t)+|\phi_{i}(x,t)|$ $\displaystyle\lesssim$
$\displaystyle\left(\frac{\varepsilon_{i}}{\varepsilon_{i}^{2}+|x|^{2}}\right)^{\frac{n-2}{2}}+e^{-c|x|/\varepsilon_{i}}+1.$

Therefore

$\displaystyle\left|u_{i}^{p}-W_{i}^{p}\right|$ $\displaystyle\lesssim$
$\displaystyle\left(u_{i}^{p-1}+W_{i}^{p-1}\right)\left(|\phi_{i}|+|a_{i}(t)|Z_{0,i}\right)$
$\displaystyle\lesssim$
$\displaystyle\left[\left(\frac{\varepsilon_{i}}{\varepsilon_{i}^{2}+|x|^{2}}\right)^{\frac{n-2}{2}}+e^{-c|x|/\varepsilon_{i}}+1\right]^{\frac{4}{n-2}}\left(1+e^{-c|x|/\varepsilon_{i}}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)$
$\displaystyle\lesssim$
$\displaystyle\left(\frac{\varepsilon_{i}^{2}}{\varepsilon_{i}^{4}+|x|^{4}}+1\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

By (17.8), we have

$\displaystyle\left|\lambda_{i}^{\prime}(t)Z_{n+1,i}(x,t)\right|$
$\displaystyle\lesssim$
$\displaystyle\varepsilon_{i}^{-2}\left(1+\frac{|x-\xi_{i}(t)|}{\varepsilon_{i}}\right)^{2-n}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)$
$\displaystyle\lesssim$
$\displaystyle\frac{\varepsilon_{i}^{n-4}}{\left(\varepsilon_{i}+|x|\right)^{n-2}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)$
$\displaystyle\lesssim$
$\displaystyle\frac{\varepsilon_{i}^{2}}{\varepsilon_{i}^{4}+|x|^{4}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Similar estimates hold for
$(-a_{i}^{\prime}+\mu_{0}a_{i}\lambda_{i}^{-2})Z_{0i}$,
$\xi_{ji}^{\prime}\cdot Z_{ji}$, $j=1,\cdots,n$ and $a_{i}\partial_{t}Z_{0i}$.
∎

######  Lemma 17.8.

For any $(x,t)\in Q_{4/5}$,

$|\nabla\phi_{i}(x,t)|\lesssim\left(\frac{\varepsilon_{i}^{2}}{\varepsilon_{i}^{3}+|x|^{3}}+1\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$
(17.12)

###### Proof.

We consider three cases separately.

Case 1. If $1/4\leq|x|\leq 4/5$, this estimate (in fact, a bound on the
Lipschitz seminorm) follows from standard interior gradient estimates.

Case 2. If $|x|\leq\varepsilon_{i}$, by looking at the inner equation (13) and
using Proposition 17.6, we obtain

$|\nabla\varphi_{i}(y,\tau)|\lesssim\varepsilon_{i}^{\frac{n-2}{2}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Scaling back to $\phi_{i}$, this is

$|\nabla\phi_{i}(x,t)|\lesssim\varepsilon_{i}^{-1}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

In fact, this gives a full Lipschitz estimate in space-time.

Case 3. Finally, we consider the remaining case where
$\varepsilon_{i}\leq|x|\leq 1/4$. Take the decomposition
$\phi_{i}=\phi_{h,i}+\phi_{n,i}$, where

$\left\\{\begin{aligned}
&\partial_{t}\phi_{h,i}-\Delta\phi_{h,i}=0,\quad&\mbox{in }~{}~{}Q_{3/4},\\\
&\phi_{h,i}=\phi_{i},\quad&\mbox{on
}~{}~{}\partial^{p}Q_{3/4}.\end{aligned}\right.$

By standard interior gradient estimates for heat equation, we have

$\|\nabla\phi_{h,i}\|_{L^{\infty}(Q_{2/3})}\lesssim\|\phi_{i}\|_{L^{\infty}(\partial^{p}Q_{3/4})}.$
(17.13)

Next, for the non-homogeneous part $\phi_{n,i}$, by the heat kernel
representation, we have

$\displaystyle|\nabla\phi_{n,i}(x,t)|$ $\displaystyle=$
$\displaystyle\left|\int_{-\frac{81}{100}}^{t}\int_{B_{\frac{3}{4}}}\nabla_{x}G(x,t;y,s)\left(\partial_{t}-\Delta\right)\phi_{i}(y,s)dyds\right|$
$\displaystyle\lesssim$
$\displaystyle\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)\int_{-\frac{9}{16}}^{t}\int_{B_{\frac{3}{4}}}\frac{|x-y|}{(t-s)^{\frac{n+2}{2}}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(\frac{\varepsilon_{i}^{2}}{\varepsilon_{i}^{4}+|y|^{4}}+1\right)$
$\displaystyle\lesssim$
$\displaystyle\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)\int_{-\frac{9}{16}}^{t}\int_{B_{\frac{3}{4}}}(t-s)^{-\frac{n+1}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(\frac{\varepsilon_{i}^{2}}{|y|^{4}}+1\right)$
$\displaystyle\lesssim$
$\displaystyle\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)\int_{-\frac{9}{16}}^{t}(t-s)^{-\frac{1}{2}}\left[\frac{\varepsilon_{i}^{2}}{\left(\sqrt{t-s}+|x|\right)^{4}}+1\right]$
$\displaystyle\lesssim$
$\displaystyle\left(\frac{\varepsilon_{i}^{2}}{|x|^{3}}+1\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Combining this estimate with (17.13), we get (17.12) in this case. ∎

Next we extend this Lipschitz estimate in spatial variables to the full
Lipschitz estimate in space-time variables. For this we need a technical lemma
transforming the spatial Lipschitz estimate into a full Lipschitz estimate in
space-time.

######  Lemma 17.9.

Given a constant $\sigma>0$, suppose $\psi\in C^{\infty}(Q_{1}^{-})$ satisfies

  1. (1)

$|\nabla\psi|\leq\sigma$ in $Q_{1}^{-}$;

  2. (2)

$|\partial_{t}\psi-\Delta\psi|\leq\sigma$ in $Q_{1}^{-}$.

Then the Lipschitz seminorm (with respect to the parabolic distance) of $\psi$
in $Q_{1/2}^{-}$ satisfies

$|\psi|_{Lip(Q_{1/2}^{-})}\lesssim\sigma.$

###### Proof.

We need only to prove that, for any $x\in B_{1/2}$ and $-1/4<t_{1}<t_{2}<0$,

$|\psi(x,t_{1})-\psi(x,t_{2})|\lesssim\sigma\sqrt{t_{2}-t_{1}}.$ (17.14)

Denote $h:=\sqrt{t_{2}-t_{1}}$ and define

$\widetilde{\psi}(y,s):=\frac{1}{h}\left[\psi(x+hy,t_{2}+h^{2}s)-\psi(x,t_{2})\right].$

It satisfies

  1. (1)

$|\nabla\widetilde{\psi}|\leq\sigma$ in $Q_{2}^{-}$;

  2. (2)

for any $(y,s)\in Q_{2}^{-}$,

$|\partial_{t}\widetilde{\psi}(y,s)-\Delta\widetilde{\psi}(y,s)|\leq h\sigma;$
(17.15)

  3. (3)

because $\widetilde{\psi}(0,0)=0$, by integrating (1) we obtain

$\sup_{y\in B_{2}}|\widetilde{\psi}(y,0)|\leq 2\sigma.$ (17.16)

Fix a function $\eta\in C_{0}^{\infty}(B_{2})$. Multiplying (17.15) and
integrating by parts, we obtain

$\displaystyle\left|\frac{d}{ds}\int_{B_{2}}\widetilde{\psi}\eta\right|$
$\displaystyle\leq$ $\displaystyle
h\sigma\int_{B_{2}}|\eta|+\int_{B_{2}}|\nabla\widetilde{\psi}||\nabla\eta|$
$\displaystyle\lesssim$ $\displaystyle\sigma.$

Combining this inequality with (17.16), we obtain

$\left|\int_{B_{2}}\widetilde{\psi}(y,s)\eta(y)\right|\lesssim\sigma\quad\mbox{
for any}~{}~{}s\in(-4,0).$

With the Lipschitz estimate in (1), this implies that

$\sup_{y\in B_{2}}|\widetilde{\psi}(y,s)|\lesssim\sigma\quad\mbox{ for
any}~{}~{}s\in(-4,0).$ (17.17)

In particular,

$|\widetilde{\psi}(0,-1)|\lesssim\sigma.$

Scaling back to $\psi$, this is (17.14). ∎

######  Proposition 17.10.

For any $(x,t)\in Q_{3/4}$,

$|\phi_{i}|_{Lip(Q_{|x|/4}(x,t))}\lesssim\left(\frac{\varepsilon_{i}^{2}}{\varepsilon_{i}^{3}+|x|^{3}}+1\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$
(17.18)

###### Proof.

When $|x|<\varepsilon_{i}$ and $1/4<|x|<3/4$, this was already covered in the
proof of Lemma 17.8, see Case 1 and Case 2 therein.

If $\varepsilon_{1}<|x|<1/4$, denote $r:=|x|/2$ and let

$\widetilde{\phi}_{i}(y,s):=\phi_{i}\left(x+ry,t+r^{2}s\right).$

By Lemma 17.7 and Lemma 17.8, it satisfies the assumptions in Lemma 17.9 with
the constant

$\sigma\lesssim\left(\frac{\varepsilon_{i}^{2}}{r^{2}}+r\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Hence this lemma implies that

$|\widetilde{\phi}_{i}|_{Lip(Q_{1/2}^{-})}\lesssim\left(\frac{\varepsilon_{i}^{2}}{r^{2}}+r\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Scaling back to $\phi_{i}$, this is (17.18) in this case. ∎

####  17.3. Estimate of time derivative

In this subsection we establish the following estimate on
$\partial_{t}\phi_{i}$. We can also obtain some estimates on
$\nabla^{2}\phi_{i}$, but we do not need it.

######  Proposition 17.11.

For any $(x,t)\in Q_{2/3}$,

$|\partial_{t}\phi_{i}(x,t)|\lesssim\left(\frac{\varepsilon_{i}^{2}}{\varepsilon_{i}^{4}+|x|^{4}}+\frac{1}{\varepsilon_{i}^{2}+|x|}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$
(17.19)

###### Proof.

Take an arbitrary sequence of points $(x_{i},t_{i})\in Q_{2/3}$. Denote
$r_{i}:=|x_{i}|/2$.

Case 1. First assume

$\limsup_{i\to+\infty}\frac{r_{i}}{\varepsilon_{i}}<+\infty.$

In this case we work in the inner coordinates introduced in Section 13. By
(13.2) and (13.3), we have the following representation formulas:

$\left\\{\begin{aligned}
&\frac{\dot{a}_{i}(\tau)-\mu_{0}a_{i}(\tau)}{\lambda_{i}(\tau)}=\int_{\mathbb{R}^{n}}Z_{0}(y)E_{K,i}(y,\tau)dy,\\\
&\frac{\dot{\xi}_{i}(\tau)}{\lambda_{i}(\tau)}=-\int_{\mathbb{R}^{n}}Z_{j}(y)E_{K,i}(y,\tau)dy,\quad
j=1,\cdots,n,\\\
&\frac{\dot{\lambda}_{i}(\tau)}{\lambda_{i}(\tau)}=-\int_{\mathbb{R}^{n}}Z_{n+1}(y)E_{K,i}(y,\tau)dy.\end{aligned}\right.$

By the form of $E_{K}$ in Section 13, and the smallness of

$\left|\frac{\dot{a}_{i}-\mu_{0}a_{i}}{\lambda_{i}}\right|+\left|\frac{\dot{\xi}_{i}}{\lambda_{i}}\right|+\left|\frac{\dot{\lambda}_{i}}{\lambda_{i}}\right|,$

the above three equations are solved as

$\left(\frac{\dot{a}_{i}(\tau)-\mu_{0}a_{i}(\tau)}{\lambda_{i}(\tau)},\frac{\dot{\xi}_{i}(\tau)}{\lambda_{i}(\tau)},\frac{\dot{\lambda}_{i}(\tau)}{\lambda_{i}(\tau)}\right)=\mathcal{J}\left(\nabla\varphi_{i}(\tau),\varphi_{i}(\tau),\frac{a_{i}(\tau)}{\lambda_{i}(\tau)}\right),$
(17.20)

where $\mathcal{J}=(\mathcal{J}_{0},\cdots,\mathcal{J}_{n+1})$ is a vector
valued, nonlinear (but smooth) integral operator. Plugging this back into
(13), we get

$\displaystyle\partial_{\tau}\varphi_{i}-\Delta_{y}\varphi_{i}$
$\displaystyle=$
$\displaystyle\left(W+\varphi+a_{i}Z_{0}\right)^{p}-W^{p}+\mathcal{J}\cdot Z$
$\displaystyle+$
$\displaystyle\left(\mathcal{J}_{1},\cdots,\mathcal{J}_{n}\right)\cdot\nabla\varphi+\mathcal{J}_{n+1}\left(y\cdot\nabla\varphi_{i}+\frac{n-2}{2}\varphi_{i}\right)$
$\displaystyle+$
$\displaystyle\frac{a_{i}}{\lambda_{i}}\left[\left(\mathcal{J}_{1},\cdots,\mathcal{J}_{n}\right)\cdot\nabla
Z_{0}+\mathcal{J}_{n+1}\left(y\cdot\nabla
Z_{0}+\frac{n}{2}Z_{0}\right)\right].$

Starting from the $L^{\infty}$ bound of $\varphi_{i}$ (by rescaling the
estimates in Proposition 17.6) and bootstrapping parabolic estimates, we get

$\|\varphi_{i}\|_{C^{2+\theta,1+\theta/2}(B_{2K}\times[\tau-1,\tau])}\lesssim\varepsilon_{i}^{\frac{n-2}{2}},\quad\mbox{for
any}~{}~{}\tau.$ (17.21)

A byproduct of this estimate is the validity of (17.19) in this case.

Case 2. Substituting (17.21) into (17.20) gives, for any $\tau_{1}<\tau_{2}$,

$\displaystyle\left\|\left(\frac{\dot{a}_{i}-\mu_{0}a_{i}}{\lambda_{i}},\frac{\dot{\xi}_{i}}{\lambda_{i}},\frac{\dot{\lambda}_{i}}{\lambda_{i}}\right)\right\|_{C^{\theta/2}([\tau_{1},\tau_{2}])}\lesssim\varepsilon_{i}^{\frac{n-2}{2}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$
(17.22)

After scaling back to the original coordinates, by noting that
$d\tau/dt\sim\varepsilon_{i}^{-2}$, this estimate is transformed into

$\displaystyle\left\|\left(a_{i}^{\prime}-\mu_{0}\frac{a_{i}}{\lambda_{i}^{2}},\xi_{i}^{\prime},\lambda_{i}^{\prime}\right)\right\|_{C^{\theta/2}([-3/4,3/4])}\lesssim\varepsilon_{i}^{\frac{n-4}{2}-\theta}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$
(17.23)

By Proposition 17.6, (17.23) and Proposition 17.10, we deduce that the terms
in the right hand side of (12.8) are uniformly bounded in
$C^{\theta,\theta/2}\left((B_{1}\setminus B_{1/8})\times(-1,1)\right)$. Hence
standard Schauder estimates for heat equation gives

$\sup_{\left((B_{2/3}\setminus
B_{1/4})\times(-4/9,4/9)\right)}|\partial_{t}\phi_{i}|\lesssim\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}.$

Case 3. In this case we assume

$\lim_{i\to+\infty}\left(r_{i}+\frac{\varepsilon_{i}}{r_{i}}\right)=0.$

Define

$\widetilde{u}_{i}(x,t):=r_{i}^{\frac{n-2}{2}}u_{i}\left(r_{i}x,t_{i}+r_{i}^{2}t\right).$

The decomposition in Section 12 can be transferred to $\widetilde{u}_{i}$ as

$\widetilde{u}_{i}=W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}+\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}+\widetilde{\phi}_{i},$
(17.24)

where

$\left\\{\begin{aligned}
&\widetilde{\xi}_{i}(t):=r_{i}^{-1}\xi_{i}(t_{i}+r_{i}^{2}t),\\\
&\widetilde{\lambda}_{i}(t):=r_{i}^{-1}\lambda_{i}(t_{i}+r_{i}^{2}t),\\\
&\widetilde{a}_{i}(t):=r_{i}^{-1}a_{i}(t_{i}+r_{i}^{2}t),\\\
&\widetilde{\phi}_{i}(x,t):=r_{i}^{(n-2)/2}\phi_{i}(r_{i}x,t_{i}+r_{i}^{2}t).\end{aligned}\right.$

Because for any $t\in[-3/4,3/4]$,

$r_{i}>2K\varepsilon_{i}>K\lambda_{i}(t),$

the orthogonal relation in Proposition 12.1 still holds between
$\widetilde{\phi}_{i}$ and
$Z_{j,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}$, $j=0,\cdots,n+1$.

Furthermore, the error equation (12.8) now reads as

$\displaystyle\partial_{t}\widetilde{\phi}_{i}-\Delta\widetilde{\phi}_{i}$
$\displaystyle=$
$\displaystyle\left(W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}+\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}+\widetilde{\phi}_{i}\right)^{p}-W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}^{p}-pW_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}^{p-1}\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}$
(17.25) $\displaystyle+$
$\displaystyle\left(-\widetilde{a}_{i}^{\prime}+\mu_{0}\frac{\widetilde{a}_{i}}{\widetilde{\lambda}_{i}^{2}},\widetilde{\xi}_{i}^{\prime},\widetilde{\lambda}_{i}^{\prime}\right)\cdot
Z_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}-\widetilde{a}_{i}\partial_{t}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}.$

By (17.11) and (17.23), we get the following estimates for those terms in the
right hand side of (17.25):

$\left\\{\begin{aligned}
&\|W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}\lesssim\left(\frac{\varepsilon_{i}}{r_{i}}\right)^{\frac{n-2}{2}};\\\
&\|\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}\lesssim\varepsilon_{i}^{\frac{n-2}{2}}e^{-c\frac{r_{i}}{\varepsilon_{i}}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right);\\\
&\|\widetilde{\phi}_{i}\|_{L^{\infty}((B_{1}\setminus
B_{1/4})\times(-4,4))}\lesssim
r_{i}^{\frac{n-2}{2}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right);\\\
&\left\|\left(\widetilde{a}_{i}^{\prime}-\mu_{0}\widetilde{a}_{i}\widetilde{\lambda}_{i}^{-2},\widetilde{\xi}_{i}^{\prime},\widetilde{\lambda}_{i}^{\prime}\right)\cdot
Z_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\right\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}\lesssim\frac{\varepsilon_{i}^{n-4}}{r_{i}^{\frac{n}{2}-3}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right);\\\
&\left\|\widetilde{a}_{i}\partial_{t}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\right\|_{C^{\theta,\theta/2}((B_{2}\setminus
B_{1/4})\times(-4,4))}\lesssim\varepsilon_{i}^{n-2}e^{-c\frac{r_{i}}{\varepsilon_{i}}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).\end{aligned}\right.$
(17.26)

These estimates imply that in $\left(B_{2}\setminus
B_{1/4}\right)\times(-4,4)$,

$\widetilde{u}_{i}\lesssim\left(\frac{\varepsilon_{i}}{r_{i}}\right)^{\frac{n-2}{2}}+r_{i}^{\frac{n-2}{2}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)$
(17.27)

and

$\displaystyle\left|\left(W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}+\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}+\widetilde{\phi}_{i}\right)^{p}-W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}^{p}-pW_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}^{p-1}\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\right|$
$\displaystyle\lesssim$
$\displaystyle\left[W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}^{p-1}+\widetilde{u}_{i}^{p-1}\right]\left(\left|\widetilde{\phi_{i}}\right|+\left|\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\right|\right)$
$\displaystyle\lesssim$
$\displaystyle\left(\frac{\varepsilon_{i}^{2}}{r_{i}^{2}}+r_{i}^{2}\right)r_{i}^{\frac{n-2}{2}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

By Proposition 17.10 and a rescaling, we know that the Lipschitz seminorm of
$\widetilde{\phi}_{i}$ t in $\left(B_{2}\setminus B_{1/4}\right)\times(-4,4)$
is bounded by

$r_{i}^{\frac{n}{2}}\left(\frac{\varepsilon_{i}^{2}}{r_{i}^{3}}+1\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

By these estimates and standard $W^{2,p}$ estimates for heat equation, we
obtain

$\displaystyle\int_{-3}^{3}\int_{B_{9/5}\setminus
B_{1/3}}|\partial_{t}\widetilde{\phi}_{i}|$ $\displaystyle\lesssim$
$\displaystyle\left[r_{i}^{\frac{n}{2}}\left(\frac{\varepsilon_{i}^{2}}{r_{i}^{3}}+1\right)+\varepsilon_{i}^{\frac{n-2}{2}}\left(\frac{\varepsilon_{i}}{r_{i}}\right)^{\frac{n-3}{2}}+r_{i}^{\frac{n-2}{2}}\left(\frac{\varepsilon_{i}^{2}}{r_{i}^{2}}+r_{i}^{2}\right)\right]$
(17.28)
$\displaystyle\quad\quad\times\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Because

$\left(\partial_{t}-\Delta\right)\partial_{t}\widetilde{u}_{i}=p\widetilde{u}_{i}^{p-1}\partial_{t}\widetilde{u}_{i},$

and $\widetilde{u}_{i}$ are uniformly bounded in $\left(B_{9/5}\setminus
B_{1/3}\right)\times(-3,3)$, we get

$\displaystyle\left\|\partial_{t}\widetilde{u}_{i}\right\|_{L^{\infty}((B_{3/2}\setminus
B_{1/2})\times(-1,1))}$ $\displaystyle\lesssim$
$\displaystyle\int_{-3}^{3}\int_{B_{9/5}\setminus
B_{1/3}}|\partial_{t}\widetilde{u}_{i}|$ $\displaystyle\lesssim$
$\displaystyle\left(\int_{-3}^{3}\int_{B_{9/5}\setminus
B_{1/3}}|\partial_{t}\widetilde{\phi}_{i}|\right)+\|\partial_{t}W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}$
$\displaystyle\quad\quad\quad\quad+\left\|\partial_{t}\left(\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\right)\right\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))},$

where we have used the expansion from (17.24),

$\partial_{t}\widetilde{u}_{i}=\partial_{t}W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}+\partial_{t}\left(\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\right)+\partial_{t}\widetilde{\phi}_{i}.$

Similar to (17.26), we also have

$\left\\{\begin{aligned}
&\|\partial_{t}W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}\lesssim\varepsilon_{i}^{n-4}r_{i}^{-\frac{n}{2}-3}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right);\\\
&\left\|\partial_{t}\left(\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\right)\right\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}\lesssim\varepsilon_{i}^{\frac{n-2}{2}}e^{-c\frac{r_{i}}{\varepsilon_{i}}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).\end{aligned}\right.$

Substituting these estimates and (17.28) into (17.3), we obtain

$\displaystyle\left\|\partial_{t}\widetilde{\phi}_{i}\right\|_{L^{\infty}((B_{3/2}\setminus
B_{1/2})\times(-1,1))}$ $\displaystyle\lesssim$
$\displaystyle\left\|\partial_{t}\widetilde{u}_{i}\right\|_{L^{\infty}((B_{3/2}\setminus
B_{1/2})\times(-1,1))}$ $\displaystyle+$
$\displaystyle\|\partial_{t}W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}+\left\|\partial_{t}\left(\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\right)\right\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}$ $\displaystyle\lesssim$
$\displaystyle\int_{-3}^{3}\int_{B_{9/5}\setminus
B_{1/3}}|\partial_{t}\widetilde{\phi}_{i}|$ $\displaystyle+$
$\displaystyle\|\partial_{t}W_{\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}+\left\|\partial_{t}\left(\widetilde{a}_{i}Z_{0,\widetilde{\xi}_{i},\widetilde{\lambda}_{i}}\right)\right\|_{L^{\infty}((B_{2}\setminus
B_{1/4})\times(-4,4))}$ $\displaystyle\lesssim$
$\displaystyle\left[r_{i}^{\frac{n}{2}}\left(\frac{\varepsilon_{i}^{2}}{r_{i}^{3}}+1\right)+\varepsilon_{i}^{\frac{n-2}{2}}\left(\frac{\varepsilon_{i}}{r_{i}}\right)^{\frac{n-3}{2}}+r_{i}^{\frac{n-2}{2}}\left(\frac{\varepsilon_{i}^{2}}{r_{i}^{2}}+r_{i}^{2}\right)\right]\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)$
$\displaystyle+$
$\displaystyle\left[\frac{\varepsilon_{i}^{n-4}}{r_{i}^{\frac{n}{2}-3}}+\varepsilon_{i}^{\frac{n-2}{2}}e^{-c\frac{r_{i}}{\varepsilon_{i}}}\right]\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Scaling back to $\phi_{i}$, we get (17.19). ∎

###  18\. Linearization of Pohozaev identity

Given a smooth function $v$ and a sphere $\partial B_{r}$, define the Pohozaev
invariant as

$\mathcal{P}_{v}(r):=\int_{\partial B_{r}}\left[\frac{|\nabla
v|^{2}}{2}-\left(\partial_{r}v\right)^{2}-\frac{n-2}{2r}v\partial_{r}v\right].$

Choose of a sequence of $r_{i}$ satisfying

$\lim_{i\to+\infty}\frac{r_{i}}{\varepsilon_{i}}=+\infty.$

Multiplying (1.1) by $x\cdot\nabla u_{i}(x,0)$ and integrating in $B_{r_{i}}$,
after integrating by parts we obtain a Pohozaev identity

$\mathcal{P}_{u_{i}(0)}(r_{i})-\int_{\partial
B_{r_{i}}}\frac{u_{i}(0)^{p+1}}{p+1}=-\frac{1}{r_{i}}\int_{B_{r_{i}}}\partial_{t}u_{i}(0)\left[x\cdot\nabla
u_{i}(0)+\frac{n-2}{2}u_{i}(0)\right].$ (18.1)

Substitute the decomposition (17.1) into this identity, and take a expansion.
Let us estimate each term in this expansion. We will see that the zeroth order
terms cancel with each other, because these terms form exactly the Pohozaev
invariant of the bubble. The next order term then gives us some information.

####  18.1. The left hand side

For the left hand side, we have the expansion

$\displaystyle\mathcal{P}_{u_{i}(0)}(r_{i})-\int_{\partial
B_{r_{i}}}\frac{u_{i}(0)^{p+1}}{p+1}$ $\displaystyle=$
$\displaystyle\mathcal{P}_{W_{i}(0)}(r_{i})-\int_{\partial
B_{r_{i}}}\frac{W_{i}(0)^{p+1}}{p+1}+\mathcal{P}_{\phi_{i}(0)}(r_{i})+\mathcal{P}_{a_{i}(0)Z_{0,i}(0)}(r_{i})$
$\displaystyle+$ $\displaystyle\underbrace{\int_{\partial B_{r_{i}}}\nabla
W_{i}(0)\cdot\nabla\phi_{i}(0)-2\partial_{r}W_{i}(0)\partial_{r}\phi_{i}(0)-\frac{n-2}{2r_{i}}\left[W_{i}(0)\partial_{r}\phi_{i}(0)+\phi_{i}(0)\partial_{r}W_{i}(0)\right]}_{\mbox{Main
order term}}$ $\displaystyle-$ $\displaystyle\frac{1}{p+1}\int_{\partial
B_{r_{i}}}\left[u_{i}(0)^{p+1}-W_{i}(0)^{p+1}\right]+\mbox{cross terms
involving }a_{i}(0)Z_{0,i}(0).$

Let us estimate each term in this expansion.

  1. (1)

Because $W_{i}(0)$ is a smooth solution of (1.9), it satisfies the standard
Pohozaev identity

$\mathcal{P}_{W_{i}(0)}(r_{i})-\int_{\partial
B_{r_{i}}}\frac{W_{i}(0)^{p+1}}{p+1}=0.$

  2. (2)

By Proposition 17.1 and Proposition 17.10, we get

$\mathcal{P}_{\phi_{i}(0)}(r_{i})=O\left(r_{i}^{n-3}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)^{2}.$

Here the factor
$\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)$
is kept for later use.

  3. (3)

By estimates on $a_{i}$ and $\lambda_{i}$ in Proposition 17.6, on $\partial
B_{r_{i}}$ we have

$\left\\{\begin{aligned} &|a_{i}(t)Z_{0,i}(t)|\lesssim
e^{-cr_{i}/\varepsilon_{i}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right),\\\
&\left|\nabla\left(a_{i}(t)Z_{0,i}(t)\right)\right|\lesssim\varepsilon_{i}^{-1}e^{-cr_{i}/\varepsilon_{i}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).\end{aligned}\right.$
(18.2)

By these estimates we get

$\left|\mathcal{P}_{a_{i}(0)Z_{0,i}(0)}(r)\right|\lesssim\varepsilon_{i}^{-2}r_{i}^{n-1}e^{-cr_{i}/\varepsilon_{i}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)^{2}.$

  4. (4)

First by integrating (17.12), for any $x\in\partial B_{r_{i}}$, we have

$\phi_{i}(x,0)=\frac{1}{|\partial B_{\rho}|}\int_{\partial
B_{\rho}}\phi_{i}(0)+O\left(\frac{\varepsilon_{i}^{2}}{r_{i}^{2}}+\rho\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Next, because $\xi_{i}(0)=0$, on $\partial B_{r_{i}}$, by Proposition 17.6 and
(17.11), we obtain

$\left\\{\begin{aligned}
&W_{i}(0)\lesssim\varepsilon_{i}^{\frac{n-2}{2}}r_{i}^{2-n},\\\
&\partial_{r}W_{i}(0)=-c(n)\varepsilon_{i}^{\frac{n-2}{2}}r_{i}^{1-n}+O\left(\varepsilon_{i}^{\frac{n+2}{2}}r_{i}^{-1-n}\right),\\\
&\nabla W_{i}(0)=\partial_{r}W_{i}(0)\frac{x}{|x|},\\\
&|\nabla\phi_{i}(t)|\lesssim\left(\varepsilon_{i}^{2}r_{i}^{-3}+1\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).\end{aligned}\right.$

Therefore the main order term equals

$\displaystyle-c(n)\left[\frac{1}{|\partial B_{\rho}|}\int_{\partial
B_{\rho}}\phi_{i}+O\left(\frac{\varepsilon_{i}^{2}}{r_{i}^{2}}+\rho\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)\right]\frac{\varepsilon_{i}^{\frac{n-2}{2}}}{r_{i}}$
$\displaystyle+$ $\displaystyle
O\left(\varepsilon_{i}^{\frac{n-2}{2}}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

  5. (5)

On $\partial B_{r_{i}}$, we have

$W_{i}(0)\lesssim\varepsilon_{i}^{\frac{n-2}{2}}r_{i}^{2-n},\quad|a_{i}Z_{0,i}(0)|\lesssim
e^{-cr_{i}/\varepsilon_{i}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Therefore

$\left|\int_{\partial
B_{r_{i}}}\left[u_{i}(t)^{p+1}-W_{i}(t)^{p+1}\right]\right|\lesssim\left(\varepsilon_{i}^{\frac{n+2}{2}}r_{i}^{-3}+r_{i}^{n-1}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

  6. (6)

Finally, by (18.2) and estimates of $W_{i}(0)$ and $\phi_{i}(0)$, the cross
terms involving $a_{i}Z_{0,i}(0)$ are of the order

$O\left(\varepsilon_{i}^{\frac{n-4}{2}}+r_{i}^{n-3}+\varepsilon_{i}^{-1}r_{i}^{n-2}\right)e^{-cr_{i}/\varepsilon_{i}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Putting these estimates together, we see that the left hand side of (18.1)
equals

$\displaystyle-c(n)\left[\frac{1}{|\partial B_{\rho}|}\int_{\partial
B_{\rho}}\phi_{i}+O\left(\frac{\varepsilon_{i}^{2}}{r_{i}^{2}}+\rho\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)\right]\frac{\varepsilon_{i}^{\frac{n-2}{2}}}{r_{i}}$
$\displaystyle+$ $\displaystyle
O\left(r_{i}^{n-3}+\varepsilon_{i}^{-2}r_{i}^{n-1}e^{-cr_{i}/\varepsilon_{i}}+\varepsilon_{i}^{\frac{n-2}{2}}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)$
$\displaystyle+$
$\displaystyle\left(\varepsilon_{i}^{\frac{n+2}{2}}r_{i}^{-3}+r_{i}^{n-1}+\varepsilon_{i}^{\frac{n-4}{2}}e^{-cr_{i}/\varepsilon_{i}}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

####  18.2. The right hand side

By (17.1), we get

$\partial_{t}u_{i}=\left(a_{i}^{\prime},\xi_{i}^{\prime},\lambda_{i}^{\prime}\right)\cdot
Z_{\ast,i}+\partial_{t}\phi_{i}+a_{i}\partial_{t}Z_{0,i}$ (18.4)

and

$\displaystyle x\cdot\nabla u_{i}+\frac{n-2}{2}u_{i}$ $\displaystyle=$
$\displaystyle\lambda_{i}Z_{n+1,i}+\left(x\cdot\nabla\phi_{i}+\frac{n-2}{2}\phi_{i}\right)$
$\displaystyle+$ $\displaystyle a_{i}\left(x\cdot
Z_{0,i}+\frac{n-2}{2}Z_{0,i}\right).$

Therefore for any $r$, we have

$\displaystyle\int_{B_{r}}\partial_{t}u_{i}(x,0)\left[x\cdot\nabla
u_{i}(x,0)+\frac{n-2}{2}u_{i}(x,0)\right]dx$ $\displaystyle=$
$\displaystyle\lambda_{i}(0)\left(a_{i}^{\prime}(0),\xi_{i}^{\prime}(0),\lambda_{i}^{\prime}(0)\right)\cdot\int_{B_{r}}Z_{\ast,i}(x,0)Z_{n+1,i}(x,0)dx$
$\displaystyle+$
$\displaystyle\left(a_{i}^{\prime}(0),\xi_{i}^{\prime}(0),\lambda_{i}^{\prime}(0)\right)\cdot\int_{B_{r}}Z_{\ast,i}(x,0)\left[x\cdot\nabla\phi_{i}(x,0)+\frac{n-2}{2}\phi_{i}(x,0)\right]dx$
$\displaystyle+$ $\displaystyle
a_{i}(0)\left(a_{i}^{\prime}(0),\xi_{i}^{\prime}(0),\lambda_{i}^{\prime}(0)\right)\cdot\int_{B_{r}}Z_{\ast,i}(x,0)\left[x\cdot
Z_{0,i}(x,0)+\frac{n-2}{2}Z_{0,i}(x,0)\right]dx$ $\displaystyle+$
$\displaystyle\lambda_{i}(0)\int_{B_{r}}Z_{n+1,i}(x,0)\partial_{t}\phi_{i}(x,0)dx$
$\displaystyle+$
$\displaystyle\int_{B_{r}}\partial_{t}\phi_{i}(x,0)\left[x\cdot\nabla\phi_{i}(x,0)+\frac{n-2}{2}\phi_{i}(x,0)\right]dx$
$\displaystyle+$ $\displaystyle
a_{i}(0)\int_{B_{r}}\partial_{t}\phi_{i}(x,0)\left[x\cdot
Z_{0,i}(x,0)+\frac{n-2}{2}Z_{0,i}(x,0)\right]dx$ $\displaystyle+$
$\displaystyle
a_{i}(0)\lambda_{i}(0)\int_{B_{r}}Z_{n+1,i}(x,0)\partial_{t}Z_{0,i}(x,0)dx$
$\displaystyle+$ $\displaystyle
a_{i}(0)\int_{B_{r}}\partial_{t}Z_{0,i}(x,0)\left[x\cdot\nabla\phi_{i}(x,0)+\frac{n-2}{2}\phi_{i}(x,0)\right]dx$
$\displaystyle+$ $\displaystyle
a_{i}(0)^{2}\int_{B_{r}}\partial_{t}Z_{0,i}(x,0)\left[x\cdot
Z_{0,i}(x,0)+\frac{n-2}{2}Z_{0,i}(x,0)\right]dx$ $\displaystyle=:$
$\displaystyle\mathrm{I}+\mathrm{II}+\mathrm{III}+\mathrm{IV}+\mathrm{V}+\mathrm{VI}+\mathrm{VII}+\mathrm{VIII}+\mathrm{IX}.$

Let us take $r=r_{i}$ and estimate each term one by one.

  1. (1)

By (15.9), Proposition 17.6 and the orthogonal relation between $Z_{0}$ and
$Z_{n+1}$,

$\left|\lambda_{i}(0)a_{i}^{\prime}(0)\int_{B_{r_{i}}}Z_{0,i}(x,t)Z_{n+1,i}(x,0)dx\right|\lesssim\varepsilon_{i}^{\frac{n-2}{2}}e^{-cr_{i}/\varepsilon_{i}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Similarly, for each $j=1,\cdots,n$,

$\left|\lambda_{i}(0)\xi_{j,i}^{\prime}(0)\int_{B_{r_{i}}}Z_{j,i}(x,t)Z_{n+1,i}(x,0)dx\right|\lesssim\varepsilon_{i}^{\frac{3n}{2}-4}r_{i}^{3-n}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

By Lemma 13.5, (15.9) and Proposition 17.6,

$\left|\lambda_{i}(0)\lambda_{i}(0)^{\prime}\int_{B_{r_{i}}}Z_{i,n+1}(x,0)^{2}dx\right|\lesssim
K^{-\frac{n-2}{2}}\varepsilon_{i}^{\frac{n-2}{2}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Putting these three estimates together we obtain

$|\mathrm{I}|\lesssim\left[e^{-cr_{i}/\varepsilon_{i}}+\left(\frac{\varepsilon_{i}}{r_{i}}\right)^{n-3}+K^{-\frac{n-2}{2}}\right]\varepsilon_{i}^{\frac{n-2}{2}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

  2. (2)

By Proposition 17.6, Proposition 17.1 and Lemma 17.8, we get

$|\mathrm{II}|\lesssim\varepsilon_{i}^{n-4}r_{i}^{2}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)^{2}.$

  3. (3)

By (15.9) and (17.8),

$|\mathrm{III}|\lesssim\varepsilon_{i}^{n-2}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)^{2}.$

  4. (4)

To estimate $\mathrm{IV}$, first note that because $r_{i}\gg\varepsilon_{i}$,
the orthogonal condition (12.2) for $\phi_{i}$ reads as

$\int_{B_{r_{i}}}\phi_{i}(x,t)\eta_{in}(x,t)Z_{n+1,i}(x,t)dx=0.$

Differentiating this equation in $t$ gives

$\displaystyle\left|\int_{B_{r_{i}}}Z_{n+1,i}(x,0)\partial_{t}\phi_{i}(x,0)\eta_{in}(x,0)dx\right|$
$\displaystyle\leq$
$\displaystyle\left|\int_{B_{r_{i}}}\phi_{i}(x,0)\partial_{t}\eta_{in}(x,0)Z_{n+1,i}(x,0)dx\right|+\left|\int_{B_{r_{i}}}\phi_{i}(x,0)\eta_{in}(x,0)\partial_{t}Z_{n+1,i}(x,0)dx\right|$
$\displaystyle\lesssim$ $\displaystyle
K^{2}\varepsilon_{i}^{n-3}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

On the other hand, a direct calculation using Proposition 17.11 gives

$\displaystyle\left|\int_{B_{r_{i}}}Z_{n+1,i}(x,0)\partial_{t}\phi_{i}(x,0)\left[1-\eta_{in}(x,0)\right]dx\right|$
$\displaystyle\lesssim$
$\displaystyle\left(K^{-2}+r_{i}\right)\varepsilon_{i}^{\frac{n-4}{2}}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

Putting these two estimates together we get

$|\mathrm{IV}|\lesssim\left[\left(K^{-2}+r_{i}\right)\varepsilon_{i}^{\frac{n-2}{2}}+K^{2}\varepsilon_{i}^{n-2}\right]\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

  5. (5)

By Proposition 17.1, Lemma 17.8 and Proposition 17.11,

$|\mathrm{V}|\lesssim\left(\varepsilon_{i}^{2}r_{i}^{n-4}+r_{i}^{n-1}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)^{2}.$

  6. (6)

By Proposition 17.6 and Proposition 17.11,

$|\mathrm{VI}|\lesssim\varepsilon_{i}^{n-2}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)^{2}.$

  7. (7)

By Proposition 17.6,

$|\mathrm{VII}|\lesssim\varepsilon_{i}^{n-2}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)^{2}.$

  8. (8)

By Proposition 17.6, Proposition 17.1 and Lemma 17.8,

$|\mathrm{VIII}|\lesssim\varepsilon_{i}^{\frac{3n}{2}-3}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)^{2}.$

  9. (9)

By Proposition 17.6,

$|\mathrm{IX}|\lesssim\varepsilon_{i}^{2n-3}\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right)^{2}.$

Putting these estimates together, we get

$\mbox{RHS}\lesssim
r_{i}^{-1}\left[\left(K^{-2}+r_{i}\right)\varepsilon_{i}^{\frac{n-2}{2}}+\varepsilon_{i}^{2}r_{i}^{n-4}+r_{i}^{n-1}\right]\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$
(18.6)

###  19\. A weak form of Schoen’s Harnack inequality

In this section, we establish a weak form of Schoen’s Harnack inequality,
which then finishes the proof of Theorem 10.1. In the Yamabe problem, this
Harnack inequality was first introduced in Schoen [77], see also Li [47] and
Li-Zhang [49] for a proof using the method of moving plane and moving sphere.
The following proof instead is mainly a consequence of the Pohozaev identity
calculation in Section 18.

Combining (18.1) and (18.6) in the previous section, we get

$\displaystyle\frac{1}{|\partial B_{\rho}|}\int_{\partial
B_{\rho}}\phi_{i}(0)$ $\displaystyle\lesssim$
$\displaystyle\left(\rho+\frac{\varepsilon_{i}^{2}}{r_{i}^{2}}+\frac{r_{i}^{n-2}}{\varepsilon_{i}^{(n-2)/2}}+\frac{\varepsilon_{i}^{2}}{r_{i}^{2}}+K^{-2}+r_{i}+\frac{r_{i}^{n-3}}{\varepsilon_{i}^{\frac{n}{2}-3}}+e^{-c\frac{r_{i}}{\varepsilon_{i}}}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$

By choosing $r_{i}=\varepsilon_{i}^{(n-1)/n}$, we obtain a constant
$\delta(n)>0$ such that

$\frac{1}{|\partial B_{\rho}|}\int_{\partial
B_{\rho}}\phi_{i}(0)\lesssim\left(\rho+K^{-2}+\varepsilon_{i}^{\delta(n)}\right)\left(\left\|\phi_{i}\right\|_{L^{\infty}(Q_{1})}+e^{c\varepsilon_{i}^{-2}}\right).$
(19.1)

On $\partial B_{\rho}$, $u_{i}(0)\to u_{\infty}(0)$ uniformly. In view of the
estimates (15.8)-(15.10), we deduce that $\phi_{i}(0)\to u_{\infty}(0)$
uniformly on $\partial B_{\rho}$, too. Passing to the limit in (19.1), we
obtain

$\frac{1}{|\partial B_{\rho}|}\int_{\partial
B_{\rho}}u_{\infty}(x,0)\lesssim\left(\rho+K^{-2}\right)\|u_{\infty}\|_{L^{\infty}(Q_{1})}\lesssim\rho+K^{-2}.$
(19.2)

Since $u_{\infty}$ is a smooth, nonnegative solution of (1.1), if $u_{\infty}$
is not identically zero, by the standard Harnack ineqaulity,

$\inf_{Q_{1/4}}u_{\infty}>0.$ (19.3)

On the other hand, if we have chosen $K$ large enough at the beginning (in
Proposition 12.1) and taken $\rho$ arbitrarily small, (19.2) contradicts
(19.3). 111We can also avoid the use of Harnack inequality, and use only the
strong maximum principle. For this approach, we need to take a sequence of
$K\to+\infty$ in Proposition 12.1. This changes the error function $\phi_{i}$,
but it does not affect the argument. This is because both (19.2) and (19.3)
involve only $u_{\infty}$, which is the weak limit of $u_{i}$ and does not
depend on the construction of $\phi_{i}$.

Hence in the setting of Section 10, we must have $u_{\infty}=0$. This finishes
the proof of Theorem 10.1.

For the study of bubble clusterings (see Section 20 below), we need a
quantitative version of the above qualitative description. If the solution
exists for a sufficiently large time, an iteration of the following
proposition backwardly in time will lead to an quantitative upper bound on the
error function, see Section 20 for details. In particular, if the solution
exists globally in time (e.g. a solution independent of time), we can recover
Schoen’s Harnack inequality.

######  Proposition 19.1.

There exist two universal constants $\varepsilon_{0}$ and $M_{0}$ so that the
following holds. Suppose $u_{\varepsilon}$ is a positive solution of (1.1) in
$Q_{1}$, satisfying (II.a’)-(I.d’) and the decomposition (17.1) with
parameters
$(a_{\varepsilon}(t),\xi_{\varepsilon}(t),\lambda_{\varepsilon}(t))$, where

$\lambda_{\varepsilon}(0)=\varepsilon,\quad\xi_{\varepsilon}(0)=0.$

If $\varepsilon\leq\varepsilon_{0}$ and

$\|\phi_{\varepsilon}\|_{L^{\infty}(Q_{1})}\geq
M_{0}\varepsilon^{\frac{n-2}{2}},$ (19.4)

then

$\|\phi_{\varepsilon}\|_{L^{\infty}(B_{1}\times(-1/4,1/8))}\leq\frac{1}{2}\|\phi_{\varepsilon}\|_{L^{\infty}(Q_{1})}.$

###### Proof.

We prove this proposition by contradiction. Assume

  * •

$u_{i}$ is a sequence of solutions satisfying (II.a)-(II.c) and (10.1);

  * •

the decomposition given in (17.1) holds, where

$\varepsilon_{i}:=\lambda_{i}(0)\to 0,\quad\xi_{i}(0)=0;$

  * •

$\phi_{i}$ satisfies

$\lim_{i\to+\infty}\frac{\|\phi_{i}\|_{L^{\infty}(Q_{1})}}{\varepsilon_{i}^{\frac{n-2}{2}}}=+\infty,$
(19.5)

but

$\|\phi_{i}\|_{L^{\infty}(B_{1}\times(-1/4,1/8))}>\frac{1}{2}\|\phi_{i}\|_{L^{\infty}(Q_{1})}.$
(19.6)

We show that this leads to a contradiction.

Step 1. Denote $\delta_{i}:=\|\phi_{i}\|_{L^{\infty}(Q_{1})}$. Let
$\widetilde{\phi}_{i}:=\phi_{i}/\delta_{i}$. In this step we prove

$\widetilde{\phi}_{i}\to 0\quad\mbox{uniformly in any compact set
of}~{}~{}Q_{1}.$ (19.7)

First by Proposition 17.10, for any $\rho>0$, $\widetilde{\phi}_{i}$ are
uniformly Lipschitz in $(B_{2/3}\setminus B_{\rho})\times(-4/9,4/9)$. Hence
after passing to a subsequence, we may assume $\widetilde{\phi}_{i}$ converges
to a limit $\widetilde{\phi}$, uniformly in any compact set of
$(B_{2/3}\setminus\\{0\\})\times(-4/9,4/9)$.

Because

$u_{i}=\phi_{i}+O\left(\varepsilon_{i}^{\frac{n-2}{2}}|x|^{2-n}\right),$

in view of (19.5), $u_{i}/\delta_{i}$ also converges uniformly to
$\widetilde{\phi}$ in any compact set of
$(B_{2/3}\setminus\\{0\\})\times(-4/9,4/9)$. As a consequence,

$\widetilde{\phi}\geq 0\quad\mbox{in
}\left(B_{2/3}\setminus\\{0\\}\right)\times(-4/9,4/9).$ (19.8)

Dividing (1.1) by $\delta_{i}$, and then letting $i\to+\infty$, by the above
uniform convergence of $u_{i}/\delta_{i}$, we get

$\partial_{t}\widetilde{\phi}-\Delta\widetilde{\phi}=0\quad\mbox{in
}\left(B_{2/3}\setminus\\{0\\}\right)\times(-4/9,4/9).$

Since $\left|\widetilde{\phi}\right|\leq 1$ in $Q_{1}$, removable singularity
theorem for heat equation implies that $\widetilde{\phi}$ is smooth and
satisfies the heat equation in $Q_{2/3}$.

By (19.8) and the strong maximum principle, either $\widetilde{\phi}\equiv 0$
or $\widetilde{\phi}>0$ everywhere in $Q_{2/3}$. We claim that the first case
must happen.

Indeed, dividing both sides of (19) by $\delta_{i}$ and letting $i\to+\infty$,
by (19.5) and the above uniform convergence of $\widetilde{\phi}_{i}$, we
obtain

$\frac{1}{|\partial B_{\rho}|}\int_{\partial
B_{\rho}}\widetilde{\phi}(x,0)\lesssim\rho+K^{-2}.$

Sending $\rho\to 0$ and $K\to+\infty$ as before, we get
$\widetilde{\phi}(0,0)=0$. By the strong maximum principle (or Harnack
inequality), $\widetilde{\phi}\equiv 0$ in $Q_{2/3}$.

Step 2. Choose a small radius $\rho$. By results obtained in Step 1, for all
$i$ large,

$|\phi_{i}|\ll\delta_{i}\quad\mbox{on}~{}~{}\partial^{p}Q_{2/3}\setminus\left(B_{\rho}\times\\{-2/3\\}\right).$
(19.9)

On $B_{\rho}\times\\{-2/3\\}$, we have the trivial bound

$|\phi_{i}|\leq\delta_{i}.$ (19.10)

As in Subsection 17.1, we obtain
Claim. If $\rho$ is sufficiently small, there exists a constant
$\sigma(\rho)\ll 1$ (independent of $\varepsilon_{i}$) such that

$|\phi_{i,1}|\leq\sigma(\rho)\delta_{i}\quad\mbox{in }Q_{7/12}.$ (19.11)

Here $\phi_{i,1}$ denotes the decomposition of the outer component taken in
Section 14.

By this claim, repeating the estimates of other terms in the decomposition of
the outer component, $\phi_{i,2}$-$\phi_{i,5}$, and the corresponding estimate
for the inner component $\phi_{in,i}$ in Section 16, we obtain

$|\phi_{i}|\lesssim\sigma(\rho)\delta_{i}\quad\mbox{in }Q_{1/2}.$ (19.12)

Step 3. By (19.12), we get

$u_{i}\lesssim\varepsilon_{i}^{\frac{n-2}{2}}+\sigma(\rho)\delta_{i}\quad\mbox{in
}\left(B_{1/2}\setminus B_{1/4}\right)\times(-1/4,1/4).$

Because $u_{i}$ satisfies the standard parabolic Harnack inequality in
$\left(B_{1}\setminus B_{1/8}\right)\times(-1,1)$, we get

$u_{i}\lesssim\varepsilon_{i}^{\frac{n-2}{2}}+\sigma(\rho)\delta_{i}\quad\mbox{in
}\left(B_{1}\setminus B_{1/2}\right)\times(-1/4,1/8).$

By the definition of $\phi_{i}$, we get

$-\varepsilon_{i}^{\frac{n-2}{2}}\lesssim\phi_{i}\lesssim\varepsilon_{i}^{\frac{n-2}{2}}+\sigma(\rho)\delta_{i}\quad\mbox{in
}\left(B_{1}\setminus B_{1/2}\right)\times(-1/4,1/8).$

Combining these two estimates with (19.12), for all $i$ sufficiently large, we
have

$\left|\widetilde{\phi}_{i}\right|\leq 1/4\quad\mbox{in
}B_{1}\times(-1/4,1/8).$

This is a contradiction with (19.6). ∎

###  20\. A conditional exclusion of bubble clustering

In this section, we work in the following settings. (This will appear in a
suitable rescalings of bubble clustering from Theorem 3.1 in Part III.)

  1. (1)

There exist two sequences $R_{i}$ and $T_{i}$, both diverging to $+\infty$ as
$i\to+\infty$;

  2. (2)

$u_{i}$ is a sequence of positive, smooth solution of (1.1) in
$\Omega_{i}:=B_{R_{i}}\times(-T_{i},T_{i})$;

  3. (3)

as $i\to+\infty$,

$\int_{-T_{i}}^{T_{i}}\int_{B_{R_{i}}}\partial_{t}u_{i}(x,t)^{2}dxdt\to 0;$
(20.1)

  4. (4)

there exists an $N\in\mathbb{N}$ such that for each $i$ and
$t\in(-T_{i},T_{i})$, in any compact set of $\mathbb{R}^{n}$, we have the
bubble decomposition

$u_{i}(x,t)=\sum_{j=1}^{N}W_{\xi_{ij}^{\ast}(t),\lambda_{ij}^{\ast}(t)}(x)+o_{i}(1),$
(20.2)

where $o_{i}(1)$ are measured in $H^{1}_{loc}(\mathbb{R}^{n})$;

  5. (5)

as $i\to+\infty$,

$\max_{j=1,\cdots,N}\sup_{-T_{i}<t<T_{i}}\lambda_{ij}^{\ast}(t)\to 0;$ (20.3)

  6. (6)

for any $t\in(-T_{i},T_{i})$,

$\min_{1\leq j\neq k\leq N}|\xi_{ij}^{\ast}(t)-\xi_{ik}^{\ast}(t)|\geq 2,$
(20.4)

and at $t=0$,

$\xi_{i1}^{\ast}(0)=0,\quad|\xi_{i2}^{\ast}(0)|=1;$ (20.5)

  7. (7)

after relabelling indices, assume for some $N^{\prime}\leq N$ and any
$j=1,\cdots,N^{\prime}$,

$\xi_{ij}^{\ast}(0)\to P_{j},\quad\mbox{as}~{}~{}i\to+\infty,$

while for any $j=N^{\prime}+1,\cdots,N$,

$|\xi_{ij}^{\ast}(0)|\to+\infty,\quad\mbox{as}~{}~{}i\to+\infty,$

By (20.5), $P_{1}=0$ and $P_{2}\in\partial B_{1}$, so $N^{\prime}\geq 2$.

The main result of this section is

######  Proposition 20.1.

Under the above assumptions, we have

$T_{i}\leq 100\max_{1\leq j\leq N^{\prime}}|\log\lambda_{ij}^{\ast}(0)|.$
(20.6)

We prove this proposition by contradiction, so assume (20.6) does not hold,
that is,

$T_{i}>100\max_{1\leq j\leq N^{\prime}}|\log\lambda_{ij}^{\ast}(0)|.$ (20.7)

With this bound in hand, we can iterate Proposition 19.1 backwardly in time,
leading to an optimal upper bound on the error function as Schoen’s Harnack
inequality in Yamabe problem. This bound allows us to define a Green function
from $u_{i}$. This will be done in Subsection 20.1. Then in Subsection 20.2,
still similar to the treatment in Yamabe problem, we employ Pohozaev identity
again to give a sign restriction on the next order term in the expansion of
this Green function at a pole. This then leads to a contradiction with the
assumption that there is one bubble located at $P_{1}$ and $P_{2}$
respectively.

####  20.1. Construction of Green function

Under the above assumptions, for any $i,j$ and $t\in(-T_{i}+1,T_{i}-1)$,
Proposition 12.1 can be applied to $u_{i}$ in $Q_{1}(\xi_{ij}^{\ast}(t),t)$.
We denote the corresponding parameters by $\xi_{ij}(t)$, $\lambda_{ij}(t)$ and
$a_{ij}(t)$, and the error function

$\phi_{ij}(x,t):=u_{i}(x,t)-W_{\xi_{ij}(t),\lambda_{ij}(t)}(x)-a_{ij}(t)Z_{0,\xi_{ij}(t),\lambda_{ij}(t)}(x).$
(20.8)

Denote $\varepsilon_{ij}:=\lambda_{ij}(0)$, $j=1,\cdots,N^{\prime}$ and
$\varepsilon_{i}=\max_{1\leq j\leq N^{\prime}}\varepsilon_{ij}$. By
Proposition 12.1 and (20.7), we get

$T_{i}>50\max_{1\leq j\leq N^{\prime}}|\log\varepsilon_{ij}|.$ (20.9)

######  Lemma 20.2.

For each $j=1,\cdots,N^{\prime}$ and any
$t\in(-50|\log\varepsilon_{ij}|,50|\log\varepsilon_{ij}|)$,

$\frac{1}{2}\varepsilon_{ij}\leq\lambda_{ij}(t)\leq 2\varepsilon_{ij}.$
(20.10)

###### Proof.

Applying Proposition 17.6 to each $\phi_{ij}$ in $Q_{1}(\xi_{ij}(t),t)$, we
obtain

$\left|\lambda_{ij}^{\prime}(t)\right|\lesssim\lambda_{ij}(t)^{\frac{n-4}{2}}.$
(20.11)

Define

$\widetilde{T}_{ij}:=\sup\left\\{T:~{}~{}T\leq
50|\log\varepsilon_{ij}|,~{}~{}\lambda_{ij}(t)\leq
2\varepsilon_{ij}\quad\mbox{in }~{}[0,T]\right\\}.$

Integrating (20.11) on $[0,\widetilde{T}_{ij}]$ leads to

$\sup_{t\in[0,\widetilde{T}_{ij}]}\lambda_{ij}(t)\leq\varepsilon_{ij}e^{C\varepsilon_{ij}^{(n-6)/2}|\log\varepsilon_{ij}|}\leq\varepsilon_{ij}e^{C\varepsilon_{ij}^{(n-6)/4}}<\frac{3}{2}\varepsilon_{ij}.$
(20.12)

Hence we must have $\widetilde{T}_{ij}=50|\log\varepsilon_{ij}|$. This gives
the upper bound of (20.10) in the positive side. The lower bound and the
negative side follow in the same way. ∎

For each $t\in(-50|\log\varepsilon_{ij}|,50|\log\varepsilon_{ij}|)$, applying
Proposition 19.1 to $Q_{1}(\xi_{ij}(t),t)$, we obtain

$\sup_{Q_{1}(\xi_{ij}(t),t)}|\phi_{ij}|\leq
M_{0}\varepsilon_{i}^{\frac{n-2}{2}}+\frac{1}{2}\sup_{Q_{1}(\xi_{ij}(t+1),t+1)}|\phi_{ij}|.$
(20.13)

For any $R>1$ fixed, an iteration of this estimate from
$t=50|\log\varepsilon_{ij}|$ to any $t\in[-R,R]$ leads to

$\sup_{B_{1}(\xi_{ij}(0))\times(-R,R)}|\phi_{ij}|\lesssim\varepsilon_{i}^{\frac{n-2}{2}}.$
(20.14)

This bound depends only on the constant $M_{0}$ in Proposition 19.1, and it is
independent of $R$.

Substituting (20.14) and estimates of parameters in Proposition 17.6 into
(20.8), we get

$\sup_{B_{1}(\xi_{ij}(0))\times(-R,R)}u_{i}\lesssim\varepsilon_{i}^{\frac{n-2}{2}}.$

In $\left(B_{7/4}(\xi_{ij}(0))\setminus
B_{3/4}(\xi_{ij}(0))\right)\times(-R,R)$, by applying standard Harnack
inequality to $u_{i}$, (20.14) is extended to

$\sup_{B_{3/2}(\xi_{ij}(0))\times(-R,R)}|\phi_{ij}|\lesssim\varepsilon_{i}^{\frac{n-2}{2}}.$
(20.15)

This estimate gives us the expansion

$u_{i}(x,t)=\varepsilon_{ij}^{-\frac{n-2}{2}}W\left(\frac{x-\xi_{ij}(t)}{\varepsilon_{ij}}\right)+O\left(\varepsilon_{i}^{\frac{n-2}{2}}\right)\quad\mbox{in}~{}~{}B_{3/2}(\xi_{ij}(0))\times(-R,R).$
(20.16)

Here we note that, by Proposition 17.6,

$|\xi_{ij}^{\prime}(t)|\lesssim\lambda_{ij}(t)^{\frac{n-4}{2}}\lesssim\varepsilon_{i}^{\frac{n-4}{2}}.$

Hence

$|\xi_{ij}(t)-\xi_{ij}(0)|\lesssim_{R}\varepsilon_{i}^{\frac{n-4}{2}},\quad\mbox{in}~{}~{}[-R,R].$
(20.17)

Assume

$\lim_{i\to+\infty}\frac{\varepsilon_{ij}}{\varepsilon_{i}}=m_{j}\in[0,1]\quad\mbox{for
any }~{}~{}j=1,\cdots,N^{\prime}.$

By the definition of $\varepsilon_{i}$, there exists at least one $j$
satisfying $m_{j}=1$.

It is directly verified that for some dimensional constant $c(n)>0$,

$\varepsilon_{i}^{-\frac{n-2}{2}}u_{i}^{p}dxdt\rightharpoonup
c(n)m_{j}^{\frac{n-2}{2}}\delta_{P_{j}}\otimes dt$ (20.18)

weakly as Radon measures in $B_{3/2}(P_{j})\times(-R,R)$.

Set

$\widehat{u}_{i}:=\varepsilon_{i}^{-\frac{n-2}{2}}u_{i}.$

It satisfies

$\partial_{t}\widehat{u}_{i}-\Delta\widehat{u}_{i}=\varepsilon_{i}^{-\frac{n-2}{2}}u_{i}^{p}=\varepsilon_{i}^{2}\widehat{u}_{i}^{p}.$
(20.19)

By (20.15), in any compact set of
$\left(B_{3/2}(P_{j})\setminus\\{P_{j}\\}\right)\times\mathbb{R}$,
$\widehat{u}_{i}$ are uniformly bounded. Because $u_{i}$ satisfies the
standard parabolic Harnack inequality in any compact set of
$\left(\mathbb{R}^{n}\setminus\cup_{j=1}^{N^{\prime}}\\{P_{j}\\}\right)\times\mathbb{R}$,
we deduce that $\widehat{u}_{i}$ are also uniformly bounded in any compact set
of
$\left(\mathbb{R}^{n}\setminus\cup_{j=1}^{N^{\prime}}\\{P_{j}\\}\right)\times\mathbb{R}$.
Then by standard parabolic regularity theory, $\widehat{u}_{i}$ converges to
$\widehat{u}_{\infty}$ smoothly in any compact set of
$\left(\mathbb{R}^{n}\setminus\cup_{j=1}^{N^{\prime}}\\{P_{j}\\}\right)\times\mathbb{R}$.

By (20.18), $\widehat{u}_{\infty}$ satisfies

$\partial_{t}\widehat{u}_{\infty}-\Delta\widehat{u}_{\infty}=c(n)\sum_{j=1}^{N^{\prime}}m_{j}^{\frac{n-2}{2}}\delta_{P_{j}}\otimes
dt\quad\mbox{in}~{}~{}\mathbb{R}^{n}\times\mathbb{R}.$ (20.20)

######  Lemma 20.3.

For each $j=1\cdots,N^{\prime}$,

$m_{j}>0.$

###### Proof.

Because there is one $m_{j}$ satisfying $m_{j}=1$, $\widehat{u}_{\infty}\neq
0$. By the strong maximum principle,

$\widehat{u}_{\infty}>0\quad\mbox{in}~{}~{}\left(\mathbb{R}^{n}\setminus\cup_{j=1}^{N^{\prime}}\\{P_{j}\\}\right)\times\mathbb{R}.$
(20.21)

By (20.16), for each $j=1,\cdots,N^{\prime}$,

$\widehat{u}_{i}\lesssim\left(\frac{\varepsilon_{ij}}{\varepsilon_{i}}\right)^{\frac{n-2}{2}}\quad\mbox{in}~{}~{}\left(B_{3/2}(P_{j})\setminus
B_{1/2}(P_{j})\right)\times(-1,1).$

If $m_{j}=0$, we would have

$\widehat{u}_{\infty}=\lim_{i\to+\infty}\widehat{u}_{i}=0\quad\mbox{in}~{}~{}\left(B_{3/2}(P_{j})\setminus
B_{1/2}(P_{j})\right)\times(-1,1).$

This is a contradiction with (20.21). ∎

######  Lemma 20.4.

For any
$(x,t)\in\left(\mathbb{R}^{n}\setminus\cup_{j=1}^{N^{\prime}}\\{P_{j}\\}\right)\times\mathbb{R}$,

$\widehat{u}_{\infty}(x,t)\equiv
c(n)\sum_{j=1}^{N^{\prime}}m_{j}^{\frac{n-2}{2}}|x-P_{j}|^{2-n}.$ (20.22)

###### Proof.

Take two arbitrary $s<t$. Take a sequence of cut-off functions $\zeta_{R}\in
C_{0}^{\infty}(\mathbb{R}^{n})$ such that

$\left\\{\begin{aligned} &0\leq\zeta_{R}\leq 1,\\\ &\zeta_{R}\equiv
1\quad\mbox{in}~{}~{}B_{R}\setminus\cup_{j=1}^{N^{\prime}}B_{1/R}(P_{j}),\\\
&\zeta_{R}\equiv
0\quad\mbox{in}~{}~{}B_{2R}^{c}\cup\cup_{j=1}^{N^{\prime}}B_{1/2R}(P_{j}).\end{aligned}\right.$

By the comparison principle, for any
$x\in\mathbb{R}^{n}\setminus\\{P_{1},\cdots,P_{N^{\prime}}\\}$,

$\displaystyle\widehat{u}_{\infty}(x,t)$ $\displaystyle\geq$
$\displaystyle\int_{\mathbb{R}^{n}}\widehat{u}_{\infty}(y,s)\zeta_{R}(y)G(x-y,t-s)dy$
$\displaystyle+$ $\displaystyle
c(n)\sum_{j=1}^{N^{\prime}}m_{j}^{\frac{n-2}{2}}\int_{s}^{t}G(x-P_{j},t-s)ds.$

Here $G$ denotes the standard heat kernel on $\mathbb{R}^{n}$.

Letting $R\to+\infty$, by the monotone convergence theorem we obtain

$\widehat{u}_{\infty}(x,t)\geq\int_{\mathbb{R}^{n}}\widehat{u}_{\infty}(y,s)G(x-y,t-s)dy+c(n)\sum_{j=1}^{N^{\prime}}m_{j}^{\frac{n-2}{2}}\int_{s}^{t}G(x-P_{j},t-s)ds.$

Letting $s\to-\infty$, we get

$\widehat{u}_{\infty}(x,t)\geq
c(n)\sum_{j=1}^{N^{\prime}}m_{j}^{\frac{n-2}{2}}|x-P_{j}|^{2-n}.$

Thus their difference

$\widehat{v}_{\infty}(x,t):=\widehat{u}_{\infty}(x,t)-c(n)\sum_{j=1}^{N^{\prime}}m_{j}^{\frac{n-2}{2}}|x-P_{j}|^{2-n}$

is a positive caloric function on $\mathbb{R}^{n}\times\mathbb{R}$. By [51] or
[90], there exists a nonnegative Radon measure $\nu$ on
$\\{(\xi,\lambda):\lambda=|\xi|^{2}\\}\subset\mathbb{R}^{n}\times\mathbb{R}$
such that

$\widehat{v}_{\infty}(x,t)=\int_{\\{\lambda=|\xi|^{2}\\}}e^{\lambda t+\xi\cdot
x}d\nu(\xi,\lambda).$

Because (20.16) holds for any $R$, $\widehat{u}_{\infty}$ is bounded in
$\left(B_{1}(P_{1})\setminus B_{1/2}(P_{1})\right)\times\mathbb{R}$, so
$\widehat{v}_{\infty}$ is also bounded $\left(B_{1}(P_{1})\setminus
B_{1/2}(P_{1})\right)\times\mathbb{R}$. This is possible only if $\nu=0$, and
(20.22) follows. ∎

By this lemam, near $P_{1}=0$, there exists a smooth harmonic function $h$
such that

$\widehat{u}_{\infty}(x)=c(n)m_{1}|x|^{2-n}+h(x).$

Because $P_{2}\in\partial B_{1}$, we deduce that

$h(0)>0.$ (20.23)

####  20.2. Local Pohozaev invariants

By the expansion of $\widehat{u}_{\infty}(x)$ near $0$, in particular,
(20.23), and a direct calculation, we find a dimensional constant $c(n)>0$
such that for all $r$ small,

$\mathcal{P}_{\widehat{u}_{\infty}}(r)=\frac{c(n)m_{1}h(0)}{r}+O(1).$ (20.24)

On the other hand, because $\widehat{u}_{\infty}$ comes from $u_{i}$, we claim
that

######  Lemma 20.5.

For all $r$ small,

$\mathcal{P}_{\widehat{u}_{\infty}}(r)\lesssim K^{-2}+r.$ (20.25)

###### Proof.

In (18.1), choose $r_{i}$ to be a fixed, small $r>0$. Multiply both sides of
(18.1) by $\varepsilon_{i}^{2-n}$. Let us analyse the convergence of both
sides.

Step 1. The left hand side of (18.1).

By the convergence of $\widehat{u}_{i}$ in
$C^{\infty}_{loc}\left(\left(\mathbb{R}^{n}\setminus\cup_{j=1}^{N^{\prime}}\\{P_{j}\\}\right)\times\mathbb{R}\right)$,
we get

$\varepsilon_{i}^{2-n}\mathcal{P}_{u_{i}(0)}(r)=\mathcal{P}_{\widehat{u}_{i}(0)}(r)\to\mathcal{P}_{\widehat{u}_{\infty}(0)}(r),\quad\mbox{as
}~{}~{}i\to+\infty$ (20.26)

and

$\varepsilon_{i}^{2-n}\int_{\partial
B_{r}}u_{i}^{p+1}\lesssim\varepsilon_{i}^{2}\to 0,\quad\mbox{as
}~{}~{}i\to+\infty.$ (20.27)

Hence the left hand side converges to $\mathcal{P}_{\widehat{u}_{\infty}}(r)$.

Step 2. The right hand side of (18.1).

For simplicity of notations, denote the parameters
$(a_{i1},\xi_{i1},\lambda_{i1})$ just by $(a_{i},\xi_{i},\lambda_{i})$, and
$\phi_{i1}$, the error function in $Q_{1}$, just by $\phi_{i}$. Plugging the
$L^{\infty}$ estimate (20.14) into Proposition 17.6, we get the following
estimates:

$\sup_{t\in(-1,1)}|a_{i}(t)|\lesssim\varepsilon_{i}^{n-1},$ (20.28)
$\sup_{t\in(-1,1)}\left(\left|a_{i}^{\prime}(t)-\mu_{0}\frac{a_{i}(t)}{\lambda_{i}(t)^{2}}\right|+\left|\xi_{i}^{\prime}(t)\right|+\left|\lambda_{i}^{\prime}(t)\right|\right)\lesssim
K^{-\frac{n-2}{2}}\varepsilon_{i}^{n-3},$ (20.29)
$|\nabla\phi_{i}(x,t)|\lesssim\frac{\varepsilon_{i}^{(n+2)/2}}{\varepsilon_{i}^{3}+|x|^{3}}+\varepsilon_{i}^{\frac{n-2}{2}}\quad\mbox{for
any}~{}~{}(x,t)\in Q_{1},$ (20.30)

and

$|\partial_{t}\phi_{i}(x,t)|\lesssim\frac{\varepsilon_{i}^{(n+2)/2}}{\varepsilon_{i}^{4}+|x|^{4}}+\frac{\varepsilon_{i}^{(n-2)/2}}{\varepsilon_{i}+|x|}\quad\mbox{for
any}~{}~{}(x,t)\in Q_{1}.$ (20.31)

The factor $K^{-(n-2)/2}$ in (20.29) comes from Lemma 13.5.

As in Subsection 18.2, we expand the right hand side of (18.1) into nine
terms, $\mathrm{I}$-$\mathrm{IX}$. Let us estimate them one by one.

  1. (1)

By (20.29) and the orthogonal relation between $Z_{0}$ and $Z_{n+1}$,

$\left|\lambda_{i}(0)a_{i}^{\prime}(0)\int_{B_{r}}Z_{0,i}(x,0)Z_{n+1,i}(x,0)dx\right|\lesssim\varepsilon_{i}^{n-2}e^{-cr/\varepsilon_{i}}.$

Similarly, for each $j=1,\cdots,n$,

$\left|\lambda_{i}(0)\xi_{i}^{\prime}(0)\int_{B_{r}}Z_{j,i}(x,0)Z_{n+1,i}(x,0)dx\right|\lesssim\varepsilon_{i}^{2n-5}r^{3-n},$

and

$\left|\lambda_{i}(0)\lambda_{i}^{\prime}(0)\int_{B_{r}}Z_{n+1,i}(0)^{2}\right|\lesssim
K^{-\frac{n-2}{2}}\varepsilon_{i}^{n-2}.$

Combining these three estimates, we obtain

$|\mathrm{I}|\lesssim\left[K^{-\frac{n-2}{2}}+\left(\frac{\varepsilon_{i}}{r}\right)^{n-3}+e^{-cr/\varepsilon_{i}}\right]\varepsilon_{i}^{n-2}.$

  2. (2)

By (20.14), (20.29) and (20.30),

$|\mathrm{II}|\lesssim\varepsilon_{i}^{2n-6}r^{2}.$

  3. (3)

By (20.28) and (20.29),

$|\mathrm{III}|\lesssim\varepsilon_{i}^{2n-4}.$

  4. (4)

As in the treatment of $\mathrm{IV}$ in Subsection 18.2, we get

$|\mathrm{IV}|\lesssim
K^{2}\varepsilon_{i}^{\frac{3n}{2}-3}+\left(K^{-2}+r\right)\varepsilon_{i}^{n-2}.$

  5. (5)

By (20.14), (20.28) and (20.29),

$|\mathrm{V}|\lesssim\varepsilon_{i}^{n}r^{n-4}+\varepsilon_{i}^{n-2}r^{n-1}.$

  6. (6)

By (20.28) and (20.31),

$|\mathrm{VI}|\lesssim\varepsilon_{i}^{2n-4}.$

  7. (7)

By (20.28) and (20.29),

$|\mathrm{VII}|\lesssim\varepsilon_{i}^{2n-4}.$

  8. (8)

By (20.28), (20.29) and (20.30),

$|\mathrm{VIII}|\lesssim\varepsilon_{i}^{3n-6}.$

  9. (9)

By (20.28) and (20.29)

$|\mathrm{IX}|\lesssim\varepsilon_{i}^{3n-6}.$

Putting these estimates together and using (20.26), we obtain (20.25). ∎

Combining this lemma with (20.24), after letting $r\to 0$, we deduce that
$h(0)=0$. This is a contradiction with (20.23), so (20.7) cannot be true. The
proof of Proposition 20.1 is thus complete.

###  Appendix A Linearization estimates around bubbles

In this appendix, we collect some estimates about the linearized equation
around the standard bubble. This is mainly used to study the inner problem in
Section 13.

Recall that the standard positive bubble is

$W(x):=\left(1+\frac{|x|^{2}}{n(n-2)}\right)^{-\frac{n-2}{2}}.$

Concerning eigenvalues and eigenfunctions for the linearized operator
$-\Delta-pW^{p-1}$, we have (see for example [11, Proposition 2.2])

######  Theorem A.1.

  * (i)

There exists one and only one negative eigenvalue for $-\Delta-pW^{p-1}$,
denoted by $-\mu_{0}$, for which there exists a unique (up to a constant),
positive, radially symmetric and exponentially decaying eigenfunction $Z_{0}$.

  * (ii)

There exist exactly $(n+1)$-eigenfunctions $Z_{i}$ in
$L^{\infty}(\mathbb{R}^{n})$ corresponding to eigenvalue $0$, given by

$\left\\{\begin{aligned} &Z_{i}=\frac{\partial W}{\partial x_{i}},\quad
i=1,\cdots,n,\\\ &Z_{n+1}=\frac{n-2}{2}W+x\cdot\nabla W.\end{aligned}\right.$

######  Remark A.2.

The following decay as $|x|\to\infty$ holds for these eigenfunctions:

$\left\\{\begin{aligned} &Z_{0}(x)\lesssim e^{-c|x|},\\\
&|Z_{i}(x)|\lesssim|x|^{1-n},\quad i=1,\cdots,n,\\\
&|Z_{n+1}(x)|\lesssim|x|^{2-n}.\end{aligned}\right.$

Throughout the paper $Z_{0}$ is normalized so that

$\int_{\mathbb{R}^{n}}Z_{0}^{2}=1.$

For any $\xi\in\mathbb{R}^{n}$ and $\lambda\in\mathbb{R}^{+}$, in accordance
with the scalings for $W$, define

$Z_{i,\xi,\lambda}(x):=\lambda^{-\frac{n}{2}}Z_{i}\left(\frac{x-\xi}{\lambda}\right).$

This scaling preserves the $L^{2}(\mathbb{R}^{n})$ norm of $Z_{i}$.

Next we study the linearized parabolic equation around $W$. The first one is a
nondegeneracy result.

######  Proposition A.3 (Nondegeneracy).

Suppose $\alpha>2$ and $\varphi\in L^{\infty}(-\infty,0;\mathcal{X}_{\alpha})$
is a solution of

$\partial_{t}\varphi-\Delta\varphi=pW^{p-1}\varphi,\quad\mbox{in
}~{}~{}\mathbb{R}^{n}\times\mathbb{R}^{-}.$ (A.1)

Then there exist $(N+2)$ constants $c_{0},\cdots,c_{n+1}$ such that

$\varphi(x,t)\equiv c_{0}e^{\mu_{0}t}Z_{0}(x)+\sum_{i=1}^{n+1}c_{i}Z_{i}(x).$

###### Proof.

By standard parabolic regularity theory and the $O(|x|^{-4})$ decay of
$W^{p-1}$ at infinity, we deduce that $\varphi\in
L^{\infty}(-\infty,0;\mathcal{X}_{\alpha}^{2+\theta})$ and
$\partial_{t}\varphi\in L^{\infty}(-\infty,0;\mathcal{X}_{2+\alpha})$. Because
$\alpha>2$, we may assume

$\int_{\mathbb{R}^{n}}\varphi(x,t)Z_{i}(x)dx\equiv 0,\quad\forall t<0,\quad
i=0,\cdots,n+1.$

Then $\partial_{t}\varphi$ satisfy these orthogonal conditions, too. Since it
is also a solution of (A.1), by the method in [11, Section 2] (in particular,
the coercivity estimate in [11, Lemma 2.3]), we deduce that
$\partial_{t}\varphi\equiv 0$. Therefore $\varphi$ is a stationary solution of
(A.1). Since $\varphi\in\mathcal{X}_{\alpha}$ and it is orthogonal to $Z_{i}$,
by Theorem A.1, $\varphi\equiv 0$. ∎

Now we state two a priori estimates for this linearized equation.

######  Lemma A.4.

Suppose $\alpha^{\prime}>\max\\{2,n/2\\}$,
$\varphi_{0}\in\mathcal{X}_{\alpha^{\prime}}$. If $\varphi_{0}$ satisfies the
orthogonal condition

$\int_{\mathbb{R}^{n}}\varphi_{0}(x)Z_{i}(x)dx=0,\quad\forall i=0,\cdots,n+1,$
(A.2)

then there exists a unique $\varphi\in
L^{\infty}_{loc}\left(\mathbb{R}^{+};\mathcal{X}_{\alpha^{\prime}}\right)$
solving the equation

$\left\\{\begin{aligned}
&\partial_{t}\varphi-\Delta\varphi=pW^{p-1}\varphi\quad\mbox{in}~{}~{}\mathbb{R}^{n}\times(0,2T_{D}),\\\
&\varphi(x,0)=\varphi_{0}(x)\quad\mbox{on}~{}~{}\mathbb{R}^{n},\end{aligned}\right.$
(A.3)

Moreover, $\varphi$ satisfies the orthogonal condition

$\int_{\mathbb{R}^{n}}\varphi(x,t)Z_{i}(x)dx=0,\quad\forall t>0,\quad\forall
i=0,\cdots,n+1.$ (A.4)

and the decay property

$\varphi(\cdot,\cdot+t)\to
0\quad\mbox{in}~{}~{}C_{loc}^{(1+\theta)/2}(\mathbb{R};C_{loc}^{1,\theta}(\mathbb{R}^{n}))~{}~{}\mbox{as}~{}~{}t\to+\infty.$
(A.5)

###### Proof.

Existence of a global solution $\varphi$ to the problem (A.3) follows from
standard parabolic theory, if we note that the heat semigroup $e^{t\Delta}$ is
uniformly bounded as an operator from $\mathcal{X}_{\alpha^{\prime}}$ into
itself.

By standard parabolic regularity theory, $\partial_{t}\varphi$,
$\nabla\varphi$ and $\Delta\varphi$ belong to
$L^{\infty}_{loc}\left(\mathbb{R}^{+};\mathcal{X}_{\alpha^{\prime}}\right)$.
The orthogonal condition (A.4) then follows by testing (A.3) with $Z_{i}$.

Testing (A.3) with $\varphi$ and then applying Theorem A.1, we obtain

$\frac{1}{2}\frac{d}{dt}\int_{\mathbb{R}^{n}}\varphi(x,t)^{2}dx=-\int_{\mathbb{R}^{n}}\left[|\nabla\varphi(x,t)|^{2}-pW(x)^{p-1}\varphi(x,t)^{2}\right]dx\leq
0.$ (A.6)

Testing (A.3) with $\partial_{t}\varphi$, we also obtain

$\frac{1}{2}\frac{d}{dt}\int_{\mathbb{R}^{n}}\left[|\nabla\varphi(x,t)|^{2}-pW(x)^{p-1}\varphi(x,t)^{2}\right]dx=-\int_{\mathbb{R}^{n}}\partial_{t}\varphi(x,t)^{2}dx\leq
0.$ (A.7)

Combining these two identities with Proposition A.3, we deduce that

$\lim_{t\to+\infty}\|\varphi(t)\|_{H^{1}(\mathbb{R}^{n})}=0.$

Then (A.5) follows by an application of standard parabolic regularity theory.
∎

The second one is inspired by [73, Lemma 5.1].

######  Lemma A.5.

If $\alpha>2$, there exists a positive constant $C(\alpha)$ so that the
following holds. Given $T>1$, assume $\varphi\in
L^{\infty}\left(0,T;\mathcal{X}_{\alpha}\right)$, $a_{i}\in L^{\infty}([0,T])$
and $E\in L^{\infty}\left(0,T;\mathcal{X}_{2+\alpha}\right)$ solve the problem

$\left\\{\begin{aligned}
&\partial_{t}\varphi-\Delta\varphi=pW^{p-1}\varphi+\sum_{i=0}^{n+1}a_{i}Z_{i}+E\quad\mbox{in}~{}~{}\mathbb{R}^{n}\times(0,T),\\\
&\varphi(x,0)=0,\end{aligned}\right.$ (A.8)

and $\varphi$ satisfies the orthogonal condition

$\int_{\mathbb{R}^{n}}\varphi(x,t)Z_{i}(x)dx=0,\quad\forall
t\in[0,T],\quad\forall i=0,\cdots,n+1.$ (A.9)

Then

$\left\\{\begin{aligned} &\sum_{i=0}^{n+1}|a_{i}(t)|\leq
C\|E(t)\|_{2+\alpha},\quad\forall t\in[0,T],\\\
&\|\varphi\|_{C^{(1+\theta)/2}\left(0,T;\mathcal{X}^{1+\theta}_{\alpha}\right)}\leq
C(\alpha)\|E\|_{L^{\infty}\left(0,T;\mathcal{X}_{2+\alpha}\right)}.\end{aligned}\right.$

###### Proof.

First, for each $i$, multiplying (A.8) by $Z_{i}$ and integrating on
$\mathbb{R}^{n}$, we obtain

$a_{i}(t)=-\frac{\int_{\mathbb{R}^{n}}E(x,t)Z_{i}(x)}{\int_{\mathbb{R}^{n}}Z_{i}(x)^{2}}.$
(A.10)

Because $\alpha>2$, this gives the estimate on $a_{i}(t)$.

For the second estimate, by standard parabolic regularity theory, it suffices
to prove

$\|\varphi\|_{L^{\infty}\left(0,T;\mathcal{X}_{\alpha}\right)}\leq
C(\alpha)\|E\|_{L^{\infty}\left(0,T;\mathcal{X}_{2+\alpha}\right)}.$ (A.11)

We will argue by contradiction. Assume there exists a sequence of $T_{k}>1$, a
sequence of $\varphi_{k}\in
L^{\infty}\left(0,T_{k};\mathcal{X}_{\alpha}\right)$, $a_{k,i}\in
L^{\infty}\left([0,T_{k}]\right)$ and $E_{k}\in
L^{\infty}\left(0,2T_{k};\mathcal{X}_{2+\alpha}\right)$, satisfying (A.8) and
(A.9), with

$\|\varphi_{k}\|_{L^{\infty}\left(0,T_{k};\mathcal{X}_{\alpha}\right)}=1$
(A.12)

but

$\|E_{k}\|_{L^{\infty}\left(0,2T_{k};\mathcal{X}_{2+\alpha}\right)}\leq\frac{1}{k}.$
(A.13)

Take a $t_{k}\in(0,T_{k})$ with $\|\varphi_{k}(t_{k})\|_{\alpha}\geq 1/2$ and
a point $x_{k}\in\mathbb{R}^{n}$ such that

$\left(1+|x_{k}|\right)^{\alpha}|\varphi_{k}(x_{k},t_{k})|\geq 1/4.$ (A.14)

We claim that

######  Sub-Lemma.

$\limsup_{k\to\infty}|x_{k}|<+\infty$.

###### Proof.

By the heat kernel representation,

$\displaystyle\varphi(x,t_{k})$ $\displaystyle=$
$\displaystyle\int_{0}^{t_{k}}\int_{\mathbb{R}^{n}}\left[4\pi(t_{k}-t)\right]^{-\frac{n}{2}}e^{-\frac{|x-y|^{2}}{4(t_{k}-t)}}$
$\displaystyle\quad\quad\quad\quad\times\left[pW(y)^{p-1}\varphi_{k}(y,t)+\sum_{i=0}^{n+1}a_{k,i}(t)Z_{i}(y)+E_{k}(y,t)\right]dydt.$

Following the calculation in [87, Appendix B], we get

$\displaystyle\left|\int_{0}^{t_{k}}\left[4\pi(t_{k}-t)\right]^{-\frac{n}{2}}e^{-\frac{|x-y|^{2}}{4(t_{k}-t)}}pW(y)^{p-1}\varphi_{k}(y,t)dydt\right|$
$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{R}^{n}}|x-y|^{2-n}\left(1+|y|\right)^{-4-\alpha}$
$\displaystyle\lesssim$ $\displaystyle\left(1+|x|\right)^{-2-\alpha}$

and

$\displaystyle\left|\int_{0}^{t_{k}}\left[4\pi(t_{k}-t)\right]^{-\frac{n}{2}}e^{-\frac{|x-y|^{2}}{4(t_{k}-t)}}\left[\sum_{i=0}^{n+1}a_{k,i}(t)Z_{i}(y)+E_{k}(y,t)\right]dydt\right|$
$\displaystyle\lesssim$
$\displaystyle\frac{1}{k}\int_{\mathbb{R}^{n}}|x-y|^{2-n}\left(1+|y|\right)^{-2-\alpha}$
$\displaystyle\lesssim$
$\displaystyle\frac{1}{k}\left(1+|x|\right)^{-\alpha}.$

Putting these two estimates together we get

$|\varphi_{k}(x_{k},t_{k})|\leq
C\left(1+|x_{k}|\right)^{-2-\alpha}+\frac{1}{k}\left(1+|x_{k}|\right)^{-\alpha}.$

Combining this inequality with (A.14) we finish the proof of this sub-lemma. ∎

Next we divide the proof into two cases.

Case 1. $t_{k}\to+\infty$.

Let

$\widetilde{\varphi}_{k}(x,t):=\varphi_{k}(x,t_{k}+t).$

An application of standard parabolic regularity theory shows that
$\widetilde{\varphi}_{k}$ are uniformly bounded in
$C^{1+\theta,(1+\theta)/2}_{loc}(\mathbb{R}^{n}\times(-\infty,0])$. After
passing to a subsequence, $\widetilde{\varphi}_{k}$ converges to a limit
$\widetilde{\varphi}_{\infty}$, which satisfies the following conditions.

  * •

Passing to the limit in (A.12) gives

$\|\widetilde{\varphi}_{\infty}\|_{L^{\infty}\left(-\infty,0;\mathcal{X}_{\alpha}\right)}\leq
1.$ (A.15)

  * •

Since $\alpha>2$, we can pass to the limit in (A.9) for $\varphi_{k}$,
obtaining

$\int_{\mathbb{R}^{n}}\widetilde{\varphi}_{\infty}(x,t)Z_{i}(x)dx=0,\quad\quad\mbox{for
any}~{}i=0,\cdots,n+1,\quad t\leq 0.$ (A.16)

  * •

Passing to the limit in (A.8) for $\widetilde{\varphi}_{k}$ and noting (A.13)
as well as the estimate on $a_{k,i}$, we see $\widetilde{\varphi}_{\infty}$ is
a solution of (A.1).

By Proposition A.3, these three conditions imply that
$\widetilde{\varphi}_{\infty}\equiv 0$.

On the other hand, passing to the limit in (A.14) leads to

$\left(1+|x_{\infty}|\right)^{\alpha}|\widetilde{\varphi}_{\infty}(x_{\infty},0)|\geq
1/4.$

This is a contradiction with the fact that $\widetilde{\varphi}_{\infty}\equiv
0$.

Case 2. $t_{k}\to t_{\infty}$.

As in the previous case, now $\varphi_{k}$ itself converges to a limit
$\varphi_{\infty}$, which solves the equation

$\left\\{\begin{aligned}
&\partial_{t}\varphi_{\infty}-\Delta\varphi_{\infty}=pW^{p-1}\varphi_{\infty}\quad\mbox{in}~{}~{}\mathbb{R}^{n}\times(0,t_{\infty}),\\\
&\varphi_{\infty}(x,0)=0.\end{aligned}\right.$

Passing to the limit in (A.12) still gives

$\|\varphi_{\infty}\|_{L^{\infty}\left(0,t_{\infty};\mathcal{X}_{\alpha}\right)}\leq
1.$ (A.17)

By standard parabolic theory, we also get $\varphi_{\infty}\equiv 0$. Then we
get the same contradiction as in Case 1. ∎

###  Appendix B Estimates on some integrals

In this appendix we give some technical integral estimates involving the heat
kernel associated to the outer equation in Section 14.

######  Lemma B.1.

Assume $0<\nu<n$, $0\leq\gamma<n-\nu$, $t>s$. Then

$\int_{\mathbb{R}^{n}}\left(t-s\right)^{-\frac{n}{2}}e^{-c\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}|y|^{-\nu}dy\lesssim\left(|x|+\sqrt{t-s}\right)^{-\nu}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}.$

###### Proof.

After a change of variables $\hat{x}:=x/\sqrt{t-s}$, $\hat{y}:=y/\sqrt{t-s}$,
this integral is transformed into

$(t-s)^{-\nu/2}\int_{\mathbb{R}^{n}}e^{-c|\hat{x}-\hat{y}|^{2}}\left(1+\frac{1}{|\hat{y}|}\right)^{\gamma}|\hat{y}|^{-\nu}d\hat{y}.$

To estimate the integral in this formula, we consider two cases separately.

Case 1. If $|\hat{x}|\leq 1$, this integral is bounded by a universal
constant.

Case 2. If $|\hat{x}|\geq 1$, we divide this integral into two parts,
$\mathrm{I}$ outside $B_{|\hat{x}|/3}(0)$, and $\mathrm{II}$ in
$B_{|\hat{x}|/3}(0)$.

Outside $B_{|\hat{x}|/3}(0)$,

$\left(1+\frac{1}{|\hat{y}|}\right)^{\gamma}|\hat{y}|^{-\nu}\lesssim\left(1+\frac{1}{|\hat{x}|}\right)^{\gamma}|\hat{x}|^{-\nu}.$

Hence

$\displaystyle\mathrm{I}$ $\displaystyle\lesssim$
$\displaystyle\left(1+\frac{1}{|\hat{x}|}\right)^{\gamma}|\hat{x}|^{-\nu}\int_{B_{|\hat{x}|/3}(0)^{c}}e^{-c|\hat{x}-\hat{y}|^{2}}d\hat{y}$
$\displaystyle\lesssim$
$\displaystyle(t-s)^{\nu/2}|x|^{-\nu}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}.$

For $\mathrm{II}$, by noting that for $\hat{y}\in B_{|\hat{x}|/3}(0)$,

$e^{-c|\hat{x}-\hat{y}|^{2}}\lesssim e^{-c|\hat{x}|^{2}/9}.$

we obtain

$\displaystyle\mathrm{II}$ $\displaystyle\lesssim$ $\displaystyle
e^{-c|\hat{x}|^{2}/9}\int_{B_{|\hat{x}|/3}(0)}|\hat{y}|^{-\nu-\gamma}d\hat{y}$
$\displaystyle\lesssim$ $\displaystyle e^{-c\frac{|x|^{2}}{t-s}}$
$\displaystyle\lesssim$ $\displaystyle(t-s)^{\nu/2}|x|^{-\nu}.$

Adding $\mathrm{I}$ and $\mathrm{II}$ together, we get

$\int_{\mathbb{R}^{n}}e^{-c|\hat{x}-\hat{y}|^{2}}\left(1+\frac{1}{|\hat{y}|}\right)^{\gamma}|\hat{y}|^{-\nu}d\hat{y}\lesssim(t-s)^{\nu/2}|x|^{-\nu}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}.\qed$

######  Lemma B.2.

Assume $t>s$, $c_{1}$, $c_{2}$, $L$ and $\lambda$ are four positive constants.
Then

$\displaystyle\int_{B_{L\lambda}^{c}}\left(t-s\right)^{-\frac{n}{2}}e^{-c_{1}\frac{|x-y|^{2}}{t-s}}\left(1+\frac{\sqrt{t-s}}{|y|}\right)^{\gamma}e^{-c_{2}\frac{|y|}{\lambda}}dy$
$\displaystyle\lesssim$ $\displaystyle
e^{-\frac{c_{2}}{3}\frac{|x|}{\lambda}}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}+e^{-\frac{c_{2}L}{2}-\frac{c_{1}}{9}\frac{|x|^{2}}{t-s}}\left(\frac{\lambda}{\sqrt{t-s}}\right)^{n-\gamma}\left(1+\frac{\lambda}{\sqrt{t-s}}\right)^{\gamma}.$

###### Proof.

After a change of variables $\hat{x}:=x/\sqrt{t-s}$, $\hat{y}:=y/\sqrt{t-s}$,
this integral is transformed into

$\int_{B_{L\lambda/\sqrt{t-s}}^{c}}e^{-c_{1}|\hat{x}-\hat{y}|^{2}}\left(1+\frac{1}{|\hat{y}|}\right)^{\gamma}e^{-c_{2}\frac{\sqrt{t-s}}{\lambda}|\hat{y}|}d\hat{y}.$

We divide this integral into two parts, $\mathrm{I}$ outside
$B_{|\hat{x}|/3}(0)$, and $\mathrm{II}$ in $B_{|\hat{x}|/3}(0)$.

Outside $B_{|\hat{x}|/3}(0)$,

$\left\\{\begin{aligned} &e^{-c_{2}\frac{\sqrt{t-s}}{\lambda}|\hat{y}|}\leq
e^{-\frac{c_{2}}{3}\frac{\sqrt{t-s}}{\lambda}|\hat{x}|},\\\
&\left(1+\frac{1}{|\hat{y}|}\right)^{\gamma}\lesssim\left(1+\frac{1}{|\hat{x}|}\right)^{\gamma}.\end{aligned}\right.$

Therefore

$\displaystyle\mathrm{I}$ $\displaystyle\lesssim$ $\displaystyle
e^{-\frac{c_{2}}{3}\frac{\sqrt{t-s}}{\lambda}|\hat{x}|}\left(1+\frac{1}{|\hat{x}|}\right)^{\gamma}\int_{B_{|\hat{x}|/3}(0)^{c}}e^{-c_{1}|\hat{x}-\hat{y}|^{2}}d\hat{y}$
$\displaystyle\lesssim$ $\displaystyle
e^{-\frac{c_{2}}{3}\frac{|x|}{\lambda}}\left(1+\frac{\sqrt{t-s}}{|x|}\right)^{\gamma}.$

In $B_{|\hat{x}|/3}(0)$,

$e^{-c_{1}|\hat{x}-\hat{y}|^{2}}\leq e^{-c_{1}|\hat{x}|^{2}/9}.$

Hence

$\displaystyle\mathrm{II}$ $\displaystyle\lesssim$ $\displaystyle
e^{-c_{1}|\hat{x}|^{2}/9}\int_{B_{|\hat{x}|/3}\setminus
B_{L\lambda/\sqrt{t-s}}}\left(1+\frac{1}{|\hat{y}|}\right)^{\gamma}e^{-c_{2}\frac{\sqrt{t-s}}{\lambda}|\hat{y}|}d\hat{y}$
$\displaystyle\lesssim$ $\displaystyle
e^{-\frac{c_{2}L}{2}-\frac{c_{1}}{9}\frac{|x|^{2}}{t-s}}\left(\frac{\lambda}{\sqrt{t-s}}\right)^{n}\left(1+\frac{\sqrt{t-s}}{\lambda}\right)^{\gamma}$
$\displaystyle\lesssim$ $\displaystyle
e^{-\frac{c_{2}L}{2}-\frac{c_{1}}{9}\frac{|x|^{2}}{t-s}}\left(\frac{\lambda}{\sqrt{t-s}}\right)^{n}\left(1+\frac{\lambda}{\sqrt{t-s}}\right)^{\gamma}.\qed$

##  Part III Energy concentration in the general case

###  3\. Setting

In this part, we still consider a sequence of _smooth, positive solutions_
$u_{i}$ to the nonlinear heat equation (1.1) (with $p=(n+2)/(n-2)$) in
$Q_{1}$, but now satisfying the following three assumptions.

(IIII.a) Weak limit:

$u_{i}$ converges weakly to $u_{\infty}$ in $L^{p+1}(Q_{1})$, and $\nabla
u_{i}$ converges weakly to $\nabla u_{\infty}$ in $L^{2}(Q_{1})$. Here
$u_{\infty}$ is a _smooth solution_ of (1.1) in $Q_{1}$.

(IIII.b) Energy concentration behavior:

there exists an $N\geq 2$ such that

$\left\\{\begin{aligned} &|\nabla u_{i}|^{2}dxdt\rightharpoonup|\nabla
u_{\infty}|^{2}dxdt+N\Lambda\delta_{0}\otimes dt,\\\
&u_{i}^{p+1}dxdt\rightharpoonup u_{\infty}^{p+1}dxdt+N\Lambda\delta_{0}\otimes
dt,\end{aligned}\right.$

weakly as Radon measures.

(IIII.c) Convergence of time derivatives:

as $i\to\infty$, $\partial_{t}u_{i}$ converges to $\partial_{t}u_{\infty}$
strongly in $L^{2}(Q_{1})$.

The only difference with the assumptions in Part II is (IIII.b), where now we
do not assume there is only one bubble, in other words, now we are in _the
higher multiplicity case_ instead of _the multiplicity one case_.

The main result in this part is

######  Theorem 3.1.

After passing to a subsequence, the followings hold for $u_{i}$.

  1. (1)

For all $i$ and any $t\in[-9/16,9/16]$, there exist exactly $N$ local maximal
point of $u_{i}(\cdot,t)$ in the interior of $B_{1}(0)$.

Denote these points by $\xi_{ij}^{\ast}(t)$ ($j=1,\cdots,N$) and let
$\lambda_{ij}^{\ast}(t):=u_{i}(\xi_{ij}^{\ast}(t),t)^{-\frac{2}{n-2}}$.

  2. (2)

Both $\lambda_{ij}^{\ast}$ and $\xi_{ij}^{\ast}$ are continuous functions on
$[-9/16,9/16]$.

  3. (3)

There exists a constant $C_{2}$ such that, for all $i$ and any $x\in B_{1}$,

$\displaystyle\min_{1\leq j\leq
N}|x-\xi_{ij}^{\ast}(t)|^{\frac{n-2}{2}}u_{i}(x,t)+\min_{1\leq j\leq
N}|x-\xi_{ij}^{\ast}(t)|^{\frac{n}{2}}|\nabla u_{i}(x,t)|$ (3.1)
$\displaystyle\quad\quad\quad\quad+\min_{1\leq j\leq
N}|x-\xi_{ij}^{\ast}(t)|^{\frac{n+2}{2}}\left(|\nabla^{2}u_{i}(x,t)|+|\partial_{t}u_{i}(x,t)|\right)\leq
C_{2}.$

  4. (4)

For each $j=1,\cdots,N$, as $i\to\infty$,

$\lambda_{ij}^{\ast}(t)\to 0,\quad\xi_{ij}^{\ast}(t)\to 0,\quad\mbox{uniformly
on }[-9/16,9/16],$

and the function

$u_{ij}^{t}(y,s):=\lambda_{ij}^{\ast}(t)^{\frac{n-2}{2}}u\left(\xi_{ij}^{\ast}(t)+\lambda_{ij}^{\ast}(t)y,t+\lambda_{ij}^{\ast}(t)^{2}s\right),$

converges to $W(y)$ in $C^{\infty}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$.

  5. (5)

For any $1\leq j\neq k\leq N$ and $t\in[-9/16,9/16]$,

$\lim_{i\to+\infty}\frac{|\xi_{ij}^{\ast}(t)-\xi_{ik}^{\ast}(t)|}{\max\left\\{\lambda_{ij}^{\ast}(t),\lambda_{ik}^{\ast}(t)\right\\}}=+\infty.$
(3.2)

  6. (6)

$u_{\infty}\equiv 0$.

######  Remark 3.2.

Item (5) says there is no bubble towering. Item (6) will be used (in an
inductive way) to prove this property.

The proof of this theorem uses an induction argument on $N$. This part is
organized in the following way.

  1. (1)

In Section 4, we establish several preliminary results.

  2. (2)

In Section 5, we find all bubbles under the inductive assumption that Theorem
3.1 holds up to $N-1$.

  3. (3)

In Section 6, we prove the Lipschitz hypothesis (10.1) in Part II, under the
assumptions (II.a)-(II.c). This also provides the first step of our inductive
argument.

  4. (4)

In Section 20, we finish the proof of Theorem 3.1 by employing Proposition
19.1 in Part II.

###  4\. Preliminaries

Before starting the bubble tree construction, we recall several technical
results.

The first one is a uniform Morrey space estimate on $u_{i}$, which is just a
special case of Corollary 3.5 in Part I

######  Lemma 4.1.

There exists a constant $M>0$ such that for each $i$ and $Q_{r}(x,t)\subset
Q_{3/4}$,

$\int_{Q_{r}(x,t)}\left(|\nabla u_{i}|^{2}+u_{i}^{p+1}\right)\leq Mr^{2}.$

The second one is the non-concentration estimate of $\partial_{t}u_{i}$ in
Lemma 11.3 from Part II, which still holds in the current setting (thanks to
(IIII.c)).

The following lemma is about the convergence of rescalings of $u_{i}$.

######  Lemma 4.2.

Given a sequence $(x_{i},t_{i})\in Q_{1/2}$, $r_{i}\to 0$, define

$u_{i}^{r_{i}}(x,t):=r_{i}^{\frac{n-2}{2}}u_{i}\left(x_{i}+r_{i}x,t_{i}+r_{i}^{2}t\right).$
(4.1)

Then we have

  1. (1)

After passing to a subsequence, $u_{i}^{r_{i}}$ converges weakly to $0$ or
$W_{\xi,\lambda}$ for some
$(\xi,\lambda)\in\mathbb{R}^{n}\times\mathbb{R}^{+}$.

  2. (2)

If the defect measure associated to $u_{i}^{r_{i}}$ is nontrivial, it must be
of the form

$\sum_{j}k_{j}\Lambda\delta_{P_{j}}\otimes dt,$

where $P_{j}$ are distinct points in $\mathbb{R}^{n}$ and

$\sum_{j}k_{j}\leq M/\Lambda,\quad k_{j}\in\mathbb{N}.$

  3. (3)

If there is no defect measure, then $u_{i}^{r_{i}}$ converges in
$C^{\infty}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$.

###### Proof.

As in Part I, $u_{i}^{r_{i}}$ converges weakly to a nonnegative solution of
(1.1) in $L^{p+1}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$. By Lemma 11.3 (non-
concentration of time derivatives), $\partial_{t}u_{i}^{r_{i}}$ converges to
$0$ in $L^{2}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$. Hence this weak limit is
independent of time. By Caffarelli-Gidas-Spruck [8], it must be $0$ or
$W_{\xi,\lambda}$ for some
$(\xi,\lambda)\in\mathbb{R}^{n}\times\mathbb{R}^{+}$. Furthermore, the defect
measure is also independent of time, that is, there exists a Radon measure
$\mu_{0}$ on $\mathbb{R}^{n}$ such that the defect measure equals
$\mu_{0}\otimes dt$. By Lemma 4.1, for any $R>0$,

$\int_{-R^{2}}^{R^{2}}\int_{B_{R}}d\mu_{0}dt\leq MR^{2}.$

Hence

$\mu_{0}(\mathbb{R}^{n})\leq M.$ (4.2)

By Lemma 9.2, there exist finitely many distinct points
$P_{j}\in\mathbb{R}^{n}$ and constants $k_{j}\in\mathbb{N}$ such that

$\mu_{0}=\sum_{j}k_{j}\Lambda\delta_{P_{j}}.$

Finally, if there is no defect measure, the smooth convergence of
$u_{i}^{r_{i}}$ follows from a direct application of the
$\varepsilon$-regularity theorem, Theorem 3.9, by noting the smoothness of its
weak limit. ∎

We also need a technical result about the blow up of each time slice of
$u_{i}$. This will be used to find the first bubble in the next section.

######  Lemma 4.3.

For each $t\in[-9/16,9/16]$,

$\lim_{i\to+\infty}\max_{B_{1}}u_{i}(x,t)=+\infty.$

###### Proof.

For each $u_{i}$ and $(x,t)\in Q_{3/4}$, define $\rho_{i}(x,t)$ to be the
unique (if it exists) solution to the equation

$\Theta_{r^{2}}(x,t;u_{i})=\varepsilon_{\ast}.$

Here $\varepsilon_{\ast}$ is the small constant in the
$\varepsilon$-regularity theorem, Theorem 3.7.

By (IIII.b), if $x\neq 0$, there is a uniform, positive lower bound for
$\rho_{i}(x,t)$. By (IIII.a) and (IIII.b), for any $s>0$ fixed,

$\lim_{i\to+\infty}\Theta_{s}(0,t;u_{i})=\left(4\pi\right)^{-\frac{n}{2}}\frac{\Lambda}{n}N+o_{s}(1).$

Therefore

$\lim_{i\to+\infty}\rho_{i}(0,t)=0.$

As a consequence, $\rho_{i}(\cdot,t)$ has an interior minimal point in
$B_{1}$, say $x_{i}(t)$. Denote $r_{i}:=\rho_{i}(x_{i}(t),t)$. Define
$u_{i}^{r_{i}}$ as in (4.1), with base point at $(x_{i}(t),t)$.

Claim. For some $(\xi,\lambda)\in\mathbb{R}^{n}\times\mathbb{R}^{+}$,
$u_{i}^{r_{i}}$ converges to $W_{\xi,\lambda}$ in
$C^{\infty}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$.

By this claim we get

$\lim_{i\to+\infty}\rho_{i}(x_{i}(t),t)^{\frac{n-2}{2}}u_{i}(x_{i}(t),t)=W_{\xi,\lambda}(0)>0,$

and the proof of this lemma is complete.

Proof of the Claim. By the definition of $r_{i}$ and the scaling invariance of
$\Theta_{s}(\cdot)$, for any $y\in B_{r_{i}^{-1}}$,

$\Theta_{1}(y,0;u_{i}^{r_{i}})\leq\varepsilon_{\ast}.$

Moreover, the equality is attained at $y=0$, that is,

$\displaystyle\varepsilon_{\ast}$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{n}}\left[\frac{|\nabla
u_{i}^{r_{i}}(y,-1)|^{2}}{2}-\frac{u_{i}^{r_{i}}(y,-1)^{p+1}}{p+1}\right]\psi\left(\frac{y-x_{i}(t)}{r_{i}}\right)^{2}G(y,1)dy$
(4.3) $\displaystyle+$
$\displaystyle\frac{1}{2(p-1)}\int_{\mathbb{R}^{n}}u_{i}^{r_{i}}(y,-1)^{2}\psi\left(\frac{x-x_{i}(t)}{r_{i}}\right)^{2}G(y,1)dy+Ce^{-cr_{i}^{-2}}.$

By Theorem 3.9, $u_{i}^{r_{i}}$ are uniformly bounded in
$C^{2}(B_{r_{i}^{-1}-1}\times(-\delta_{\ast},\delta_{\ast}))$. This then
implies that there is no defect measure appearing when we apply Lemma 4.2 to
$u_{i}^{r_{i}}$. In fact, if the defect measure is nontrivial, by Lemma 4.2,
it has the form

$\sum_{j}k_{j}\Lambda\delta_{P_{j}}\otimes dt,\quad
P_{j}\in\mathbb{R}^{n},~{}~{}k_{j}\in\mathbb{N}.$

Then by Lemma 9.3, for a.e. $t\in(-\delta_{\ast},\delta_{\ast})$,
$u_{i}^{r_{i}}(\cdot,t)$ should develop nontrivial Dirac measures at $P_{j}$.
This is a contradiction with the uniform
$C^{2}(B_{r_{i}^{-1}-1}\times(-\delta_{\ast},\delta_{\ast}))$ regularity of
$u_{i}^{r_{i}}$.

Since there is no defect measure, Lemma 4.2 implies that $u_{i}^{r_{i}}$
converges to $u_{\infty}$ in
$C^{\infty}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$. In view of the Liouville
theorem in Caffarelli-Gidas-Spruck [8], it suffices to show that
$u_{\infty}\neq 0$. (By the strong maximum principle, either $u_{\infty}\equiv
0$ or $u_{\infty}>0$ everywhere.)

First by the monotonicity formula (Proposition 3.2), we have

$\displaystyle\varepsilon_{\ast}$ $\displaystyle\leq$
$\displaystyle\int_{-2}^{-1}(-s)^{\frac{p+1}{p-1}}\int_{\mathbb{R}^{n}}\left[\frac{|\nabla
u_{i}^{r_{i}}(y,s)|^{2}}{2}-\frac{u_{i}^{r_{i}}(y,s)^{p+1}}{p+1}\right]\psi\left(\frac{y-x_{i}(t)}{r_{i}}\right)^{2}G(y,-s)$
$\displaystyle+$
$\displaystyle\frac{1}{2(p-1)}\int_{-2}^{-1}(-s)^{\frac{2}{p-1}}\int_{\mathbb{R}^{n}}u_{i}^{r_{i}}(y,s)^{2}\psi\left(\frac{x-x_{i}(t)}{r_{i}}\right)^{2}G(y,-s)+Ce^{-cr_{i}^{-2}/4}.$

By the $C^{\infty}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$ convergence of
$u_{i}^{r_{i}}$, the Morrey space bound in Lemma 4.1 for $u_{i}^{r_{i}}$ and
the exponential decay of $G(y,-s)$ as $|y|\to+\infty$, we can let
$i\to+\infty$ in the above inequality to get

$\displaystyle\varepsilon_{\ast}$ $\displaystyle\leq$
$\displaystyle\int_{-2}^{-1}(-s)^{\frac{p+1}{p-1}}\int_{\mathbb{R}^{n}}\left[\frac{|\nabla
u_{\infty}(y,s)|^{2}}{2}-\frac{u_{\infty}(y,s)^{p+1}}{p+1}\right]G(y,-s)$
$\displaystyle+$
$\displaystyle\frac{1}{2(p-1)}\int_{-2}^{-1}(-s)^{\frac{2}{p-1}}\int_{\mathbb{R}^{n}}u_{\infty}(y,s)^{2}G(y,-s).$

Hence it is impossible that $u_{\infty}\equiv 0$. ∎

###  5\. Bubble tree construction

In this section, we construct bubbles by finding local maximal points of
$u_{i}(\cdot,t)$. The construction is divided into six steps. During the
course of this construction, we will also prove Theorem 3.1 except the last
point (6), under the inductive assumption that this theorem holds when the
multiplicity is not larger than $N-1$ ($N\geq 2$).

Step 1. Construction of the first maximal point. By (IIII.a) and Lemma 4.3,
for each $t$, $\max_{B_{1}}u_{i}(x,t)$ is attained at an interior point, say
$\xi_{i1}^{\ast}(t)$. Denote

$\lambda_{i1}^{\ast}(t):=u_{i}\left(\xi_{i1}^{\ast}(t),t\right)^{-\frac{2}{n-2}}.$

By Lemma 4.3,

$\lim_{i\to+\infty}\lambda_{i1}^{\ast}(t)=0.$

Let

$u_{i1}(y,s):=\lambda_{i1}^{\ast}(t)^{\frac{n-2}{2}}u_{i}\left(\xi_{i1}^{\ast}(t)+\lambda_{i1}^{\ast}(t)y,t+\lambda_{i1}^{\ast}(t)^{2}s\right).$

It satisfies the following conditions.

  * •

By Lemma 4.1, for any $R>0$,

$\int_{Q_{R}}\left[|\nabla u_{i1}|^{2}+u_{i1}^{p+1}\right]\leq MR^{2}.$ (5.1)

  * •

By Lemma 11.3, for any $R>0$,

$\lim_{i\to+\infty}\int_{Q_{R}}|\partial_{t}u_{i1}|^{2}=0.$ (5.2)

  * •

At $s=0$,

$\max_{|y|\leq\lambda_{i1}^{\ast}(0)^{-1}/2}u_{i1}(y,0)=u_{i1}(0,0)=1.$ (5.3)

Using these conditions we show

######  Lemma 5.1.

As $i\to+\infty$, $u_{i1}$ are uniformly bounded in
$C^{\infty}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$, and it converges to $W$ in
$C^{\infty}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$.

###### Proof.

We use Lemma 4.2 to study the convergence of $u_{i1}$. We need only to show
that there is no defect measure appearing. Once this is established, $u_{i1}$
would converge in $C^{\infty}_{loc}(\mathbb{R}^{n}\times\mathbb{R})$. Then by
(5.3), this limit must be $W$.

Assume by the contrary that the defect measure is nonzero. By Lemma 4.2, it
has the form

$\sum_{j}k_{j}\Lambda\delta_{P_{j}}\otimes dt,\quad
P_{j}\in\mathbb{R}^{n},~{}~{}k_{j}\in\mathbb{N}.$

Then applying Lemma 4.3 to $u_{i1}$ at $t=0$, we obtain

$\lim_{i\to+\infty}\max_{B_{1}(P_{j})}u_{i1}(x,0)=+\infty.$

This is a contradiction with (5.3). ∎

By this lemma, there exists a sequence $R_{i1}\to+\infty$ such that

$\lim_{i\to+\infty}\sup_{y\in
B_{R_{i1}}}|y|^{\frac{n-2}{2}}u_{i1}(y,0)<+\infty.$ (5.4)

Step 2. Iterative construction. Suppose
$\xi_{i1}^{\ast}(t)$,$\cdots$,$\xi_{i,j-1}^{\ast}(t)$ have been constructed.
If

$\max_{x\in B_{1}}\min_{1\leq k\leq
j-1}|x-\xi_{ik}^{\ast}(t)|^{\frac{n-2}{2}}u_{i}(x,t)$

are uniformly bounded in $B_{1}$, we stop at this step. Otherwise,

$\lim_{i\to\infty}\max_{x\in B_{1}}\min_{1\leq k\leq
j-1}|x-\xi_{ik}^{\ast}(t)|^{\frac{n-2}{2}}u_{i}(x,t)=+\infty.$ (5.5)

By (IIII.a), this function has an interior maximal point, say
$\xi_{ij}^{\ast}(t)$. Define $\lambda_{ij}^{\ast}(t)$, $u_{ij}$ as in Step 1.
The estimates (5.1) and (5.2) still hold for $u_{ij}$, while (5.3) now reads
as

$\left\\{\begin{aligned} &u_{ij}(0,0)=1,\\\ &u_{ij}(y,0)\leq\frac{\min_{1\leq
k\leq j-1}|\xi_{ij}^{\ast}(t)-\xi_{ik}^{\ast}(t)|^{\frac{n-2}{2}}}{\min_{1\leq
k\leq
j-1}|\xi_{ij}^{\ast}(t)-\xi_{ik}^{\ast}(t)+\lambda_{ij}^{\ast}(t)y|^{\frac{n-2}{2}}},\end{aligned}\right.$
(5.6)

the right hand side of which converges to $1$ uniformly in any compact set of
$\mathbb{R}^{n}$, by noting that we have

$\min_{1\leq k\leq
j-1}|\xi_{ij}^{\ast}(t)-\xi_{ik}^{\ast}(t)|^{\frac{n-2}{2}}u_{i}\left(\xi_{ij}^{\ast}(t),t\right)=+\infty.$

(This follows by letting $x=\xi_{ij}^{\ast}(t)$ in (5.5).)

We can argue as in Step 1 to deduce that $u_{ij}$ converges to $W$ in