From the equation (<ref>) and  the decomposition (<ref>), $\tm_v$ satisfies the equation

\begin{equation}\label{eq:mvtil}
\pa_s \tilde{m}_v =  \As_0 \tilde{m}_v + \tilde \eta \; \tr \pa_{\tr} \tm_v + \frac{\pa_{\tr} \big[(\tilde P_1 - \tilde{N}_1) \tilde{m}_v\big]}{\tr} + \frac{\pa_{\tr}\big(\tilde{m}_v^2\big)}{2\tr} + \Ec_1(\tr, s),
\end{equation}

where $\As_0$ is defined as in (<ref>) but here acting on the variable $\tr$ instead of $r$ (the reader should bear in mind that $\tm_v$ is a function of the $\tr$ variable), and we write for simplicity

$$\tilde\eta = \frac{\tnu_s}{\tnu} - \beta \tnu^2 = \Oc(\nu^2),$$

and the error is given by

\begin{align}
\Ec_1 &= \big(-\pa_s + \As_0\big) \big(\tilde P_1 - \tilde{N}_1\big) +  \tilde\eta \; \tr \pa_{\tr} \Big(Q + \tilde P_1 - \tilde{N}_1\Big) + \frac{\pa_{\tr} (\tilde{P}_1 - \tilde{N}_1)^2}{2\tr} + \tilde N_0(v^\perp).\label{def:errorEc1}
\end{align}

For a fixed small constant $\zeta_*$ with $0 < \zeta_* \ll 1$, we introduce

\begin{equation}
\tilde{m}_v^*(\tr, s) = \chi_{\frac{\zeta_*}{\tnu}}(\tr) \tilde{m}_v(\tr, s),
\end{equation}

where $\chi_{\frac{\zeta_*}{\tnu}}$ is defined as in (<ref>). We write from (<ref>) the equation for  $\tilde{m}_v^*$,

\begin{equation} \label{eq:mwtilstar}
\pa_s \tilde{m}_v^* =  \As_0 \tilde{m}_v^* +  \tilde\eta \; \tr \pa_{\tr} \tm_v^*   + \frac{\pa_{\tr} \big((\tilde P_1 - \tilde{N}_1) \tilde{m}^*_v\big)}{\tr} + \frac{\pa_{\tr}\big(\tilde{m}^*_v \tilde{m}_v\big)}{2\tr} + \chi_{\frac{\zeta_*}{\tnu}}\Ec_1 + T(\tm_v),\nonumber
\end{equation}

\begin{align} \label{def:Tmv}
T(\tm_v) &=  - \Big(\pa_{\tr}^2  + \pa_s \big)\chi_{\frac{\zeta_*}{\tnu}} \tilde{m}_v  + \pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}} \left[ -2\pa_{\tr} + \frac{1}{\tr} -  \frac{\big[2Q + 2\tilde P_1 - 2\tilde{N}_1 + \tm_v\big]}{2\tr}  + \tilde \eta \; \tr   \right] \tilde{m}_v.
\end{align}

Thanks to the orthogonality condition (<ref>), we can use the coercivity estimate (see Lemma <ref>)

\begin{equation}\label{est:coer1}
-\int \tm_v^* \As_0 \tm_v^* \frac{\omega_0}{\tr}d\tr \geq \delta_1 \int \left(|\pa_{\tr}\tm_{v}^*|^2  + \frac{|\tm_{v}^*|^2}{1 + \tr^2}\right) \frac{\omega_0}{\tr} d\tr ,
\end{equation}

and (see Lemma <ref>)

\begin{equation}\label{est:coer2}
\int |\As_0 \tm_{v}^*|^2 \frac{\omega_0}{\tr}d\tr \geq \delta_0 \int \left(|\pa_{\tr}^2\tm_{v}^*|^2 +  \frac{|\pa_{\tr}\tm_{v}^*|^2}{ \langle \tr \rangle^2} + \frac{|\tm_v^*|^2}{ \langle \tr \rangle^4(1 + \ln^2 \la \tr \ra )} \right) \frac{\omega_0}{\tr} d\tr.
\end{equation}

Since the support of $\tm_v^*$ is $\tr \leq \frac{2\zeta_*}{\tnu}$, we have by (<ref>) the control

\begin{equation}
\int |\As_0 \tm_{v}^*|^2 \frac{\omega_0}{\tr}d\tr \geq \frac{C\tnu^2}{\zeta_*^2}\left(-\int \tm_v^* \As_0 \tm_v^* \frac{\omega_0}{\tr}d\tr \right).
\end{equation}

Thanks to these coercivity estimates, we are able to establish the following monotonicity formula to control the inner norm.

We have for $\tau_0$ large enough

\begin{equation}\label{eq:2ndEne}
\frac{d}{d\tau}\| \As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}}^2 \leq C \left(\| \As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}}^2  + \frac{1}{\zeta_*^2}\| m_\e\|^2_{H^2 (\zeta_* \leq \zeta \leq \zeta^*)} + \frac{\nu^4}{\zeta_*^2|\ln \nu|^2}\right),
\end{equation}

\begin{equation}\label{eq:1stEne}
\frac{1}{2}\frac{d}{d\tau} \left[- \tnu^2\int \tm_v^* \As_0 \tm_v^* \frac{\omega_0}{\tr} \right]   \leq -\frac{1}{2} \|\As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}} +   \frac{C'}{\zeta_*^2}\| m_\e\|^2_{H^2 (\zeta_* \leq \zeta \leq \zeta^*)} + \frac{C'\nu^4}{ \zeta_*^2|\ln \nu|^2},
\end{equation}

where $C$ and $C'$ are independent from the constants $\kappa,K, K', K''$ and $\zeta_*$.

Step 1 Energy estimate for the second order derivative. We first prove (<ref>). We integrate equation (<ref>) against $\As_0^2 \tm_v^*$ in $L^2_{\frac{\omega_0}{\tr}}$,  and using the self-adjointness of $\As_0$ in $L^2_{\frac{\omega_0}{\tr}}$ and the fact that $\int f\As_0 f \frac{\omega_0}{\tr}d\tr  \leq 0$ (note that $T_0$ is in the kernel of $\As_0$ which is strictly positive on $(0, +\infty)$, so a standard Sturm-Liouville argument yields the non-negativity of $-\As_0$) to write the energy identity

\begin{align}
&\frac{1}{2}\frac{d}{ds} \int |\As_0 \tm_v^*|^2 \frac{\omega_0}{\tr}d\tr \leq \int \As_0\big[\chi_{\frac{\zeta_*}{\tnu}} \Ec_1   + \Fc(\tm_v) + T(\tm_v)\big]\As_0 \tm_v^* \frac{\omega_0}{\tr} d\tr, \label{id:Ener1}
\end{align}

$$\Fc(\tm_v) =  \tilde \eta \; \tr \pa_{\tr} \tm_v^*  + \frac{\pa_{\tr} \big((\tilde P_1 - \tilde{N}_1) \tilde{m}^*_v\big)}{\tr} + \frac{\pa_{\tr}\big(\tilde{m}^*_v \tilde{m}_v\big)}{2\tr}.$$

The error term: We claim the following estimate for the error term

\begin{equation}\label{est:Ec1}
\int \left|\As_0 \big(  \Ec_1(\tr,s) \chi_{\frac{\zeta_*}{\tnu}}(\tr)\big) \right|^2\frac{\omega_0(\tr)}{\tr}d\tr\leq C \frac{\nu^8}{|\ln \nu|^2},
\end{equation}

for some positive constant $C$ which is independent from $K$, $K'$ and $K''$. Since the proof of (<ref>) is technical and a bit lengthy, we postpone it to the end of this section and continue the proof of the lemma. We have by Cauchy-Schwarz inequality, and

\begin{align*}
\left|\int \As_0\big(\chi_{\frac{\zeta_*}{\tnu}} \Ec_1\big) \As_0 \tm_v^* \frac{\omega_0}{\tr} d\tr  \right| \lesssim \frac{\nu^4}{|\ln \nu|} \|\As_0 \tm_v^* \|_{L^2_\frac{\omega_0}{\tr}} \lesssim \nu^2\|\As_0 \tm_v^* \|_{L^2_\frac{\omega_0}{\tr}}^2 + \frac{\nu^6}{|\ln \nu|^2}.
\end{align*}

The scaling term: From the definition of $\As_0$, we compute

\begin{equation}\label{eq:commutatorA0rdr}
\As_0 (\tr \pa_{\tr} f) = \tr \pa_{\tr} \As_0 f + [\As_0, \tr \pa_{\tr}] f,
\end{equation}

\big[\As_0, r\pa_r\big] = 2\As_0  -r U\pa_r - \frac{2(r^2 - 1)}{r^2 + 1} U.

We then write by integration by parts and the coercivity estimate (<ref>),

\begin{align*}
&\left|\tilde\eta\int \As_0\big(\tr \pa_{\tr} \tm_v^*\big) \As_0 \tm_v^* \frac{\omega_0}{\tr}d\tr\right|= \left|\tilde\eta \int \Big[\tr \pa_{\tr} \As_0 \tm_v^* + 2\As_0 \tm_v^* - rU \pa_{\tr} \tm_v^* - \frac{2(\tr^2 - 1)}{ \langle \tr \rangle ^2 } U \tm_v^* \Big]\As_0 \tm_v^* \frac{\omega_0}{\tr}d\tr \right|\\
& \lesssim \nu^2 \left[ \int |\As_0 \tm_v^*|^2 \pa_{\tr}(\omega_0) d\tr +  \|\As_0 \tm_v^* \|_{L^2_\frac{\omega_0}{\tr}} + \int \left(\frac{|\pa_{\tr} \tm_v^*|^2}{\langle \tr \rangle^2} + \frac{|\tm_v^*|^2}{\langle \tr \rangle^8}\right) \frac{\omega_0}{\tr}d\tr \right] \lesssim \nu^2\|\As_0 \tm_v^* \|_{L^2_\frac{\omega_0}{\tr}}.
\end{align*}

The small linear and nonlinear term: By definition, we have

\begin{equation}\label{def:commAsfg}
\As_0 (fg) = g\As_0 f + [\As_0, g]f \quad \textup{with} \quad [\As_0, g] =  2\pa_{\tr}g \pa_{\tr} - \Big[1 - Q \Big]\frac{\pa_{\tr}g}{r},
\end{equation}

\begin{equation}\label{def:commAsdr}
\As_0(\pa_{\tr} f) = \pa_{\tr} \As_0 f + [\As_0, \pa_{\tr}] f,
\end{equation}

$$ [\As_0, \pa_{\tr}] = \left(\frac{3\tr^4 - 6\tr^2 - 1}{8\tr^2}\right)U \pa_{\tr} + \frac{4\tr}{(1  +\tr^2)}U.$$

This gives the formula

\begin{align*}
\As_0 \Big(\frac{\pa_{\tr} \big(\tm_v^* F \big)}{\tr}\Big) &= \As_0 \left( \frac{F}{\tr} \pa_{\tr} \tm_v^* + \frac{\pa_{\tr} F}{\tr} \tm_v^*  \right) = \As_0 (\pa_{\tr} \tm_v^*) \frac{F}{\tr} + \big[\As_0, \frac{F}{\tr} \big] \pa_{\tr} \tm_v^* + \As_0 \tm_v^* \frac{\pa_{\tr} F}{\tr} + \big[ \As_0, \frac{\pa_{\tr} F}{\tr} \big] \tm_v^*\nonumber \\
& = \Big[ \pa_{\tr} \As_0 \tm_v^* + \big[\As_0, \pa_{\tr}\big] \tm_v^* \Big] \frac{F}{\tr}+ \big[\As_0, \frac{F}{\tr} \big] \pa_{\tr} \tm_v^* + \As_0 \tm_v^* \frac{\pa_{\tr} F}{\tr} + \big[ \As_0, \frac{\pa_{\tr} F}{\tr} \big] \tm_v^*. \label{eq:comFmv}
\end{align*}

We then write by integration by parts

\begin{align*}
\int \As_0 \Big(\frac{\pa_{\tr} \big(\tm_v^* F \big)}{\tr}\Big) \As_0 \tm_v^* \frac{\omega_0}{\tr}d\tr &= \int |\As_0 \tm_v^*|^2 \left[-\frac{1}{2} \frac{\tr}{\omega_0} \pa_{\tr}\left(\frac{F \omega_0}{\tr^2} \right) + \frac{\pa_{\tr} F}{\tr}\right] \frac{\omega_0}{\tr}d\tr \\
& \quad + \int \left[\big[\As_0, \pa_{\tr}\big] \tm_v^* \frac{F}{\tr}+ \big[\As_0, \frac{F}{\tr} \big] \pa_{\tr} \tm_v^* + \big[ \As_0, \frac{\pa_{\tr} F}{\tr} \big] \tm_v^*\right] \As_0 \tm_v^* \frac{\omega_0}{\tr}d\tr.
\end{align*}

By the definition of the commutators, we observe

$$\left|-\frac{1}{2} \frac{\tr}{\omega_0} \pa_{\tr}\left(\frac{F \omega_0}{\tr^2} \right) + \frac{\pa_{\tr} F}{\tr}\right| \lesssim \frac{|\pa_{\tr} F|}{\tr} + \frac{|F|}{\tr^2},$$

\begin{align*}
&\left|\big[\As_0, \pa_{\tr}\big] \tm_v^* \frac{F}{\tr}+ \big[\As_0, \frac{F}{\tr} \big] \pa_{\tr} \tm_v^* + \big[ \As_0, \frac{\pa_{\tr} F}{\tr} \big] \tm_v^*\right|\\
&  \quad \lesssim \sum_{i = 0}^1 \frac{|\pa_{\tr}^i F|}{\tr^{2 - i}} |\pa_{\tr}^2 \tm_v^*| +\sum_{i = 0}^2 \frac{|\pa_{\tr}^i F|}{\tr^{3 - i}}  |\pa_{\tr} \tm_v^*|  + \left(\sum_{i = 1}^2 \frac{|\pa_{\tr}^i F|}{\tr^{4 - i}}  + \frac{|F|}{\tr \la\tr\ra^5}\right) | \tm_v^*|.
\end{align*}

Now, consider $F = 2\tilde{P}_1 - 2\tilde{N}_1 + \tm_v$, we estimate from the definition of $\tilde{P}_1$ and $\tilde{N}_1$ and the pointwise bound (<ref>) for $k = 0, 1$ and $\tr \leq \frac{2\zeta_*}{\tnu}$,

\begin{align*}
\frac{|\pa_{\tr} F|}{\tr} + \frac{|F|}{\tr^2} \lesssim \nu^2, \quad \big|(\tr \pa_{\tr})^k F\big| \lesssim \nu^2 |\ln \la \tr \ra |.
\end{align*}

Hence, we have by Cauchy-Schwarz and (<ref>),

\begin{align*}
\left|\int \As_0 \Big(\frac{\pa_{\tr} \big(\tm_v^* F \big)}{\tr}\Big) \As_0 \tm_v^* \frac{\omega_0}{\tr}d\tr \right| \lesssim \nu^2\|\As_0 \tm_v^*\|^2_{L^2_\frac{\omega_0}{\tr}}.
\end{align*}

The cut-off term: From the expression (<ref>) of $T(\tm_v)$, we compute

\begin{align*}
\As_0 T(\tm_v) = -2\As_0\big( \pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}} \pa_{\tr} \tm_v \big) + \As_0 \big(F \tm_v \big)&= -2\pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}} \pa_{\tr} \As_0 \tm_v + F \As_0 \tm_v -2\pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}} \big[\As_0, \pa_{\tr}\big] \tm_v \\
& \qquad + \big[\As_0, \pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}}\big] \pa_{\tr} \tm_v + \big[\As_0, F \big] \tm_v,
\end{align*}

$$F =  - \Big(\pa_{\tr}^2  + \pa_s \big)\chi_{\frac{\zeta_*}{\tnu}}  + \pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}} \left[ \frac{1}{\tr} -  \frac{\big[2Q + 2\tilde P_1 - 2\tilde{N}_1 + \tm_v\big]}{2\tr}  + \tilde \eta \; \tr   \right].$$

By definition, we have

\begin{equation}\label{est:chisr}
\left|(\tr\pa_{\tr} + \tr^2 \pa_{\tr}^2)\chi_{\frac{\zeta_*}{\tnu}} \right| \lesssim \mathbf{1}_{\{\frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}}, \quad \left|\pa_s\chi_{\frac{\zeta_*}{\tnu}} \right| \lesssim \frac{\nu^2}{|\ln \nu|} \mathbf{1}_{\{\frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}},
\end{equation}

which gives the estimate

\begin{equation}\label{est:Fcut}
|\pa_{\tr}^k F| \lesssim \left(\frac{1}{\tr^{2 + k}} + \left|\frac{\tnu_s}{\tnu} \right| \frac{1}{\tr^k}\right)\mathbf{1}_{\{\frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}} \lesssim \frac{1}{\tr^{2 + k}}\mathbf{1}_{\{\frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}}.
\end{equation}

Since the cut-off term is localized in the zone $\tr \sim \frac{\zeta_*}{\tnu}$, we can use the mid-range control (<ref>) to obtain a better estimate. More precisely, we use the decomposition <ref>, i.e.

$$\tm_v = Q(\tr) - Q(r) + \tilde{P}_1(\tr) - P_1(r) - \tilde{N}_1(\tr) - P_2(r) +  m_q.$$

A direct computation and $\nu \sim \tnu$ ( see Lemma <ref>) yield

\begin{align*}
k = 0,1,2, \quad \big|(\tr \pa_{\tr})^k\big[Q(\tr) - Q(r)\big]\big| \lesssim \frac{1}{|\ln \nu|  \langle \tr \rangle^2 }.
\end{align*}

By the definition of $P_1$ and $\tilde{P}_1$ and $P_2$, we have the rough bound for $\frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}$,

\begin{align*}
k = 0,1,2, \quad \big|(\tr \pa_{\tr})^k\big[\tilde{P}_1(\tr) - P_1(r) - \tilde{N}_1(\tr) - P_2(r)\big]\big| \lesssim \frac{\nu^2}{|\ln \nu|}.
\end{align*}

Combining with  the mid-range estimate (<ref>) yields

\begin{equation}\label{est:As0mvmidrange}
\int |\As_0 \tm_v|^2 \mathbf{1}_{\{\frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}} \frac{\omega_0}{\tr}d\tr \lesssim \frac{\nu^4}{|\ln \nu|^2} + \|m_\e\|_{H^2 (\zeta_* \leq \zeta \leq \zeta^*)}^2.
\end{equation}

Using this estimate and integration by parts, we estimate

\begin{align*}
&\left|\int \left(-2\pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}} \pa_{\tr} \As_0 \tm_v + F \As_0 \tm_v \right) \As_0 \tm_v^* \frac{\omega_0}{\tr}d\tr \right|= \left| \int |\As_0 \tm_v|^2 \left[\frac{\tr}{\omega_0} \pa_{\tr} \Big(\pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}} \frac{\omega_0}{\tr} \Big) +   F \right]  \frac{\omega_0}{\tr}d\tr \right|\\
& \qquad  \lesssim \int |\As_0 \tm_v|^2 \left(\frac{1}{\tr^2} + \left| \frac{\tnu_s}{\tnu}\right| \right) \mathbf{1}_{\{\frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}} \frac{\omega_0}{\tr}d\tr   \lesssim \frac{\nu^6}{ \zeta_*^2 |\ln \nu|^2} + \frac{\nu^2}{\zeta_*^2}\|m_\e\|_{H^2 (\zeta_* \leq \zeta \leq \zeta^*)}^2.
\end{align*}

We also have by the commutator formulas (<ref>) and (<ref>),

\begin{align*}
&\left|-2\pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}} \big[\As_0, \pa_{\tr}\big] \tm_v + \big[\As_0, \pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}}\big] \pa_{\tr} \tm_v + \big[\As_0, F \big] \tm_v \right| \lesssim  \sum_{i = 0}^2 \frac{|\pa_{\tr}^i \tm_v|}{\tr^{4 - i}} \mathbf{1}_{\{\frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}}.
\end{align*}

from which and Cauchy-Schwarz and (<ref>), we obtain

\begin{align*}
&\left|\int \left( -2\pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}} \big[\As_0, \pa_{\tr}\big] \tm_v + \big[\As_0, \pa_{\tr}\chi_{\frac{\zeta_*}{\tnu}}\big] \pa_{\tr} \tm_v + \big[\As_0, F \big] \tm_v  \right) \As_0 \tm_v \frac{\omega_0}{\tr}d\tr\right|\\
& \lesssim  \left(\int |\As_0 \tm_v|^2 \mathbf{1}_{\{\frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}} \frac{\omega_0}{\tr}d\tr \right)^\frac{1}{2} \left( \int \frac{1}{\tr^4} \Big[ |\pa_{\tr}^2 \tm_v|^2 + \frac{|\pa_{\tr} \tm_v|^2}{\tr^2} + \frac{|\tm_v|^2}{\tr^4}  \Big] \frac{\omega_0}{\tr} d\tr \right)^\frac{1}{2}\\
&\quad  \lesssim \frac{\nu^2}{\zeta_*^2} \left( \frac{\nu^4}{|\ln \nu|^2} +  \| m_\e\|_{H^2(\zeta_* \leq \zeta \leq \zeta^*)} \right).
\end{align*}

Injecting the above estimates into (<ref>) and using $\frac{ds}{d\tau} = \frac{1}{\tnu^2}$ and $\nu \sim \tnu$ yields (<ref>).

Step 2 Energy estimate for the first order derivative. We now prove (<ref>). In order to handle the term $\|\As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}}^2$ appearing in (<ref>), we use the second energy identity by integrating equation (<ref>) against $-\tnu^2 \As_0 \tm_v^*$ in $L^2_{\frac{\omega_0}{\tr}}$ to write

\begin{align*}
\frac{1}{2}\frac{d}{ds}\left[ -\tnu^2\int \tm_v^* \As_0 \tm_v^* \frac{\omega_0}{\tr} d\tr\right] & = -\tnu^2\int |\As_0 \tm_v^*|^2 \frac{\omega_0}{\tr}d\tr  - \tnu^2 \int \As_0 \big(\chi_{\frac{\zeta_*}{\tnu}} \Ec_1 \big) \tm_v^* \frac{\omega_0}{\tr}d\tr \\
& \quad - \tnu^2\int\Big[\Fc(\tm_v) + T(\tm_v) +\frac{\tnu_s}{\tnu} \tm_v^*  \Big]\As_0 \tm_v^* \frac{\omega_0}{\tr} d\tr.
\end{align*}

Since the support of $\tm_v^*$ is in the interval $0 \leq \tr \leq \frac{2\zeta_*}{\tnu}$, we have by Hardy inequality (<ref>) and (<ref>),

\begin{equation*}
\int \frac{|\tm_v^*|^2}{1 + \tr^2} \frac{\omega_0}{\tr}d\tr \lesssim \int|\pa_{\tr}\tm_v^*|^2 \frac{\omega_0}{\tr}d\tr \lesssim \frac{\zeta_*^2}{\tnu^2}\int \frac{|\pa_{\tr}\tm_v^*|^2}{1 + \tr^2} \frac{\omega_0}{\tr}d\tr \lesssim \frac{\zeta_*^2}{\tnu^2}\int |\As_0 \tm_v^*|^2 \frac{\omega_0}{\tr}d\tr,
\end{equation*}

from which and Cauchy-Schwarz and (<ref>), we estimate

\begin{align*}
& \left|\tnu^2\int\As_0\big( \chi_{\frac{\zeta_*}{\tnu}} \Ec_1 \big)\tm_v^* \frac{\omega_0}{\tr} d\tr    \right| \lesssim \tnu^4\int \frac{|\tm_v^*|^2}{1 + \tr^2} \frac{\omega_0}{\tr} d\tr + \frac{\zeta_*^2}{\tnu^2} \int |\As_0 \big(\chi_{\frac{\zeta_*}{\tnu}} \Ec_1 \big)|^2\frac{\omega_0}{\tr} d\tr\\
& \quad  \lesssim  \zeta_*^2 \tnu^2  \int \frac{|\pa_{\tr} \tm_v^*|^2}{1 + \tr^2} \frac{\omega_0}{\tr} d\tr + \frac{\zeta_*^2}{\tnu^2} \frac{\nu^8}{|\ln \nu|^2}  \lesssim \zeta_*^2\tnu^2 \|\As_0 \tm_v^*\|^2_{L^2_\frac{\omega_0}{\tr}} + \frac{\zeta_*^2 \nu^6}{|\ln \nu|^2}.
\end{align*}

As for the scaling term, we simply estimate by using Cauchy-Schwarz and (<ref>),

\begin{align*}
\left|\tnu^2 \tilde \eta \int \tr \pa_{\tr} \tm_v^* \As_0 \tm_v^*  \frac{\omega_0}{\tr} d\tr \right| &\lesssim \tnu^2 |\tilde \eta|\|\As_0 \tm_v^* \|^2_{L^2_\frac{\omega_0}{\tr}} \left( \int |\tr \pa_{\tr} \tm_v^*|^2  \frac{\omega_0}{\tr} d\tr \right)^{\frac{1}{2}} \lesssim \tnu^2 \zeta_*^2 \|\As_0 \tm_v^* \|^2_{L^2_\frac{\omega_0}{\tr}}.
\end{align*}

By the definition of $\tilde{P}_1$, $\tilde{N}_1$ and (<ref>), we have the rough bound for $\tr \leq \frac{2\zeta_*}{\tnu}$ and $k = 0, 1$,

$$ \big|(\tr \pa_{\tr})^k ((2\tilde{P}_1 - 2\tilde{N}_1 + \tm_v) \big| \lesssim \nu^2 |\ln \la \tr \ra|,$$

from which and (<ref>), we obtain

\begin{align*}
&\left|\tnu^2\int \frac{\pa_{\tr} \big[(2\tilde{P}_1 - 2\tilde{N}_1 + \tm_v) \tm_v^* \big]}{2\tr} \As_0 \tm_v^* \frac{\omega_0}{\tr} d\tr  \right| \lesssim \tnu^4 |\ln \nu|^C \|\As_0 \tm_v^* \|^2_{L^2_\frac{\omega_0}{\tr}}.
\end{align*}

To estimate for the cut-off term, we use (<ref>) and (<ref>) to bound

$$|T(\tm_v)| = \big|-2\pa_{\tr} \chi_{\frac{\zeta_*}{\tnu}} \pa_{\tr} \tm_v  + F \tm_v\big| \lesssim \left[\frac{1}{\tr} |\pa_{\tr} \tm_v| + \left(\frac{1}{\tr^2} + \frac{\tnu_s}{\tnu}\right)|\tm_v|  \right] \mathbf{1}_{\{ \frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}}.$$

We then use Cauchy-Schwarz and (<ref>) and $|\tnu_s/\tnu| \lesssim \nu^2/|\ln \nu|$ to estimate

\begin{align*}
& \left|\tnu^2\int T(\tm_v) \As_0\tm_v^* \frac{\omega_0}{\tr}d\tr \right| \lesssim \tnu^2 \left( \int |\As_0  \tm_v|^2 \mathbf{1}_{\{ \frac{\zeta_8}{\tnu} \leq \tr\leq \frac{2 \zeta_*}{\tnu} \}} \frac{\omega_0}{\tr} d\tr \right)^\frac{1}{2} \left(\int \frac{1}{\tr^2} \Big[|\pa_{\tr} \tm_v|^2 + \frac{|\tm_v|^2}{\tr^2} \Big] \mathbf{1}_{\{ \frac{\zeta_*}{\tnu} \leq \tr \leq \frac{2\zeta_*}{\tnu}\}} \frac{\omega_0}{\tr}d\tr   \right)^\frac{1}{2} \\
&\lesssim \frac{\nu^2}{\zeta_*^2} \left( \frac{\nu^4}{|\ln \nu|^2} + \| m_\e\|_{H^2(\zeta_* \leq \zeta \leq \zeta^*)} \right).
\end{align*}

Since $|\tnu_s/\tnu| \lesssim \nu^2/|\ln \nu|^2$, we use the bootstrap estimate (<ref>) to bound the last term

$$\left|\tnu \tnu_s \int \tm_v^* \As_0 \tm_v^* \frac{\omega_0}{\tr}d\tr  \right| \lesssim \frac{\nu^6}{|\ln \nu|^2}.$$

Summing up these estimates and taking $\zeta_*$ small enough and using $\frac{ds}{d\tau} = \frac{1}{\tnu^2}$ and $\tnu \sim \nu$ yield the desired formula (<ref>). This concludes  the proof of Lemma <ref> assuming (<ref>).

Step 3 Bound for the error. We finally prove (<ref>). We recall from Proposition <ref> the identity

$$(\As_0 - \beta \tnu^2 \tr \pa_{\tr}) \phi_{n}(\tr) = 2\beta \tnu^2\big(1 -n +\tilde{\alpha}_n(\tnu)\big) \phi_{n}(\tr). $$

We then rewrite the error term by using the relation $\tr\pa_{\tr} Q=8T_0 \sim 8\phi_0$ and $ds/d\tau=1/\tnu^2$,

\begin{align*}
\Ec_1(\tr,s) & = \Big[-a_{1,\tau}+ 2\beta \tilde{\alpha}_1(\tnu) a_1 \Big]\phi_1+\Big[a_{1,\tau}-2\beta (1 + \tilde{\alpha}_0(\tnu))a_1+8\eta\Big] \phi_0\\
& \quad -  \frac{8 \tnu_s}{\tnu}\tilde \phi_0 +\frac{a_1}{\tnu^2}\left(\frac{\tnu_s}{\tnu} \tr\pa_{\tr} (\phi_1-\phi_0)-\pa_s(\phi_1-\phi_0)\right) + \tilde{\Ec}_1(\tr,s),
\end{align*}

where  we used the definition of $\tilde{N}_1$, i.e, $\As_0 \tilde{N}_1 = \frac{\pa_{\tr}\tilde{P}_1^2}{2\tr} + 8\tnu^2\tilde \phi_0(\tr)$ with $\tilde{\phi}_0 = \phi_0 - T_0$ to write

$$\tilde{\Ec}_1(\tr,s) =  \big(\pa_s - \eta \tr \pa_{\tr}\big) \tilde{N}_1  + \frac{\pa_{\tr} (2\tilde{P}_1 \tilde N_1 - \tilde N_1^2 )}{2\tr}+ \tilde N_0(v^\perp).$$

We recall from Lemma <ref> the modulation equations

\begin{align*}
&a_{1, \tau} - 2\beta a_1 \tilde{\alpha}_1(\nu) = \Oc\left(\frac{\nu^2}{|\ln \nu|^2} \right),\qquad \left(a_{1,\tau}- 2\beta(1 + \tilde\alpha_0(\nu))a_1 + 8\nu^2(\frac{\nu_\tau}{\nu}-\beta)\right) = \Oc\left(\frac{\nu^2}{|\ln \nu|^2} \right),
\end{align*}

from which and from the estimate (<ref>), we obtain

\begin{align*}
&|a_{1,\tau}-2\beta a_1\tilde \alpha_1(\tnu)| = \Oc\left(\frac{\nu^2}{|\ln \nu|^2} \right),\qquad \left| \left(a_{1,\tau}+ 2\beta(1 + \tilde\alpha_0(\tnu))a_1 - 8\tnu^2(\frac{\tnu_\tau}{\tnu}-\beta)\right) \right|= \Oc\left(\frac{\nu^2}{|\ln \nu|}\right).
\end{align*}

From (<ref>) and (<ref>), the identities $\As_0 T_0=0$ and $\As_0 T_1=-T_0$ and $\tnu \sim \nu$, one has the bound for $r\leq \zeta^*/\tnu$,

|\As_0 \phi_0|+|\As_0 \phi_1|\lesssim \tnu^2 \frac{\tr^2}{ \langle \tr\rangle ^4},

and hence, we have the estimate

\begin{align*}
&\int_0^{\frac{\zeta^*}{\tnu}} |\As_0 \left(\Big[-a_{1,\tau}+ 2\beta \tilde{\alpha}_1(\tnu) a_1 \Big]\phi_1+\Big[a_{1,\tau}-2\beta (1 + \tilde{\alpha}_0(\tnu))a_1+8\eta\Big] \phi_0\right) |^2 \frac{\omega_0}{\tr}d\tr  \lesssim \frac{\nu^8}{|\ln \nu|^4} \int_0^{\frac{\zeta^*}{\tnu}} \frac{1}{1+\tr}d\tr\lesssim \frac{\nu^8}{|\ln \nu|^3}.
\end{align*}

From (<ref>) and $|\tnu_s/ \tnu| \lesssim \nu^2/|\ln \nu|$, we have the estimate

\left|\frac{\tnu_s}{\tnu} \right|^2\int_0^{\frac{\zeta^*}{\tnu}} |\As_0 \tilde \phi_0|^2\frac{\omega_0}{\tr}d\tr\lesssim \frac{\nu^{8}}{|\ln \nu|^2}.

By writing

\pa_s (\phi_1-\phi_0)=\frac{\tnu_s}{\tnu}(\tnu \pa_{\tnu} (\phi_1-\phi_0)+\frac{\beta_s}{\beta}(\beta\pa_\beta (\phi_1-\phi_0))

and using (<ref>) and (<ref>),  we have the bound

|\As_0 (\pa_s (\phi_1-\phi_0))|\lesssim \frac{\nu^4}{|\ln \nu|}\frac{\tr^2}{\langle \tr \rangle^4}.

We also have from (<ref>) and (<ref>) and $|\tnu_s/\tnu|\lesssim 1/|\ln \nu|$ the bound

\left| \frac{\tnu_s}{\tnu}\As_0 (\tr\pa_{\tr} (\phi_1-\phi_0))\right|\lesssim \frac{\nu^2}{|\ln \nu|} \frac{\tr^2}{\langle \tr \rangle^4}.

Hence, there holds the estimate

\begin{align*}
& \int_0^{\frac{\zeta^*}{\tnu}} |\As_0 \frac{a_1}{\tnu^2}\left(\frac{\tnu_s}{\tnu}\tr\pa_{\tr} (\phi_1-\phi_0)-\pa_s(\phi_1-\phi_0)\right)  |^2\frac{\omega_0}{\tr}d\tr \lesssim \frac{\nu^{8}}{|\ln \nu|^2}\int_0^{\frac{\zeta^*}{\tnu}} \left(\frac{1}{(1+\tr)^3}+\frac{1}{|\ln \nu|^2(1+\tr)}\right)\lesssim \frac{\nu^8}{|\ln \nu|^2}.
\end{align*}

As for the term $\tilde{\Ec}_1$, we use the definitions of $\tilde{N}_1$ and $\tilde P_1$ and $\tnu \sim \nu$ and $|\tnu_s/\tnu| \lesssim \nu^2 /|\ln \nu|$ to bound for $\tr \leq \frac{2\zeta_*}{\tnu}$,

$$|\As_0 \tilde{\Ec}_1(\tr, s)| \lesssim \tnu^4 \left(\left|\frac{\tnu_s}{\tnu} \right| + \tnu^2  \right)\frac{\ln \la \tr \ra}{\la \tr \ra^2} \lesssim \nu^6 \frac{\ln \la \tr \ra}{\la \tr \ra^2},$$

which gives

$$\int_0^{\frac{2\zeta_*}{\tnu}} |\As_0 \tilde{\Ec}_1|^2 \frac{\omega_0}{\tr}d\tr \lesssim \nu^{12} |\ln \nu|^3 \lesssim \frac{\nu^8}{|\ln \nu|^2}.$$

The collection of the above estimates concludes the proof of (<ref>).

§.§ Nonradial energy estimates

We begin with estimating the parameter $x^*$. We claim the following.
We have for $\tau_0$ large enough

\begin{equation} \label{est:xstarrefined}
\left|\frac{x^*_\tau}{\mu} \right|^2 \leq \frac{C}{\nu^2} \int_{|y| \leq \frac{2\zeta_*}{\nu}} \frac{|\nabla q^\perp|^2}{U} dy  + Ce^{-2\kappa\tau},
\end{equation}

where $C > 0$ is independent from the bootstrap constants $\kappa,N,K, K', K''$. In particular, we have the estimate

\begin{equation}\label{est:xstar}
\left|\frac{x^*_\tau}{\mu} \right| \lesssim \frac{e^{-\kappa \tau}}{\nu^2}.
\end{equation}

We multiply equation (<ref>) by $\pa_{z_i} U_\nu \rho_0$ with $i =1,2$ and $\rho_0 = e^{-\beta |z|^2/2}$, and integrate over $\Rb^2$ and use the orthogonality condition (<ref>) to write

\begin{align*}
&\frac{x^*_{i,\tau}}{\mu} \int\pa_{z_i} \big( U_\nu + \Psi_\nu \big) \pa_{z_i} U_\nu \rho_0 dz   = \int \left[- \Ls^z \e^\perp +\nabla \cdot \Big(\Gc(\e)   - \frac{x^*_\tau}{\mu} \e^0\Big) - N^\perp(\e^\perp)\right] \pa_{z_i} U_\nu \rho_0 dz.
\end{align*}

Using the definition of $U_\nu$ and $\Psi_\nu$, we have

$$\int|\pa_{z_i} U_\nu|^2 e^{-\beta\frac{|z|^2}{4}} dz = \frac{c_0}{\nu^4}\int \frac{r^3}{(1 + r^2)^6}e^{-\beta \nu^2 \frac{r^2}4} dr \sim \frac{c_0'}{\nu^4},$$

$$\left|\int \pa_{z_i}\Psi_\nu \pa_{z_i} U_\nu e^{-\beta\frac{|z|^2}{4}} dz\right| = \Oc(\nu^{-2}).$$

We write the linear operator as $\Ls_0 q^\perp = \nabla \cdot (U \nabla \Ms q^\perp)$, then by integration by parts and Cauchy-Schwarz and the decay of $U$ we estimate

\begin{align*}
&\left|\int \Ls^z \e^\perp \pa_{z_i} U_\nu \rho_0 dz \right| = \left|\frac{1}{\nu^5}\int \nabla \cdot \big(U \nabla \Ms q^\perp \big) \pa_{y_i} U \rho dy - \frac{\beta}{\nu} \int \nabla \cdot(y q^\perp) \pa_{y_i}U \rho dy \right|\\
& =\left| \frac{1}{\nu^5}\int U \nabla \Ms q^\perp \cdot \nabla (\pa_{y_i} U \rho) dy + \frac{\beta}{\nu}\int q^\perp y \cdot \nabla (\pa_{y_i} U \rho) dy  \right| \\
& \quad \lesssim \frac{1}{\nu^5} \left(\int_{|y| \leq \frac{\zeta_*}{2\nu}} U |\nabla \Ms q^\perp|^2 \rho dy   \right)^\frac{1}{2} \left(\int_{|y| \leq \frac{\zeta_*}{2\nu}} U |\nabla (\pa_{y_i} U \rho)|^2 \rho^{-1}dy \right)^\frac{1}{2} \\
& \qquad + \frac{1}{\nu^6}\left(\nu^2 \int_{|y| \geq \frac{\zeta_*}{2\nu}} \frac{|q^\perp|^2}{U} \rho dy \right)^\frac{1}{2} \left(\int_{|y| \geq \frac{\zeta_*}{2\nu}} \left|\Ms \Big[\nabla\cdot (U \nabla (\pa_{y_i} U \rho)) \Big]\right|^2 \rho^{-1} dy \right)^\frac{1}{2}  + e^{-\kappa \tau} \\
& \qquad  + \frac{1}{\nu^2}\left(\nu^2 \int \frac{|q^\perp|^2}{U} \rho dy \right)^\frac{1}{2} \left(\int \big|y \cdot \nabla (\pa_{y_i} U \rho)\big|^2 \rho^{-1} dy \right)^\frac{1}{2} \\
& \qquad \lesssim \frac{1}{\nu^5}\left(\int_{|y| \leq \frac{2\zeta_*}{\nu}} U|\nabla \Ms q^\perp|^2 \rho \right)^\frac{1}{2} + \frac{1}{\nu^6}\|\e^\perp\|_0 \, \nu^{10} + \frac{1}{\nu^2}\|\e^\perp\|_0  \lesssim \frac{1}{\nu^5}\left(\int_{|y| \leq \frac{2\zeta_*}{\nu}} \frac{|\nabla q^\perp|^2}{U} \rho \right)^\frac{1}{2} + \frac{e^{-\kappa \tau}}{\nu^2}.
\end{align*}

We write by the definition $\e^0 = \frac{\pa_\zeta m_\e}{\zeta}$ and use (<ref>),  (<ref>) and the decay of $U$,

\begin{align*}
&\left|\frac{x_{i,\tau}^*}{\mu} \int \pa_{z_i} \e \pa_{z_i}U_\nu \rho_0 dz \right| = \left|\frac{x_{i,\tau}^*}{\mu} \int \left(\frac{\pa_\zeta m_\e}{\zeta} + \e^\perp \right) \pa_{z_i}\big( \pa_{z_i}U_\nu \rho_0\big) dz \right| \\
& \quad \lesssim \left|\frac{x_{i,\tau}^*}{\mu} \right| \frac{1}{\nu^4} \left(\int |\pa_\zeta m_\e|^2 \frac{1 }{\zeta U_\nu} d\zeta \right)^\frac{1}{2} +  \left|\frac{x_{i,\tau}^*}{\mu} \right|\frac{e^{-\kappa\tau}}{\nu^3}\\
& \quad \quad \lesssim \left|\frac{x_{i,\tau}^*}{\mu} \right| \left(\frac{1}{\nu^4} \|\tilde{m}_v\|_\inn + \frac{1}{\nu^2|\ln \nu|}\int_{\zeta \geq \frac{\zeta_*}{2}} \zeta^C \rho_0 d\zeta + \frac{e^{-\kappa\tau}}{\nu^3} \right) \lesssim \left|\frac{x_{i,\tau}^*}{\mu} \right| \frac{1}{\nu^2 |\ln \nu|}.
\end{align*}

We have by Cauchy-Schwarz,

\begin{align*}
&\left |\int \nabla \cdot \Gc(\e^\perp) \pa_{z_i}U_\nu \rho_0 dz\right| = \frac{1}{\nu^5} \left|\int \nabla \cdot \Gc(q^\perp) \pa_{y_i}U \rho dy \right|\\
& \lesssim \left(\int \Big| \nabla \cdot \Gc(q^\perp)\Big|^2 \rho dy \right)^\frac{1}{2}\left(\int |\pa_{y_i}U|^2 \rho dy  \right)^\frac{1}{2} \lesssim \frac{1}{\nu^5}\left(\int \Big|\nabla \cdot \Gc(q^\perp) \Big|^2 \rho_0 dz \right)^\frac{1}{2},
\end{align*}

where we recall from the definition of $\Gc(\e^\perp)$,

$$\nabla \cdot \Gc(q^\perp) = \nabla q^\perp \cdot \nabla \Phi_{\Phi + q^0}  - 2q^\perp (\Psi + q^0) + \nabla (\Psi + q^0)\cdot \nabla \Phi_{q^\perp},$$

and $\Psi$ is given by

$$\Psi = \frac{a_1}{\nu^2} \frac{\pa_r}{r}(\phi_1 - \phi_0) + \sum_{n = 2}^N \frac{a_n}{\nu^2} \frac{\pa_r}{r}\phi_n.$$

By the definition of $\phi_n$, we have the estimate

\begin{equation}\label{est:Psir}
|\Psi(r)| + |r \pa_r \Psi(r)| \lesssim \nu^2 \la r\ra^{-2} + \frac{\la r \ra ^{-4}}{|\ln \nu|^2}\mathbf{1}_{\{r \leq \frac{\zeta_*}{2\nu}\}} + \frac{\nu^2}{|\ln \nu|^2} \la \nu r \ra^C \mathbf{1}_{\{r \geq \frac{\zeta_*}{2\nu}\}},
\end{equation}

from which and (<ref>) and (<ref>), we obtain the bound

\begin{equation*}
\int \Big|\nabla \cdot \Gc(q^\perp)\Big|^2 \rho dy \lesssim \int_{|y| \leq \frac{2\zeta_*}{\nu}} \left(\frac{|\nabla q^\perp|^2}{U} + \frac{|q^\perp|^2}{\sqrt{U}} \right) dy + e^{-2\kappa \tau}.
\end{equation*}

By using (<ref>), (<ref>) and (<ref>), we estimate the contribution from $N^\perp(\e^\perp)$,

\begin{equation*}
\int \Big|N^\perp (\e^\perp)\Big|^2 \rho_0 dz \lesssim e^{-2\kappa \tau}.
\end{equation*}

The collection of the above estimates yields (<ref>) and concludes the proof of Lemma <ref>.

We are now in the position to derive the energy estimate for the nonradial part. Let us begin with $\|\e^\perp\|_0$ by using the coercivity given in Proposition <ref>.
We have

\begin{equation} \label{ene:ebot0}
\frac{1}{2}\frac{d}{d\tau} \| \e^\perp\|_0^2 \leq - \delta' \| \e^\perp\|_0^2   + Ce^{-2\kappa \tau},
\end{equation}

where $\delta', C > 0$ are independent from the bootstrap constants $K, K', K''$.

We multiply equation (<ref>) by $\nu^2 \sqrt{\rho_0}\Ms^z (\e^\perp \sqrt{\rho_0})$ with $\rho_0(z) = e^{-\beta |z|^2/2}$ and integrate over $\Rb^2$ to write

\begin{align*}
\frac{1}{2} \frac{d}{d\tau} \| \e^\perp(\tau)\|_0^2  &=\nu^2\int \Ls^z \e^\perp \sqrt{\rho_0} \Ms^z (\e^\perp \sqrt{\rho_0}) dz  + \frac{1}{2}\int_{\Rb^2} \frac{d}{d\tau}(\nu^2U_\nu^{-1}) |\e^\perp|^2 \rho_0  dz \\
&  - \nu^2 \int \big[\nabla \cdot \Gc(\e^\perp) - \frac{x^*_\tau}{\mu} \cdot \nabla (W + \e^0) - N^\perp(\e^\perp)  \big] \sqrt{\rho_0}\Ms^z (\e^\perp \sqrt{\rho_0}) dz.
\end{align*}

We have the coercivity estimate (<ref>) for the linear part, which using (<ref>) with $\alpha=1$ gives:

\begin{align}
\label{est:dissp} \nu^2\int \tilde \Ls^z \e^\perp \sqrt{\rho_0} \Ms^z (\e^\perp \sqrt{\rho_0}) dz &= \int \tilde \Ls q^\perp \sqrt{\rho} \Ms (q^\perp \sqrt{\rho}) dy \leq - \delta_0\int \frac{|\nabla q^\perp|^2}{U} \rho dy
\end{align}

\begin{align}
\label{eq:controle0bynq} -\int \frac{|\nabla q^\perp|^2}{U} \rho dy\leq -C_0 \nu^2\int \frac{|q^\perp|^2}{U}\rho dy=-C_0\nu^2\int \frac{|\e^\perp|^2}{U_\nu}\rho_0 dz =-C_0 \|\e\|^2_0 .
\end{align}

The time derivative term: We write by the definition of $U_\nu$,

$$\frac{d}{d\tau}(\nu^2U_\nu^{-1}) = \frac{\nu_\tau}{\nu} \nu \pa_\nu[( |z^2| + \nu^2)^2] = \frac{\nu_\tau}{\nu} \frac{4\nu^2}{U_\nu} \frac{\nu^2}{|z|^2 + \nu^2},$$

from which and $\left|\frac{\nu_\tau}{\nu} \right| \lesssim \frac{1}{|\ln \nu|}$  and (<ref>), we have the estimate

\begin{align*}
\left|\int_{\Rb^2} \frac{d}{d\tau}(\nu^2U_\nu^{-1}) |\e^\perp|^2 \rho_0  dz\right| & \lesssim \frac{\nu^2}{|\ln \nu|}\int \frac{|\e^\perp|^2}{U_\nu}\rho_0 dz \lesssim \frac{1}{|\ln \nu|}\int \frac{|\nabla q^\perp|^2}{U}\rho dy.
\end{align*}

The small linear and nonlinear terms: We write by integration by parts and  Cauchy-Schwarz,

\begin{align*}
&\left|\nu^2\int \nabla \cdot \Gc(\e^\perp) \sqrt{\rho_0} \Ms^z (\e^\perp \sqrt{\rho_0}) dz\right| = \left|\nu^2 \int \Gc(\e) \cdot \Big[ \nabla \Ms^z(\e^\perp \sqrt{\rho_0}) - \frac \beta 2 z  \Ms^z(\e^\perp \sqrt{\rho_0}) \Big]\sqrt{\rho_0}dz \right|\\
&\quad \lesssim \frac{\nu^2}{\sqrt{|\ln \nu|}}\int \left(U_\nu |\nabla \Ms^z(\e^\perp \sqrt{\rho_0})|^2 + U_\nu|\Ms^z(\e^\perp \sqrt{\rho_0})|^2 \right) dz  + \nu^2 \sqrt{|\ln \nu|} \int \frac{(1 + |z|^2)|\Gc(\e)|^2}{U_\nu} \rho_0 dz.
\end{align*}

Making a change of variables and using Hardy inequality (<ref>) yields

\begin{align*}
\nu^2\int U_\nu |\nabla \Ms^z(\e^\perp \sqrt{\rho_0})|^2 dz = \int U|\nabla \Ms(q^\perp \sqrt{\rho})|^2 dy \lesssim \int \frac{|\nabla q^\perp|^2}{U} \rho dy.
\end{align*}

We also have by (<ref>),

\begin{align}\label{est:UMeperp}
\nu^2\int U_\nu |\Ms^z(\e^\perp \sqrt{\rho_0})|^2 dz \lesssim \nu^2 \int \frac{|\e^\perp|^2}{U_\nu} \rho_0 dz \lesssim \int \frac{|\nabla q^\perp|^2}{U}\rho dy.
\end{align}

We recall

$$\Psi_\nu + \e^0 = \frac{\pa_\zeta}{\zeta}(P_\nu + m_\e), \quad \nabla \Phi_{\Psi_\nu + \e^0} = -z \frac{(P_\nu + m_\e)}{\zeta^2},$$

where we have from the definition of $P_\nu$ and (<ref>),  (<ref>) and (<ref>) the bound

$$|\zeta \pa_\zeta(P_\nu + m_\e)| \lesssim  \frac{1}{\sqrt{|\ln \nu|}} \mathbf{1}_{\{\zeta \leq \frac{\zeta_*}{2}\}} +  \nu^2 |\ln \nu| \zeta^{\frac{1}{4}} \mathbf{1}_{\{\zeta \geq \frac{\zeta_*}{2}\}}. $$

We then estimate by using (<ref>),

\begin{align*}
& \nu^2 \sqrt{|\ln \nu|}\int \frac{(1 + |z|^2)|\e^\perp \nabla \Phi_{\Psi_\nu + \e^0} |^2}{U_\nu} \rho_0 dz\lesssim  \frac{\nu^2}{\sqrt{|\ln \nu|}} \int_{|z| \leq \frac{\zeta_*}{2}} \frac{|\e^\perp|^2}{U_\nu} \rho_0 dz + \nu^6|\ln \nu|^\frac{5}{2} \int_{|z| \geq \frac{\zeta_*}{2}} \frac{|\e^\perp|^2}{U_\nu}|z|^\frac{5}{2} \rho_0 dz\\
& \quad \lesssim \frac{\nu^2}{\sqrt{|\ln \nu|}} \int  \frac{|q^\perp|^2}{U} \rho dy + \nu^6|\ln \nu|^\frac{5}{2} \nu^{\frac{5}{2}}\int \frac{|q^\perp|^2}{U} (1 + |y|^\frac{5}{2}) \rho dy  \lesssim \frac{1}{\sqrt{|\ln \nu|}} \int \frac{|\nabla q^\perp|^2}{U}\rho dy.
\end{align*}

We recall $P_{1, \nu} = \frac{a_1}{\nu^2}(\phi_1(r) - \phi_0(r))$ and the bound

$$\left|\Psi_{1, \nu} \right| = \left|\frac{\zeta \pa_\zeta P_{1, \nu}}{\zeta^2} \right| \lesssim \frac{|y|^2}{1 + |y| ^4}, \quad \forall y \in \Rb^2,$$

so that we write

\begin{align*}
&\nu^2 \sqrt{|\ln \nu|}\int \frac{(1 + |z|^2)|\Psi_{1,\nu}\nabla \Phi_{\e^\perp}|^2}{U_\nu} \rho_0 dz \lesssim
\nu^4 \sqrt{|\ln \nu|} \int (1 + \nu^2 |y|^2) \big( | \nabla \Phi_{q^\perp \chi_{\frac{\zeta_*}{2\nu}}}|^2 + | \nabla \Phi_{q^\perp ( 1 - \chi_{\frac{\zeta_*}{2\nu}} )}|^2 \big)\rho  dy.
\end{align*}

We estimate by using (<ref>) and (<ref>),

\begin{align*}
&\nu^4 \sqrt{|\ln \nu|} \int (1 + \nu^2 |y|^2) | \nabla \Phi_{q^\perp \chi_{\frac{\zeta_*}{2\nu}}}|^2 \rho dy\\
& \quad  \lesssim \nu^4 \sqrt{|\ln \nu|} \left[\int_{|y| \leq \frac{\zeta_*}{\nu}} |q^\perp|^2 \langle y \rangle^4 dy + \nu^2\int_{|y| \leq \frac{\zeta_*}{\nu}} |q^\perp|^2 \langle y \rangle ^6dy   \right]  \int \frac{1}{1 + |y|^4} \rho dy \lesssim \nu^2 \sqrt{|\ln \nu|} \int \frac{|\nabla q^\perp|^2}{U} \rho dy.
\end{align*}

Using (<ref>) and noting that $w^\perp = \e^\perp$, we estimate

\begin{align*}
&\nu^4 \sqrt{|\ln \nu|} \int (1 + \nu^2 |y|^2) | \nabla \Phi_{q^\perp ( 1 - \chi_{\frac{\zeta_*}{2\nu}})}(y)|^2 \rho  dy  = \nu^4 \sqrt{|\ln \nu|} \int (1 + |z|^2) |\nabla \Phi_{\e^\perp (1 - \chi_{\frac{\zeta_*}{2}}) }(z)|^2 \rho_0 dz\\
& \quad \lesssim \nu^4 \sqrt{|\ln \nu|} \int_{|z| \leq 1}|\nabla \Phi_{\e^\perp (1 - \chi_{\frac{\zeta_*}{2}}) }(z)|^2 dz + \nu^4 \sqrt{|\ln \nu|} \int_{|z| \geq 1} \| \e^\perp\|_\out^2 |z|^\frac{3}{2} \rho_0 dz \lesssim e^{-2\kappa \tau}.
\end{align*}

We have by the definition of $P_{2, \nu}$ and (<ref>) and (<ref>),

\begin{align*}
|\Psi_{2, \nu} + \e^0| &= \left| \frac{\zeta\pa_\zeta (P_{2, \nu} + m_\e)}{\zeta} \right| \lesssim \frac{1}{\nu^2 |\ln \nu|} \frac{|y|^2 \ln \la |y| \ra}{1 + |y|^4} \mathbf{1}_{\{|z| \leq \frac{\zeta_*}{2}\}} + \frac{1}{|\ln \nu|} |z|^{-2 + \frac{1}{4}}\mathbf{1}_{\{|z| \geq \frac{\zeta_*}{2}\}},
\end{align*}

from which and a similar estimate as given above, we obtain

\begin{align*}
\nu^2 \sqrt{|\ln \nu|}\int \frac{(1 + |z|^2)|(\Psi_{2,\nu} + \e^0)\nabla \Phi_{\e^\perp}|^2}{U_\nu} \rho_0 dz \lesssim \frac{1}{\sqrt{|\ln \nu|}} \int \frac{|\nabla q^\perp|^2}{U}\rho dy  + e^{-2\kappa \tau}.
\end{align*}

We also have by (<ref>) and (<ref>) and (<ref>) the rough estimate

\begin{align*}
&\nu^2 \sqrt{|\ln \nu|}\int \frac{\left|\e^\perp \nabla \Phi_{\e^\perp}\right|^2 +\left| \frac{x^*_\tau}{\mu} \e^\perp \right|^2 }{U_\nu} \rho_0 dz  \lesssim \frac{e^{-4\kappa \tau}}{\nu^{12}} \sqrt{|\ln \nu|} \int (1 + |z|^C)\rho_0 dz + \left|\frac{x^*_\tau}{\mu} \right|^2\sqrt{|\ln \nu|} \| \e^\perp\|_0^2\lesssim e^{-2\kappa \tau},
\end{align*}

and from (<ref>) and (<ref>),

\begin{align*}
&\nu^2 \sqrt{|\ln \nu|} \left|\frac{x^*_\tau}{\mu} \right|^2 \int \frac{|\e^0|^2 }{U_\nu} \rho_0 dz = \sqrt{|\ln \nu|} \left|\frac{x^*_\tau}{\mu} \right|^2 \int_0^{+\infty} |\pa_\zeta m_\e|^2 \frac{\omega_\nu}{\zeta} d\zeta\\
& \quad \lesssim \sqrt{|\ln \nu|} \left|\frac{x^*_\tau}{\mu}\right| \left[\int_{\zeta \leq \frac{\zeta_*}{2}} |\pa_\zeta m_\e|^2 \frac{\omega_\nu}{\zeta} d\zeta +  \int_{\zeta \geq \frac{\zeta_*}{2}} |\pa_\zeta m_\e|^2 \frac{\omega_\nu}{\zeta} d\zeta \right]\\
& \qquad \lesssim \sqrt{|\ln \nu|} \left|\frac{x^*_\tau}{\mu}\right| \left[ \| \tilde m_v\|^2_\inn + \frac{\nu^2}{|\ln \nu|}\int_{\zeta \geq \frac{\zeta_*}{2}} \zeta^C e^{-\beta \zeta^2/2}d\zeta  \right] \lesssim e^{-2\kappa \tau}.
\end{align*}

The difference $(\Ls^z - \tilde{\Ls}^z)\e^\perp$: By the definition $\tilde{\Phi}_{q^\perp} = \frac{1}{\sqrt{\rho}} \Phi_{q^\perp \sqrt{\rho}}$, we have

\begin{align*}
&\Delta (\Phi_{q^\perp} - \tilde{\Phi}_{q^\perp}) = \beta \nu^2\left(1 + \frac{\beta \nu^2 |y|^2}{4} \right)\tilde{\Phi}_{q^\perp} -b\nu^2 y \cdot \nabla \tilde{\Phi}_{q^\perp},
\end{align*}

from which and an integration by parts and Cauchy-Schwarz inequality, we write

\begin{align*}
&\left|\nu^2 \int_{\Rb^2} \big(\Ls^z - \tilde \Ls^z \big)\e^\perp \sqrt{\rho_0} \Ms^z(\e^\perp \sqrt{\rho_0}) dz\right| \ = \ \left|\nu^2 \int_{\Rb^2} \nabla U_\nu \cdot \nabla (\Phi_{\e^\perp} - \tilde\Phi_{\e^\perp}) \sqrt{\rho_0} \Ms^z(\e^\perp \sqrt{\rho_0}) dz\right| \\
& \quad = \left|\int_{\Rb^2} \nabla U \cdot \nabla (\Phi_{q^\perp} - \tilde\Phi_{q^\perp}) \sqrt{\rho} \Ms(q^\perp \sqrt{\rho}) dy\right| \ \leq \ \left(\int U|\nabla \Ms(q^\perp \sqrt{\rho})|^2 dy\right)^\frac{1}{2} \left(\int U|\nabla (\Phi_{q^\perp} - \tilde{\Phi}_{q^\perp})|^2 \rho dy \right)^\frac{1}{2}\\
& \qquad + \nu^2\left(\int U|\Ms(q^\perp \sqrt{\rho})|^2 dy\right)^\frac{1}{2} \left(\int U \Big(|y \cdot \nabla (\Phi_{q^\perp} - \tilde{\Phi}_{q^\perp})|^2 + |\langle \nu y \rangle^ 2 \Phi_{q^\perp \sqrt{\rho}}|^2 \Big) \rho dy \right)^\frac{1}{2}\\
&\quad \lesssim \frac{1}{50}\int \frac{|\nabla q^\perp|^2}{U}\rho dy + C \int U \langle \nu y \rangle^ 2 |\nabla (\Phi_{q^\perp} - \tilde{\Phi}_{q^\perp})|^2 \rho dy + C\nu^2 \int U\langle \nu y \rangle^4|\Phi_{q^\perp \sqrt{\rho}}|^2 dy.
\end{align*}

We bound the last term by using (<ref>) with $\alpha = 1, 3$ and (<ref>),

\begin{equation*}
\nu^2 \int U\langle \nu y \rangle^ 4 |\Phi_{q^\perp \sqrt{\rho}}|^2 dy \lesssim \nu^2 \int |q^\perp|^2 \langle y \rangle^ 2\rho dy + \nu^6 \int |q^\perp|^2 \langle y \rangle^ 6\rho dy \lesssim \nu^2 \int \frac{|\nabla q^\perp|^2}{U}\rho dy.
\end{equation*}

For $|y| \geq \frac{\zeta_*}{2\nu}$, we use (<ref>), (<ref>) and  (<ref>) to estimate

\begin{align*}
&\int_{|y| \geq \frac{\zeta_*}{2\nu}} U\langle \nu y \rangle^ 2|\nabla(\Phi_{q^\perp} - \tilde\Phi_{q^\perp})|^2 \rho dy\\
& \lesssim \int_{|y| \geq \frac{\zeta_*}{2\nu}} U\langle \nu y \rangle^ 2 \left(| \nabla \Phi_{q^\perp \chi_{\frac{\zeta_*}{2\nu}}}|^2\rho  + | \nabla \Phi_{q^\perp ( 1 - \chi_{\frac{\zeta_*}{2\nu}})}|^2\rho + \nu^4|y|^2 |\Phi_{q^\perp \sqrt{\rho}}|^2  + |\nabla \Phi_{q^\perp \sqrt{\rho}}|^2 \right) dy\\
& \qquad \lesssim \nu^2 \int \frac{|\nabla q^\perp|^2}{U} \rho dy + \nu^2\|\e^\perp\|_\out^2.
\end{align*}

For $|y| \leq \frac{\zeta_*}{2\nu}$, we write by the definition $\tilde{\Phi}_{q^\perp} = \frac{1}{\sqrt{\rho}} \Phi_{q^\perp \sqrt{\rho}}$,

\begin{align*}
&\int_{|y| \leq \frac{\zeta_*}{2\nu}} U(1 + \nu^2 |y|^2)|\nabla(\Phi_{q^\perp} - \tilde\Phi_{q^\perp})|^2 \rho dy  \lesssim \int_{|y| \leq \frac{\zeta_*}{2\nu}}\frac{| \nabla \Phi_{q^\perp (1 - \sqrt{\rho})}|^2 + \nu^4 ( |y\Phi_{q^\perp \sqrt{\rho}}|^2 + ||y|^2 \nabla \Phi_{q^\perp \sqrt{\rho}}|^2}{1 + |y|^4} dy.
\end{align*}

The last two terms are estimated by using (<ref>) with $\alpha = 1$ and (<ref>),

\begin{align*}
&\int_{|y| \leq \frac{\zeta_*}{2\nu}} \frac{\nu^4(| y \Phi_{q^\perp \sqrt{\rho}}|^2 + ||y|^2\nabla \Phi_{q^\perp \sqrt{\rho}})|^2}{\langle y \rangle^4} dy \lesssim \nu^2 \left(\int |q^\perp|^2\langle y \rangle^2 \rho dy\right) \int_{|y| \leq \frac{\zeta_*}{2\nu}} \frac{1}{\langle y \rangle^4} dy   \lesssim \nu^2 \int \frac{|\nabla q^\perp|^2}{U} \rho dy.
\end{align*}

For the remaining term, we split into two parts

\begin{align*}
&\int_{|y| \leq \frac{\zeta_*}{2\nu}} U(1 + \nu^2|y|^2)|\nabla \Phi_{q^\perp (1 - \sqrt{\rho})}|^2 dy \lesssim \int_{|y| \leq \frac{\zeta_*}{2\nu}}  \frac{|\nabla \Phi_{q^\perp (1 - \sqrt{\rho}) \chi_{\frac{\zeta_*}{\nu}}}|^2 + |\nabla \Phi_{q^\perp (1 - \sqrt{\rho}) ( 1 - \chi_{\frac{\zeta_*}{\nu}})}|^2}{1 + |y|^4} dy.
\end{align*}

Since $(1 - \sqrt{\rho}) \chi_{\frac{\zeta_*}{\nu}}(x) \lesssim \zeta_*^2\mathbf{1}_{|x| \leq \frac{2\zeta_*}{\nu}} $, we estimate by using (<ref>) with $\alpha = 1$ and (<ref>),

\begin{align*}
&\int_{|y| \leq \frac{\zeta_*}{2\nu}}  \frac{|\nabla \Phi_{q^\perp (1 - \sqrt{\rho}) \chi_{\frac{\zeta_*}{\nu}}}(y)|^2}{1 + |y|^4} dy  \lesssim \zeta^2_* \int_{|y|\leq \frac{\zeta_*}{\nu}} |q^\perp|^2 (1 + |y|^2) dy \lesssim \zeta^2_* \int \frac{|\nabla q^\perp|^2}{U}\rho dy.
\end{align*}

Since $(1 - \sqrt{\rho})(1 - \chi_{\frac{\zeta_*}{\nu}}(x)) \leq \mathbf{1}_{|x| \geq \frac{\zeta_*}{\nu}}$,  and $\frac{1}{|x - y|} \lesssim \frac{\nu}{\zeta_*}$ for $|x| \geq \frac{\zeta^*}{\nu}$ and $|y| \leq \frac{\zeta_*}{2\nu}$, we estimate by using the outer norm (<ref>),

\begin{align*}
&\int_{|y| \leq \frac{\zeta_*}{2\nu}}  \frac{|\nabla \Phi_{q^\perp (1 - \sqrt{\rho})(1 - \chi_{\frac{\zeta_*}{\nu}})}(y)|^2}{1 + |y|^4} dy \lesssim \int_{|y| \leq \frac{\zeta_*}{2\nu}} \frac{dy}{1 + |y|^4} \left(\int_{|x| \geq \frac{\zeta_*}{\nu}} \frac{|q^\perp|}{|x - y|}dx \right)^2\\
& \qquad  \lesssim \nu^2 \left(\int_{|z| \geq \zeta_*} \frac{|\e^\perp|}{|z|} dz  \right)^2 \lesssim \nu^2 \|\e^\perp\|^2_\out \left(\int_{|z| \geq \zeta_*} \frac{dz}{|z|^{(3 - \frac{1}{4} + \frac{1}{p})p'}} \right)^\frac{2}{p'} \lesssim \frac{\nu^2}{\sqrt{\zeta_*}} \|\e^\perp\|^2_\out.
\end{align*}

Gathering these obtained estimates and using the bootstrap bound (<ref>) yields

\begin{align*}
\left|\nu^2 \int_{\Rb^2} \big(\Ls^z - \tilde \Ls^z \big)\e^\perp \sqrt{\rho_0} \Ms^z(\e^\perp \sqrt{\rho_0}) dz\right| \lesssim \zeta_*^2 \int \frac{|\nabla q^\perp|^2}{U}\rho dy + e^{-2\kappa \tau}.
\end{align*}

The error transport term: It remains to estimate the error term by writing

\begin{align*}
\frac{x_\tau^*}{\mu}\nu^2 \cdot \int \nabla W \sqrt{\rho_0} \Ms^z(\e^\perp \sqrt{\rho_0}) dz = \frac{x_\tau^*}{\mu} \nu \cdot  \int \nabla V  \sqrt{\rho} \Ms(q^\perp \sqrt{\rho}) dy,
\end{align*}

where $V(r, \tau) = \frac{1}{\nu^2} W(\zeta, \tau)$ is the approximate profile in the blowup variables defined from (<ref>), i.e,

$$V = U + \Psi, \quad \Psi = \frac{a_1}{\nu^2} \frac{\pa_r}{r}(\phi_1 - \phi_0) + \sum_{n = 2}^N \frac{a_n}{\nu^2} \frac{\pa_r}{r}\phi_n.$$

By the definition of $\phi_n$, we have the estimate

\begin{equation*}
|\Psi(r)| + |r \pa_r \Psi(r)| \lesssim \nu^2 \la r\ra^{-2} + \frac{\la r \ra ^{-4}}{|\ln \nu|^2}\mathbf{1}_{\{r \leq \frac{2\zeta_*}{\nu}\}} + \frac{\nu^2}{|\ln \nu|^2} \la \nu r \ra^C \mathbf{1}_{\{r \geq \frac{2\zeta_*}{\nu}\}}.
\end{equation*}

We use $\Ms \nabla U = 0$ and (<ref>) to estimate the leading term

\begin{align*}
&\left|\frac{x_\tau^*}{\mu} \nu \int \nabla U \sqrt{\rho}\Ms (q^\perp \sqrt{\rho}) dy\right| = \left|\int \Ms \big(\nabla U (\sqrt{\rho} -1) \big) q^\perp \sqrt{\rho} dy\right|\\
&  \lesssim \left|\frac{x_\tau^*}{\mu}\right| \nu \left[ \int_{|y| \leq \frac{\zeta_*}{2\nu}} \nu^2 |y| |q^\perp| \sqrt{\rho} dy + \int_{|y| \geq \frac{\zeta_*}{2\nu}} |\Ms \big(\nabla U (\sqrt{\rho} -1) \big) q^\perp \sqrt{\rho}| dy   \right]\\
& \lesssim \left|\frac{x_\tau^*}{\mu}\right| \nu^2 \left(\int_{|y| \leq \frac{\zeta_*}{2\nu}} |q^\perp|^2 (1 + |y|^2)\rho dy \right)^\frac{1}{2} \\
& \quad +  \left|\frac{x_\tau^*}{\mu} \right|\nu \left(\int_{|y| \geq \frac{\zeta_*}{2\nu}} |q^\perp|^2 (1 + |y|^2)\rho dy \right)^\frac{1}{2} \left(\int_{|y| \geq \frac{\zeta_*}{2\nu}} \frac{|\Ms (\nabla U (\sqrt{\rho} - 1))|^2}{1 + |y|^2} dy \right)^\frac{1}{2}\\
& \lesssim \left|\frac{x_\tau^*}{\mu}\right| \nu^2 \left(\int |q^\perp|^2(1 + |y|^2)\rho dy\right)^\frac{1}{2} \lesssim \nu \int \frac{|\nabla q^\perp|^2}{U}\rho dy + Ce^{-2\kappa \tau}.
\end{align*}

Using (<ref>) and (<ref>) and (<ref>), we estimate

\begin{align*}
&\left|\frac{x_\tau^*}{\mu} \nu \int \nabla \Psi \sqrt{\rho}\Ms (q^\perp \sqrt{\rho}) dy  \right|\\
& \lesssim \left|\frac{x_\tau^*}{\mu} \nu \right|\left(\int_{|y| \leq \frac{\zeta_*}{2}} |\nabla \Psi|^2 (1 + |y|^6) \rho \right)^\frac{1}{2} \left(\int_{|y|\leq \frac{\zeta_*}{2\nu}}|q^\perp|^2 (1  +|y|^2) \rho \right)^\frac{1}{2} + \frac{e^{-2\kappa \tau}}{|\ln \nu|^2}\\
& \lesssim \frac{1}{|\ln \nu|^2}\left(\int\frac{|\nabla q^\perp|^2}{U} \rho dy + \int |q^\perp|^2(1 +|y|^2) \rho dy \right) + \frac{e^{-2\kappa \tau}}{|\ln \nu|^2} \leq \frac{1}{|\ln \nu|^2} \int \frac{|\nabla q^\perp|^2}{U}\rho dy + Ce^{-2\kappa \tau}.
\end{align*}

The collection of all the above estimates, using the control (<ref>) and the coercivity (<ref>), yields (<ref>). This concludes the proof of Lemma <ref>.

We are now going to establish the monotonicity formula to control the inner norm (<ref>). The basic idea is that inside the blowup zone $|y| \leq \frac{\zeta_*}{2\nu}$ for $\zeta_* \ll 1$, the dynamic of (<ref>) resembles $\pa_s q^\perp = \Ls_0 q^\perp$ which allows us to use the special structure of $\Ls_0$, namely that $\Ls_0 f = \nabla \cdot(U \nabla \Ms f)$. We recall from (<ref>), the equation satisfied by $q^\perp$,

\begin{equation}\label{eq:qperp}
\pa_s q^\perp = \Ls_0 q^\perp + \eta \Lambda q^\perp - \nabla \cdot \Gc(q^\perp) + \frac{\nu x^*_\tau}{\mu} \cdot \nabla (V + q^0) + N^\perp(q^\perp),
\end{equation}

where $\Gc(q^\perp)$ and $N^\perp(q^\perp)$ are defined as in (<ref>), and $V$ is the approximate solution in the blowup variables given by (<ref>), namely that

\begin{equation*}
V = U +  \Psi,
\end{equation*}

where we recall

$$\Psi =  \Psi_1 + \Psi_2, \quad \Psi_1 = \frac{a_1}{\nu^2}(\varphi_1(r) - \varphi_0(r)), \quad \Psi_2 = \sum_{n = 2}^N \frac{a_n}{\nu^2}\varphi_n(r),$$

and $\varphi_n$ is the radial eigenfunction of $\Ls_0$, namely that

$$\varphi_n(r) = - \frac{\pa_r \phi_n}{r}, \quad \pa_r \varphi_n = -\frac{\phi_n}{r},$$

with $\phi_n$'s are the eigenfunctions of $\As$, and we write for short

$$\eta = \frac{\nu_s}{\nu} - \beta \nu^2 = \Oc(\nu^2), \quad \Lambda f = \nabla \cdot(yf).$$

We define

$$q^\perp_* = \chi_{\frac{\zeta_*}{\nu}}q^\perp,$$

where $\chi_{\frac{\zeta_*}{\nu}}$ is defined as in (<ref>) and $q^\perp_*$ solves the equation

\begin{align}\label{eq:qbotstar}
\pa_s q^\perp_* &= \Ls_0 q^\perp_* + \chi_{\frac{\zeta_*}{\nu}} \Big[\eta \nabla \cdot(y q^\perp)  - \nabla \cdot \Gc(q) + \frac{\nu x^*_\tau}{\mu} \cdot \nabla (V + q^0) + N^\perp(q^\perp) \Big] - [\Ls_0, \chi_{\frac{\zeta_*}{\nu}}] q^\perp + \pa_s \chi_{\frac{\zeta_*}{\nu}} q^\perp,
\end{align}

Consider the decomposition

\begin{equation}\label{dec:qstarci}
q_*^\bot = c_1 \pa_1 U_1 + c_2 \pa_2 U_2 + \tilde{q}_*^\bot \quad \textup{with} \quad c_i = \frac{\int q_*^\bot \pa_i U dy}{\int |\pa_i U|^2  dy } = \Oc(\|\e^\perp\|_0),
\end{equation}

which produces the orthogonality condition

$$\int \tilde{q}_*^\perp \pa_1 U dy = \int \tilde{q}_*^\perp \pa_0 U dy = 0.$$

From Lemma <ref>, we have the coercivity

\begin{equation}\label{est:coerMqtil}
-\int \Ls_0 \tilde{q}_*^\perp \Ms \tilde{q}_*^\perp  dy = \int U|\nabla \Ms \tilde{q}_*^\perp|^2 dy \geq \delta_2 \int \frac{|\nabla \tilde{q}_*^\perp|}{U} dy,
\end{equation}

and the positivity since $\int \Ls_0 \tq_*^\perp dy = 0$,

\begin{equation}\label{eq:posMLqtil}
\int \Ls_0 \tq_*^\perp \Ms \Ls_0 \tq_*^\perp dy \geq 0.
\end{equation}

Recall that $\Ms \pa_1 U = \Ms \pa_2 U = 0$, hence, we have

\begin{equation}\label{id:MsLsqqtil}
\Ms q_*^\perp = \Ms \tilde{q}_*^\perp, \quad  \Ls_0 q_*^\perp = \Ls_0 \tilde{q}_*^\perp.
\end{equation}

Thanks to this coercivity, we are able to establish the monontonicity formula to control $\| q^\perp\|_\inn$.

\begin{align*}
\frac{d}{d\tau} \left[\int \tilde{q}^\perp_* \Ms \tilde{q}_*^\perp  dy - \int \Ls_0 \tilde q^\perp_* \Ms \tilde{q}_*^\perp dy  \right] & \leq  -\delta_2'\left[ \frac{1}{\zeta_*^2}\int \tilde{q}^\perp_* \Ms \tilde{q}_*^\perp  dy - \int \Ls_0 \tilde q^\perp_* \Ms \tilde{q}_*^\perp dy  \right] \nonumber\\
& \qquad \qquad + \frac{C}{\nu^2} \Big(\|\e^\perp\|^2_0 +  \|\e^\perp\|^2_{H^1(\zeta_* \leq |z| \leq \zeta^*)} +  e^{-2\kappa \tau} \Big).
\end{align*}

We multiply equation (<ref>) by $\Ms \tilde{q}_*^\perp$ and $-\Ms \Ls \tilde{q}_*^\perp$ and integrate over $\Rb^2$ and  use the relation (<ref>) and (<ref>) and (<ref>) to write the energy identity

\begin{align*}
&\frac{1}{2}\frac{d}{ds}\left[\int \tilde{q}^\perp_* \Ms \tilde{q}_*^\perp  dy  - \int \Ms \tilde{q}_*^\perp \Ls_0\tilde{q}_*^\perp dy \right] \\
&\leq  - \delta_2 \int U |\nabla \Ms \tilde{q}_*^\perp|^2 dy  + \int \left(\chi_{\frac{\zeta_*}{\nu}}\Big[\eta \nabla \cdot(y q^\perp)  - \nabla \cdot \Gc(q^\perp) + \frac{\nu x^*_\tau}{\mu} \cdot \nabla (V+q^0) + N^\perp(q^\perp) \Big] \right) \\
& \qquad  \times  \Ms\Big(\tilde{q}_*^\perp +\Ls_0 \tilde{q}_*^\perp\Big)dy  - \int \left( [\Ls_0, \chi_{\frac{\zeta_*}{\nu}}] q^\perp - \pa_s \chi_{\frac{\zeta_*}{\nu}} q^\perp  \right) \Ms\Big(\tilde{q}_*^\perp +\Ls_0 \tilde{q}_*^\perp\Big)dy.
\end{align*}

Since we only use the positivity of $\Ms$, all the terms involving $\Ms \Ls_0 \tq_*^\perp$ are treated by using commutator formulas with $\Ms$ in order to reduce the order of derivatives so that they can be controlled by $-\delta_2\int U |\nabla \Ms \tilde{q}_*^\perp|^2 dy$. In particular, we shall use the following formula: for any well localized function $f$ without radial component,

\begin{equation}\label{def:comMLam}
[\Ms, \Lambda] f =  \frac{y \cdot\nabla U}{U^2}f + 2 \Phi_f \quad \textup{with} \quad \Lambda f = \nabla \cdot (y f),
\end{equation}

where we used the identity $y\cdot \nabla \Phi_f = \Phi_{y \cdot \nabla f} + 2\Phi_f$, and

\begin{equation}\label{eq:comMfgdev}
[\Ms, \nabla] f = \frac{\nabla U}{U^2}f, \quad [\Ms, g] f = g\Phi_f - \Phi_{gf}.
\end{equation}

The scaling term: We write $\chi_{\frac{\zeta_*}{\nu}} \nabla \cdot (y q^\perp) = \nabla \cdot (y q^\perp_*) - y\cdot \nabla \chi_{\frac{\zeta_*}{\nu}} q^\perp$, and use (<ref>) and the relation $\Ms q_*^\perp = \Ms \tq_*^\perp$ and the structure $\Ls_0 f = \nabla \cdot(U \nabla \Ms f)$ and integration by parts, to estimate

\begin{align*}
&\left|\eta\int\nabla \cdot(y q_*^\perp)\Ms\Ls_0\tq_*^\perp  dy\right|  \lesssim  \nu^2 \left|\int \nabla \cdot(y \Ms \tq_*^\perp) \Ls_0\tq_*^\perp  dy \right| + \nu^2 \left|\int \Big(\frac{y \cdot\nabla U}{U^2}q_*^\perp + 2 \Phi_{q_*^\perp}\Big) \Ls_0 \tq_*^\perp dy  \right|\\
& \lesssim \nu^2\int U|\nabla \Ms \tq_*^\perp|^2 \left(1 + \frac{|\nabla \cdot (y U)|}{U} + \frac{|\nabla U|}{U} \right)dy + \nu^2 \int U \left|\nabla\Big(\frac{y \cdot\nabla U}{U^2}q_*^\perp + 2 \Phi_{q_*^\perp}\Big)  \right|^2 dy\\
& \lesssim \nu^2\int U|\nabla \Ms \tq_*^\perp|^2 dy + \nu^2 \int \left(\frac{|\nabla q_*^\perp|^2}{U} + \frac{|\nabla U|^2}{U} |q_*^\perp|^2 + U|\nabla \Phi_{q_*^\perp}|^2 \right)dy\\
& \lesssim \nu^2 \int U|\nabla \Ms \tq_*^\perp|^2 dy + \nu^2  \int \frac{|\nabla \tq_*^\perp|^2}{U}dy + \nu^2 (|c_1|^2 + |c_2|^2) \lesssim \nu^2 \int U|\nabla \Ms \tq_*^\perp|^2 dy + \nu^2 \|\e^\perp\|_0^2.
\end{align*}

where we used in the last line the decomposition (<ref>) and the coercivity (<ref>).

Since $y\cdot \nabla \chi_{\frac{\zeta_*}{\nu}} \lesssim \mathbf{1}_{\{ \frac{\zeta_*}{\nu} \leq |y| \leq \frac{2\zeta_*}{\nu} \}}$, we use the midrange bootstrap (<ref>) and integration by parts to estimate

\begin{align*}
\left|\eta \int y\cdot \nabla \chi_{\frac{\zeta_*}{\nu}} q^\perp \Ms \Ls_0 \tq_*^\perp dy  \right| &\lesssim \nu^2 \int U|\nabla \Ms \tq_*^\perp|^2 dy + \nu^2 \int |\nabla \Ms \big( y\cdot \nabla \chi_{\frac{\zeta_*}{\nu}} q^\perp \big) | ^2 \\
& \quad \lesssim \nu^2 \int U|\nabla \Ms \tq_*^\perp|^2 dy + \nu^2 \|\e^\perp\|_{H^1(\zeta_* \leq |z| \leq \zeta^*)}^2.
\end{align*}

We also estimate by integration by parts and Cauchy-Schwarz and $\big|y\cdot \nabla \chi_{\frac{\zeta_*}{\nu} }\big| \leq \mathbf{1}_{\{\frac{\zeta_*}{\nu} \leq |y| \leq \frac{2\zeta_*}{\nu}\}}$,

\begin{align*}
&\left|\eta\int\chi_{\frac{\zeta_*}{\nu}} \nabla \cdot(y q^\perp)\Ms\tq_*^\perp  dy\right| =  \left|\eta\int \Big(\nabla \cdot(y q_*^\perp) - y\cdot \nabla \chi_{\frac{\zeta_*}{\nu} }q^\perp \Big)\Ms\tq_*^\perp  dy\right| \\
&\quad \leq \frac{\delta_2}{100}\int U|\nabla \Ms \tq_*^\perp|^2dy + \nu^4\int \frac{|y q_*^\perp|^2}{U}dy +  \nu^2\int_{\frac{\zeta_8}{\nu} \leq |y| \leq \frac{2\zeta_*}{\nu}} \big|q^\perp\Ms\tq_*^\perp\big| dy\\
& \quad \lesssim \frac{\delta_2}{100}\int U|\nabla \Ms \tq_*^\perp|^2dy  + \| \e^\perp\|_0^2 +  \|\e^\perp\|_{H^1(\zeta_* \leq |z| \leq \zeta^*)}^2.
\end{align*}

Summing up these estimates yields

\begin{align*}
&\left|\eta\int\chi_{\frac{\zeta_*}{\nu}} \nabla \cdot(y q^\perp)\Ms \big( \tq_*^\perp + \Ls_0\tq_*^\perp  \big)dy\right|   \leq \frac{\delta_2}{50} \int U|\nabla \Ms \tq_*^\perp|^2 dy + \|\e^\perp\|_{H^1(\zeta_* \leq |z| \leq \zeta^*)}^2 + \|\e^\perp\|_0^2.
\end{align*}

The $\Gc(q^\perp)$ and $N^\perp$ terms: We write by integration by parts and Cauchy-Schwarz

\begin{align*}
&\left|\int \Big( \nabla \cdot \Gc(q^\perp) - N^\perp(q^\perp) \Big) \Ms \tq_*^\perp dy \right| \leq \frac{\delta_2}{100}\int U|\Ms \nabla \tq_*^\perp|^2 dy + C\int_{|y| \leq \frac{2\zeta_*}{\nu}} \frac{|\Gc(q^\perp)|^2 + |N^\perp(q^\perp)|^2}{U} dy\\
& \leq \left(\frac{\delta_2}{100} + \frac{1}{|\ln \nu|^2}\right)\int U|\Ms \nabla \tq_*^\perp|^2 dy + C e^{-2\kappa \tau},
\end{align*}

where we use (<ref>) and the definition (<ref>) of $\Gc(q^\perp)$ and $N^\perp(q^\perp)$ and (<ref>), (<ref>)  to estimate

\begin{align*}
&\int_{|y| \leq \frac{2\zeta_*}{\nu}} \frac{|\Gc(q^\perp)|^2 + |N^\perp(q^\perp)|^2}{U} dy \lesssim \int_{|y| \leq \frac{2\zeta_*}{\nu}} \left(\frac{|q^\perp|^2}{U} \big|\Phi_{\Psi + q^0}\big|^2 + |\nabla \Phi_{q^\perp}|^2 \frac{|\Psi + q^0|^2}{U}  + |q^\perp \nabla q^\perp|^2\right) dy\\
& \quad \lesssim \frac{1}{|\ln \nu|^2} \int_{|y| \leq \frac{2\zeta_*}{\nu}} |q^\perp|^2 (1 + |y|^2) dy + e^{-2\kappa \tau}  \lesssim  \frac{1}{|\ln \nu|^2}\int \frac{|\nabla \tq_*^\perp|^2}{U}dy + \frac{1}{|\ln \nu|^2}\|\e^\perp\|_{H^1(\zeta_* \leq |z| \leq \zeta^*)} + e^{-2\kappa \tau}\\
& \quad  \lesssim \frac{1}{|\ln \nu|^2}\int U|\Ms \nabla \tq_*^\perp|^2 dy + C e^{-2\kappa \tau}.
\end{align*}

Similarly, we write by integration by parts and the definition of $\Gc$ and $N^\perp$ and Cauchy-Schwarz inequality,

\begin{align*}
&\left|\int \Ms \Big(\nabla \cdot \Gc(q^\perp) - N^\perp(q^\perp) \Big)\Ls_0 \tq_*^\perp dy\right| = \left|\int U \nabla \Big(\Ms \nabla \cdot \Gc(q^\perp) - \Ms N^\perp(q^\perp)\Big) \cdot \nabla  \Ms \tq_*^\perp dy  \right|\\
&\leq \frac{\delta_2}{100}\int U |\nabla \Ms \tq_*^\perp|^2 dy + \left|\int U \nabla \Ms \Big(\nabla q^\perp \cdot \nabla \Phi_{\Psi + q^0}  - 2 q^\perp(\Psi + q^0)\Big) \cdot \nabla \Ms \tq_*^\perp dy\right|\\
&  \quad  + C\int_{|y| \leq \frac{2\zeta_*}{\nu}} U |\nabla \Ms \nabla(\Psi + q^0) \cdot \nabla \Phi_{q^\perp})|^2 + C \int_{|y| \leq \frac{2\zeta_*}{\nu}} U | \nabla \Ms \nabla \cdot(q^\perp \nabla \Phi_{q^\perp})|^2 dy.
\end{align*}

Using the commutator formula (<ref>) and (<ref>) and the relation (<ref>), we obtain the estimate

\begin{align*}
&\left|\int U \nabla \Ms \Big(\nabla q^\perp \cdot \nabla \Phi_{\Psi + q^0}  - 2 q^\perp(\Psi + q^0)\Big) \cdot \nabla \Ms \tq_*^\perp dy\right|\\
& \quad \leq  \int U |\nabla \Ms \tq_*^\perp|^2 \left( |\Psi + q_0| + \frac{1}{2}\left|\frac{\nabla \cdot (U \nabla \Phi_{\Psi + q^0})}{U} \right| \right)dy \\
& \qquad + \left|\int U\nabla \left( [\Ms, \Psi + q^0] q^\perp + [\Ms, \nabla \Phi_{\Psi + q^0}]\cdot \nabla q^\perp + [\Ms, \nabla]q^\perp \cdot \nabla \Phi_{\Psi + q^0} \right) \cdot \nabla \Ms \tq_*^\perp dy \right|\\
&  \leq \left( \frac{1}{|\ln \nu|^2} + \frac{\delta_2}{100}\right) \int U|\nabla \Ms \tq_*^\perp|^2\\
& \qquad  + C\int_{|y| \leq \frac{2\zeta_*}{\nu}} U \Big|\nabla\left( [\Ms, \Psi + q^0] q^\perp + [\Ms, \nabla \Phi_{\Psi + q^0}]\cdot \nabla q^\perp + [\Ms, \nabla]q^\perp \cdot \nabla \Phi_{\Psi + q^0} \right)  \Big|^2 dy \\
&\qquad \qquad \quad  \leq  \frac{\delta_2}{50}\int U|\nabla \Ms \tq_*^\perp|^2 + \frac{C}{|\ln \nu^2|} \left(\int|\nabla \tq_*^\perp|^2(1 + |y|^4)  + \|\e^\perp\|^2_{H^1 (\zeta_* \leq |z| \leq \zeta^*)}\right).
\end{align*}

From (<ref>), (<ref>) and (<ref>) and (<ref>) with $\alpha = 1/2$,  we estimate

\begin{align*}
&\int_{|y| \leq \frac{2\zeta_*}{\nu}} U |\nabla \Ms \nabla(\Psi + q^0) \cdot \nabla \Phi_{q^\perp})|^2 + \int_{|y| \leq \frac{2\zeta_*}{\nu}} U | \nabla \Ms \nabla \cdot(q^\perp \nabla \Phi_{q^\perp})|^2 dy\\
& \quad  \lesssim \frac{1}{|\ln \nu|^2} \int |\tq_*^\perp|^2(1 + |y|^2) dy + \frac{1}{|\ln \nu|^2} \|\e^\perp\|^2_{H^1 (\zeta_* \leq |z| \leq \zeta^*)} +   e^{-2\kappa \tau}.
\end{align*}

Summing up these estimates yields

\begin{align*}
&\left|\int \Big[\nabla \cdot \Gc(q^\perp) - N^\perp(q^\perp) \Big]\Ms \big( \tq_*^\perp + \Ls_0\tq_*^\perp  \big)dy\right|  \leq \frac{\delta_2}{50} \int U|\nabla \Ms \tq_*^\perp|^2 dy + \|\e^\perp\|_{H^1(\zeta_* \leq |z| \leq \zeta^*)}^2 + e^{-2\kappa \tau}.
\end{align*}

The error terms: From (<ref>) and (<ref>), we note

$$|\Psi + q^0| + |r\pa_r (\Psi + q^0)| \lesssim \frac{1}{|\ln \nu|^2} \frac{1}{1 + r^4}, \quad \textup{for} \quad  r \leq \frac{2\zeta_*}{\nu},$$

from which and the fact that $\Ms \nabla U = 0$ and (<ref>), we obtain the estimate

\begin{align*}
& \left| \frac{\nu x_\tau^*}{\mu} \cdot \int \nabla (V + q^0) \Ms \big( \tq_*^\perp + \Ls_0\tq_*^\perp  \big)dy\right| = \left| \frac{\nu x_\tau^*}{\mu} \cdot \int \Ms \nabla (\Psi + q^0) \big( \tq_*^\perp + \Ls_0\tq_*^\perp  \big)dy\right|\\
& \qquad \quad \lesssim \frac{1}{|\ln \nu|^2}\int \frac{|\nabla \tq_*^\perp|^2}{U}dy + \frac{C}{|\ln \nu|^2} \|\e^\perp\|^2_{H^1(\zeta_* \leq |z| \leq \zeta^*)} + Ce^{-2\kappa \tau}.
\end{align*}

We also have the estimate from (<ref>) and Cauchy-Schwarz,

\begin{align*}
&\left| \frac{\nu x_\tau^*}{\mu} \cdot \int \nabla q^\perp \Ms\tq_*^\perp dy\right| \leq \left| \frac{\nu x_\tau^*}{\mu} \right|\left(\int_{|y| \leq \frac{2\zeta_*}{\nu}} |q^\perp|^2 U dy\right)^\frac{1}{2}\left(\int U| \nabla \Ms \tq_*^\perp|^2 dy \right)^\frac{1}{2}\\
& \quad \leq \frac{\delta_2}{50} \int U| \nabla \Ms \tq_*^\perp|^2 dy + C\left| \frac{\nu x_\tau^*}{\mu} \right|^2 \frac{\|\e^\perp\|_0^2}{\nu^2} \leq \frac{\delta_2}{50} \int U| \nabla \Ms \tq_*^\perp|^2 dy + e^{-2\kappa\tau}.
\end{align*}

Similarly, we use the commutator formula (<ref>) to obtain the estimate

\begin{align*}
&\left| \frac{\nu x_\tau^*}{\mu} \cdot \int \nabla q^\perp \Ms\Ls_0\tq_*^\perp dy\right|\leq \frac{\delta_2}{50} \int U| \nabla \Ms \tq_*^\perp|^2 dy + e^{-2\kappa\tau}.
\end{align*}

The commutator terms: Since the support of $[\Ls_0, \chi_\frac{\zeta_*}{\nu}] q^\perp$ and $\pa_s \chi_\frac{\zeta_*}{\nu}$ is on $\frac{\zeta_*}{\nu} \leq |y| \leq \frac{2\zeta_*}{\nu}$, we use the midrange bootstrap (<ref>) and Cauchy-Schwarz to obtain the estimate

\begin{align*}
&\left|\int \left([\Ls_0, \chi_\frac{\zeta_*}{\nu}] q^\perp -  \pa_s \chi_\frac{\zeta_*}{\nu} q^\perp\right)\Ms \big( \tq_*^\perp + \Ls_0\tq_*^\perp  \big)dy\right|  \leq \frac{\delta_2}{50} \int U|\nabla \Ms \tq_*^\perp|^2 dy + \|\e^\perp\|_{H^1(\zeta_* \leq |z| \leq \zeta^*)}^2.
\end{align*}

A collection of all the estimates and using the fact that

$$-\int \frac{|\nabla \tq_*^\perp|^2}{U}dy \lesssim - \int |\tq_*^\perp|^2(1  + |y|^2) dy \lesssim -\frac{\nu^2}{\zeta_*^2}\int \tq_*^\perp \Ms \tq_*^\perp dy,$$

and $\frac{ds}{d\tau} = \frac{1}{\nu^2}$ yield the conclusion of Lemma <ref>.

§.§ Analysis in the exterior zone

In the exterior zone, we derive an energy estimate for the $\| \hat w \|_{\out}$ norm of the full higher order perturbation (<ref>). From (<ref>) and (<ref>), the radial and non radial parts $\hat w^0$ and $\hat w^\perp$ of $\hat w$ solve the following equations:

\begin{equation} \label{id:eqhatw}
\partial_\tau \hat w^0+\beta \nabla.(z\hat w^0)-\Delta \hat w^0=e.\nabla \hat w^0 +f\hat w^0+g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^0} +h^0+ N^0(\hat w^\perp),
\end{equation}

\begin{equation} \label{id:eqhatwperp}
\partial_\tau \hat w^\perp+\beta \nabla.(z\hat w^\perp)-\Delta \hat w^\perp=e.\nabla \hat w^\perp +f\hat w^\perp+g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp} +h^\perp+ N^\perp(\hat w^\perp),
\end{equation}

e=-\nabla\left(\Phi_{U_\nu}+\Phi_{\Psi_{\nu,1}}\right), \quad f=2(U_\nu+\Psi_{\nu,1}), \quad g=-\nabla (U_\nu+\Psi_{\nu,1}),

and from (<ref>), (<ref>), (<ref>) and (<ref>),

\begin{align*}
h^0&= \frac{1}{2}\Big(-\pa_\tau f + \Delta f - \nabla \cdot (f \nabla \Phi_f) - \beta \nabla \cdot(z f)\Big) \\
& = \frac{\nu_\tau}{\nu}\nabla .(zU_\nu)-\frac{\nu_\tau}{\nu}\nu\partial_\nu \Psi_{\nu,1}-\frac{\beta_\tau}{\beta}\beta \pa_\beta \Psi_{\nu,1}  +(\alpha_1 a_1-a_{1,\tau})(\varphi_1-\varphi_0) \\
& \quad +2\beta\left((\tilde \alpha_1-\tilde \alpha_0)a_1-a_1-4\nu^2\right)\varphi_0 +\beta\left(8\nu^2\varphi_0- \nabla .(zU_\nu)\right) -\nabla (\Psi_{\nu,1}).\nabla \Phi_{\Psi_{\nu,1}}+\Psi_{\nu,1}^2+g.\nabla \Phi_{\chi_{\frac{\zeta^*}{2}}\hat w^0},
\end{align*}

\begin{eqnarray*}
h^\perp &=&\frac{x^*_\tau}{\mu}.\nabla \left(U_\nu+\Psi_{\nu,1}\right)+g.\nabla \Phi_{\chi_{\frac{\zeta^*}{2}}\hat w^\perp}.
\end{eqnarray*}

From the pointwise estimates (<ref>), (<ref>), (<ref>) on $U$, $\varphi_0$ and $\varphi_1$, the Poisson formula for radial functions, we have the following estimates for $|z|\geq 1$ and $k=0,1,2$:

\begin{equation} \label{bd:estimationefgexterior}
|\partial_\zeta^k e(\zeta)|\lesssim \zeta^{-1+\delta-k}, \quad|\partial_\zeta^k f(\zeta)|\lesssim \nu^2|\log \nu|\zeta^{-2+\delta-k}, \quad |\partial_\zeta^k g(\zeta)|\lesssim \nu^2\zeta^{-3+\delta-k}.
\end{equation}

We recall that $|\nu_\tau /\nu|\lesssim |\log \nu |^{-1}$ and $|\beta_\tau/\beta|\lesssim |\log \nu|^{-3}$ from (<ref>), (<ref>) and (<ref>). Hence from (<ref>), (<ref>) and (<ref>) we infer that for the forcing term for $|z| \geq 1$ and $k=0,1,2$:

\begin{equation} \label{bd:estimationhexterior}
|\partial_\zeta^k h^0(\zeta)|\lesssim \frac{\nu^2}{|\log \nu|} \zeta^{-2+\delta-k}, \quad |\nabla^k h^\perp(\zeta)|\lesssim  e^{-\kappa \tau} \nu \zeta^{-2+\delta-k}.
\end{equation}

There exists a universal constant $C>0$ independent of the constants $N,\kappa,K,K',K'',\zeta^*$ such that for $\tau_0$ large enough:

\begin{equation} \label{bd:lyapunovext0}
\frac{d}{d\tau} \| \hat w^0 \|_{\out}^{2p} \leq 2p\left(-\frac{\beta}{4}+\frac{C}{\zeta^*}\right) \| \hat w^0\|_{\out}^{2p}+C\frac{K^{'2p}\nu^{4p}}{|\log \nu|^{2p}}+ C\| \hat w^0 \|_{\out}^{2p-1}\frac{K\nu^2}{|\log \nu|}.
\end{equation}

\begin{equation} \label{bd:lyapunovextperp}
\frac{d}{d\tau} \| \hat w^\perp \|_{\out}^{2p} \leq 2p\left(-\frac{\beta}{4}+\frac{C}{\zeta^*}\right) \| \hat w^\perp \|_{\out}^{2p}+CK^{'2p}e^{-2p\kappa \tau}+ C\| \hat w^\perp \|_{\out}^{2p-1}e^{-\kappa \tau}.
\end{equation}

Step 1 (Linear energy estimate). We claim that for any function $\bar w$, for any vector field $\bar e$ and potential $\bar f$ satisfying $|\bar f|\lesssim \zeta^{*-1}$ and $|\bar e|+\zeta|\nabla \bar e|\lesssim 1$ a constant $C>0$ exists such that

\begin{align}\label{id:linearenergyout}
& \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-1}\left(-\beta \nabla.(z\bar w)+\Delta \bar w+\bar e.\nabla \bar w +\bar f\bar w \right)\frac{dz}{\zeta^2}\\
\nonumber  & \quad \leq -(2p-1)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-2} |\nabla \bar w|^2 \frac{dz}{\zeta^2} \\
& \quad +\left(-\frac{\beta}{4}+\frac{C}{\zeta^*}\right)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p}\frac{dz}{\zeta^2}+C \| \bar w\|_{L^{2p}(\frac{\zeta^*}{4}\leq |z|\leq \zeta^*)}^{2p}.\nonumber
\end{align}

We now prove this estimate. Integrating by parts, as derivatives of $\chi_{\frac{\zeta^*}{4}}$ have support in $\{\frac{\zeta^*}{4}\leq |z|\leq \zeta^*\}$, we obtain

\begin{align*}
\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) &\zeta^{(2-\frac 14)2p}\bar w^{2p-1}(-\beta \nabla.(z\bar w))\frac{dz}{\zeta^2}  =-\frac{\beta}{4}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p}\frac{dz}{\zeta^2}-\beta \int \partial_\zeta \chi_{\frac{\zeta^*}{4}} \zeta^{(2-\frac 14)2p}\bar w^{2p}\frac{dz}{\zeta^2},
\end{align*}

\begin{align*}
&\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-1} \Delta \bar w\frac{dz}{\zeta^2} \\
&= -(2p-1)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-2} |\nabla \bar w|^2 \frac{dz}{\zeta^2}   +\frac{1}{2p} \int \bar w^{2p} \Delta \left(\left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p-2 } \right)dz\\
& \quad\leq -(2p-1)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-2} |\nabla \bar w|^2 \frac{dz}{\zeta^2} \\
& \qquad +\frac{1}{2p} \int \bar w^{2p} \left(\left(1-\chi_{\frac{\zeta^*}{4}} \right) \Delta \zeta^{(2-\frac 14)2p-2 }-2\nabla \chi_{\frac{\zeta^*}{4}}.\nabla \zeta^{(2-\frac 14)2p-2 }-\Delta \chi_{\frac{\zeta^*}{4}}\zeta^{(2-\frac 14)2p-2 } \right)dz\\
& \quad \leq  -(2p-1)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-2} |\nabla \bar w|^2 \frac{dz}{\zeta^2} +  \frac{C}{|\zeta^*|^2} \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p}\frac{dz}{\zeta^2}+\Oc\left(\| \bar w\|_{L^{2p}(\frac{\zeta^*}{4}\leq |z|\leq \zeta^*)}^{2p}\right).
\end{align*}

Using the bounds $|\bar f|\lesssim \zeta^{*-1}$ and $|\bar e|+\zeta|\nabla \bar e|\lesssim 1$ yield

\begin{align*}
&\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-1} e.\nabla \bar w\frac{dz}{\zeta^2} = -\frac{1}{2p}\int \bar w^{2p} \left(\left(1-\chi_{\frac{\zeta^*}{4}} \right) \nabla.\left(\zeta^{(2-\frac 14)2p-2} e\right)-\nabla \chi_{\frac{\zeta^*}{4}}.e \zeta^{(2-\frac 14)2p-2} \right)dz\\
& \qquad  \leq \frac{C}{\zeta^*}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p}\frac{dz}{\zeta^2}+\Oc\left(\| \bar w\|_{L^{2p}(\frac{\zeta^*}{4}\leq |z|\leq \zeta^*)}^{2p}\right),
\end{align*}

\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-1} f\bar w\frac{dz}{\zeta^2}\leq  \frac{C}{\zeta^*}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p} \frac{dz}{\zeta^2}.

Summing the four identities above proves the linear estimate (<ref>).

Step 2 (Preliminary estimates). From the bound (<ref>), we have the estimate

\begin{align}
&\left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} \left(|h^0|+\zeta|\nabla h^0|\right)^{2p}\frac{dz}{\zeta^2}\right)^{\frac{1}{2p}}   \lesssim \frac{\nu^2}{|\log \nu|}\left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(\delta-\frac 14)2p}\frac{dz}{\zeta^2}\right)^{\frac{1}{2p}}\lesssim \frac{\nu^2}{|\log \nu|},\label{bd:interexth0}
\end{align}

for $\delta $ small enough. Similarly, using the bound (<ref>), the estimate (<ref>), and the bootstrap bounds (<ref>), we obtain

\begin{align}
&\left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\left(|g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})(\hat w^0)}|+\zeta |\nabla (g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})(\hat w^0)})|\right)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{1}{2p}}  \leq C\| \hat w^0 \|_{\out} \nu^2 \leq \frac{\nu^2}{|\log \nu|}, \label{bd:interextpoisson0}
\end{align}

for $\tau_0$ large enough. Similarly, we estimate  from (<ref>) for the nonradial part,

\begin{equation} \label{bd:interexthperp}
\left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} \left(|h^\perp|+\zeta|\nabla h^\perp|\right)^{2p}\frac{dz}{\zeta^2}\right)^{\frac{1}{2p}}\lesssim \nu^2 \left| \frac{x^*_\tau}{\mu} \right|\lesssim e^{-\kappa \tau},
\end{equation}

and using the bound (<ref>), the estimate (<ref>), and the bootstrap bound (<ref>):

\begin{align}
&\left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\left(|g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})(\hat w^\perp)}|+\zeta |\nabla (g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})(\hat w^\perp)})|\right)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{1}{2p}} \lesssim \| \hat w^\perp \|_{\out} \nu^2\leq e^{-\kappa \tau},\label{bd:interextpoissonperp}
\end{align}

for $\tau_0$ large enough.

Step 3 (The energy estimate for $\hat w^0$). We claim that there holds the energy estimate:

\begin{align}\label{id:lyapunovout1}
\frac{d}{d\tau} \left(\frac{1}{2p}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^0)^{2p}\frac{dz}{\zeta^2}\right)  &\leq \left(-\frac{\beta}{4}+\frac{C}{\zeta^*}\right) \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^0)^{2p}\frac{dz}{\zeta^2}\\
&   +C\frac{K^{'2p}\nu^{4p}}{|\log \nu|^{2p}}+ C\left(\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^0)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{2p-1}{2p}} \frac{\nu^2}{|\log \nu|}.\nonumber
\end{align}

We compute from (<ref>),

\begin{align}\label{id:energyestimatehatwout}
&\frac{d}{d\tau} \left(\frac{1}{2p}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^0)^{2p}\frac{dz}{\zeta^2}\right)\\
& = \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^0)^{2p-1}  \left(-\beta \nabla.(z \hat w^0)+\Delta \hat w^0+e.\nabla \hat w^0 +f\hat w^0+g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{4}})\hat w^0} +h^0+ N^0(\hat w^\perp) \right)\frac{dz}{\zeta^2}. \nonumber
\end{align}

For the last term there holds from (<ref>) and (<ref>),

\left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\left|N^0(\hat w^\perp)\right|^{2p} \frac{dz}{\zeta^2}\right)^{\frac{1}{2p}}\lesssim \frac{e^{-\kappa\tau}}{\nu^C} \| \hat w^\perp \|_{\out}\leq \frac{\nu^2}{|\log \nu|},

for $\tau_0$ large enough. Hence, from Hölder, the above bound and the estimates (<ref>), (<ref>), we get

\begin{align*}
&\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^0)^{2p-1}\left(g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{4}})\hat w^0} +h^0+ N^0(\hat w^\perp) \right)\frac{dz}{\zeta^2}\\
& \quad \leq \left(\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^0)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{2p-1}{2p}}  \times \left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} \left|g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{4}})\hat w^0} +h^0+N^0(\hat w^\perp)\right|^{2p}\frac{dz}{\zeta^2}\right)^{\frac{1}{2p}} \\
&\quad\lesssim  \left(\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^0)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{2p-1}{2p}} \frac{\nu^2}{|\log \nu|},
\end{align*}

for $\tau_0$ large enough. We inject the above identity in (<ref>), and use the linear estimate (<ref>) with the bounds (<ref>) and (<ref>) for the boundary terms and (<ref>) for the potential and vector field, which yields (<ref>).

Step 4 (The energy estimate for $\nabla (\hat w^0)$). Let $\hat w^i=\zeta \partial_{z_i}\hat w$ for $i=1,2$. We claim the energy estimate for $i=1,2$:

\begin{align}\label{id:lyapunovout2}
&\frac{d}{d\tau} \left(\frac{1}{2p}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p}\frac{dz}{\zeta^2}\right)  \leq  \left(-\frac{\beta}{4}+\frac{C}{\zeta^*}\right) \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p}\frac{dz}{\zeta^2}+\frac{C}{\zeta^*}\| \hat w^0 \|_{\out}^{2p}\\
& \qquad \qquad   \qquad  \qquad   \qquad  \qquad   \qquad   \qquad  \quad +C\frac{K^{'2p}\nu^{4p}}{|\log \nu|^{2p}}+ C\left(\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{2p-1}{2p}} \frac{K\nu^2}{|\log \nu|},\nonumber
\end{align}

which we now prove. From (<ref>) and the commutator relations $[\zeta \partial_{z_i},z.\nabla]=0$ and $[\zeta \partial_{z_i},\Delta]=\zeta^{-2}\zeta\partial_{z_i}-2\zeta^{-2}z.\nabla (\zeta \partial_{z_i})$, we infer the evolution equation for $\hat w^i$:

\begin{align*}
& \partial_{\tau} \hat w^i = -\beta \nabla .(z\hat w^i)+\Delta \hat w^i +\left( e-\frac{2z}{\zeta^2}\right).\nabla \hat w^i +\left(f+\zeta^{-2}-\frac{z.e}{\zeta^2}\right) \hat w^i  + \Fc,
\end{align*}

\begin{align*}
\Fc = \zeta \partial_{z_i}e.\nabla \hat w^0 +\zeta\partial_{z_i}f\hat w^0+\zeta\partial_{z_i}(g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^0})+\zeta\partial_{z_i}h^0+\zeta \partial_{z_i}N^0(\hat w^\perp),
\end{align*}

giving the energy estimate with $e'= e- 2\zeta^{-2}z$ and $f'=f+\zeta^{-2}-\zeta^{-2} z.e$:

\begin{align} \label{id:energyestimatehatwiout}
&\frac{d}{d\tau} \left(\frac{1}{2p}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p}\frac{dz}{\zeta^2}\right)\\
&= \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p-1}
\Big(-\beta \nabla.(z\hat w^i)+\Delta \hat w^i+e'.\nabla \hat w^i +f'\hat w^i+\Fc \Big)\frac{dz}{\zeta^2}.\nonumber
\end{align}

We apply for the linear part the estimate (<ref>) with the bounds (<ref>) and (<ref>) for the boundary terms and (<ref>)  for the vector field $e'$ and the potential $f'$:

\begin{align*}
&\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p-1} \left(-\beta \nabla.(z\hat w^i)+\Delta \hat w^i+e'.\nabla \hat w^i +f'\hat w^i\right)\frac{dz}{\zeta^2}\\
& \qquad \leq  -(2p-1)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-2} |\nabla \bar w|^2 \frac{dz}{\zeta^2} \\
& \qquad \quad + \left(-\frac{\beta}{4}+\frac{C}{\zeta^*}\right)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p}\frac{dz}{\zeta^2}+C\frac{K^{'2p}\nu^{4p}}{|\log \nu|^{2p}}.
\end{align*}

We compute now the remaining terms. From Hölder, the bounds $|\zeta \nabla e|\lesssim 1$ and $|\nabla f|\lesssim \zeta^{-2}$ from (<ref>), for any function $\bar w$:

\begin{equation} \label{bd:estimationcommutatorinterext}
\left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} \left(| \partial_{z_i}e||\zeta \nabla \bar w|+| \zeta \bar w \partial_{z_i}f|\right)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{1}{2p}}\lesssim \frac{1}{\zeta^*}\| \bar w \|_{\out}.
\end{equation}

Hence, using the above bound for $\bar w=\hat w^0$ and Hölder:

\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p-1} \left( \zeta \partial_{z_i}e.\nabla \hat w^0+\zeta\partial_{z_i}f\hat w^0 \right)\frac{dz}{\zeta^2}\leq \frac{C}{\zeta^*} \| \hat w^0 \|_{\out}.

Again, from Hölder, the bounds (<ref>) and (<ref>)

\begin{align*}
& \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p-1} \left(\zeta\partial_{z_i}(g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{4}})\hat w^0})+\zeta\partial_{z_i}h^0 \right)\frac{dz}{\zeta^2}\\
& \leq  \left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{2p-1}{2p}} \left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\left| \zeta\partial_{z_i}(g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^0})+ |\zeta\partial_{z_i}h^0| \right|^{2p} \frac{dz}{\zeta^2}\right)^{\frac{1}{2p}} \\
&\leq   C \| \hat w^0\|_{\out}^{2p-1} \frac{\nu^2}{|\log \nu|} ,
\end{align*}

for $\nu$ small enough, where we used (<ref>) and (<ref>). For the last term we integrate by parts, use Hölder and the bootstrap bounds in Definition <ref>:

\begin{align*}
& \left|\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p-1}\zeta \partial_{z_i}N^0(\hat w^\perp)\frac{dz}{\zeta^2}\right|\\
& \quad \leq C \frac{e^{-\kappa \tau}}{\nu} \left(\left(\frac{K'\nu^2}{|\log \nu|}\right)^{2p}+\| \hat w^0\|_{\out}^{2p-1}\| \hat w^\perp\|_{\out}^{2p-1}\right)\\
&\quad  +C \frac{e^{-\kappa \tau}}{\nu} \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} |\nabla \hat w^i|^2(\hat w^i)^{2p-2}\frac{dz}{\zeta^2}+C \frac{e^{-\kappa \tau}}{\nu} \| \hat w^0\|_{\out}^{2p-1}\| \hat w^\perp\|_{\out}^{2p-1}\\
& \quad \leq  C \frac{e^{-\kappa \tau}}{\nu} \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} |\nabla \hat w^i|^2(\hat w^i)^{2p-2}\frac{dz}{\zeta^2}+C\left(\frac{K'\nu^2}{|\log \nu|}\right)^{2p},
\end{align*}

where we used Young inequality on the last line and took $\tau_0$ large enough. The collection of above inequalities yields (<ref>).

Step 5 (End of the proof for $\hat w^0$). We sum the identities (<ref>) and (<ref>) for $i=1,2$, concluding the proof of (<ref>).

Step 6 (The energy estimate for $\hat w^\perp$). This step is very similar to Step 3 so we shall give less details. We claim that there holds the energy estimate

\begin{align}\label{id:lyapunovout1perp}
&\frac{d}{d\tau} \left(\frac{1}{2p}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^\perp)^{2p}\frac{dz}{\zeta^2}\right)\\
\nonumber & \leq  \left(-\frac{\beta}{4}+\frac{C}{\zeta^*}\right) \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^\perp)^{2p}\frac{dz}{\zeta^2} +C\frac{K^{'2p}\nu^{4p}}{|\log \nu|^{2p}}+ C\left(\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^0)^{\perp} \frac{dz}{\zeta^2}\right)^{\frac{2p-1}{2p}} \frac{\nu^2}{|\log \nu|}.
\end{align}

We compute from (<ref>),

\begin{align} \label{id:energyestimatehatwoutperp}
&\frac{d}{d\tau} \left(\frac{1}{2p}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^\perp)^{2p}\frac{dz}{\zeta^2}\right)\\
\nonumber & =\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^\perp)^{2p-1}  \left(-\beta \nabla.(z \hat w^\perp)+\Delta \hat w^\perp+e.\nabla \hat w^\perp +f\hat w^\perp+g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{4}})\hat w^\perp} +h^\perp+ N^\perp(\hat w^\perp) \right)\frac{dz}{\zeta^2}.\nonumber
\end{align}

For the last term there holds from (<ref>), (<ref>) and (<ref>),

\left( \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\left| N^\perp(\hat w^\perp) \right|^{2p} \frac{dz}{\zeta^2}\right)^{\frac{1}{2p}}\lesssim \frac{e^{-\kappa \tau}}{\nu^C} \| \hat w^\perp \|_{\out}  \leq \frac{1}{\zeta^*}\| \hat w^\perp \|_{\out},

for $\tau_0$ large enough. Therefore from Hölder, the above bound and the bounds (<ref>), (<ref>):

\begin{align*}
&\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^\perp)^{2p-1}\left(g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{4}})\hat w^\perp} +h^\perp+ N^\perp(\hat w^\perp) \right)\frac{dz}{\zeta^2}\\
& \qquad \lesssim \left(\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^\perp)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{2p-1}{2p}} e^{-\kappa \tau}.
\end{align*}

We inject the above identity into (<ref>) and use the linear estimate (<ref>) with the bounds (<ref>) for the boundary terms and (<ref>) for the potential and vector field to obtain (<ref>).

Step 7 (The energy estimate for $\nabla (\hat w^\perp)$). This step is very similar to Step 4. Let $\hat w^i=\zeta \partial_{z_i}\hat w^\perp$ for $i=1,2$. We claim the energy estimate for $i=1,2$:

\begin{align}\label{id:lyapunovout2perp}
\frac{d}{d\tau} \left(\frac{1}{2p}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p}\frac{dz}{\zeta^2}\right) &\leq \left(-\frac{\beta}{4}+\frac{C}{\zeta^*}\right) \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p}\frac{dz}{\zeta^2}+\frac{C}{\zeta^*}\| \hat w^0 \|_{\out}^{2p}\\
& +C\frac{K^{'2p}\nu^{4p}}{|\log \nu|^{2p}}+ C\left(\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p} \frac{dz}{\zeta^2}\right)^{\frac{2p-1}{2p}} \frac{K\nu^2}{|\log \nu|}.\nonumber
\end{align}

The evolution equation for $\hat w^i$ is

\begin{align*}
\partial_{\tau} \hat w^i & = -\beta \nabla .(z\hat w^i)+ \Delta \hat w^i + \left( e-\frac{2z}{\zeta^2}\right).\nabla \hat w^i +\left(f+\zeta^{-2}-\frac{z.e}{\zeta^2}\right) \hat w^i  + H,
\end{align*}

\begin{align*}
H = \zeta \partial_{z_i}e.\nabla \hat w^0 +\zeta\partial_{z_i}f\hat w^0+\zeta\partial_{z_i}(g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp})+\zeta\partial_{z_i}\big[h^\perp + N^\perp(\hat w^\perp)\big].
\end{align*}

Giving the energy estimate with $e'= e- 2\zeta^{-2}z$ and $f'=f+\zeta^{-2}-\zeta^{-2} z.e$, we write

\begin{align*}
&\frac{d}{d\tau} \left(\frac{1}{2p}\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p}\frac{dz}{\zeta^2}\right)\\
&  = \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p-1}\times \left(-\beta \nabla.(z\hat w^i)+\Delta \hat w^i+e'.\nabla \hat w^i +f'\hat w^i+H \right)\frac{dz}{\zeta^2}.
\end{align*}

We apply for the linear part the estimate (<ref>) with the bounds (<ref>) and (<ref>) for the boundary terms and (<ref>)  for the vector field $e'$ and the potential $f'$, and the bound (<ref>) with $\bar w =w^\perp$:

\begin{align*}
&\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p-1} \left(-\beta \nabla.(z\hat w^i)+\Delta \hat w^i+e'.\nabla \hat w^i +f'\hat w^i+ \zeta \partial_{z_i}e.\nabla \hat w^\perp+\zeta\partial_{z_i}f\hat w^\perp \right)\frac{dz}{\zeta^2}\\
& \qquad \leq  -(2p-1)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}\bar w^{2p-2} |\nabla w^\perp|^2 \frac{dz}{\zeta^2} \\
& \qquad \quad + \left(-\frac{\beta}{4}+\frac{C}{\zeta^*}\right)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p}\frac{dz}{\zeta^2}+CK^{'2p}e^{-\kappa\tau}.
\end{align*}

From Hölder and the bounds (<ref>) and (<ref>):

\begin{align*}
\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p-1} &\left(\zeta\partial_{z_i}(g.\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{4}})\hat w^\perp})+\zeta\partial_{z_i}h^\perp \right)\frac{dz}{\zeta^2}  \leq C \| \hat w^\perp \|_{\out}^{2p-1} e^{-\kappa \tau}.
\end{align*}

We integrate by parts, use Hölder and the bootstrap bounds (<ref>), (<ref>), (<ref>), (<ref>),

\begin{align*}
& \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p}(\hat w^i)^{2p-1}\zeta \partial_{z_i}N^\perp(\hat w^\perp) \frac{dz}{\zeta^2}\\
& \quad = -\int (\hat w^i)^{2p-1} N^\perp(\hat w^\perp) \left(\partial_{z_i} \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p-1}+\left(1-\chi_{\frac{\zeta^*}{4}} \right) \partial_{z_i} \zeta^{(2-\frac 14)2p-1}\right)dz\\
& \qquad -(2p-1)\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} \partial_{z_i}\hat w^i(\hat w^i)^{2p-2}\zeta N^\perp(\hat w^\perp) \frac{dz}{\zeta^2}\\
%& \quad \leq C \left(\left|\frac{x_\tau}{\mu}\right| + e^{-\kappa\tau} \right) \left(\| \nabla \hat w \|_{L^{2p}(\frac{\zeta^*}{4}\leq |z|\leq \frac{\zeta^*}{2})}\| \nabla \hat w^\perp \|_{L^{2p}(\frac{\zeta^*}{4}\leq |z|\leq \frac{\zeta^*}{2})}^{2p-1}+\| \hat w \|_{\out}\| \hat w^\perp \|_{\out}^{2p-1}  \right)\\
%& \qquad +C\left(\left|\frac{x_\tau}{\mu}\right| + e^{-\kappa \tau} \right) \left(\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} |\nabla \hat w^i|^2(\hat w^i)^{2p-2}\frac{dz}{\zeta^2}\right)^{\frac 12}\left(\int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} (\hat w^i)^{2p-2} |\zeta \nabla \hat w|^2\frac{dz}{\zeta^2}\right)^{\frac 12}\\
& \quad \leq C \frac{e^{-\kappa \tau}}{\nu}\left( \frac{K'\nu^2}{|\log \nu|} \left(K'e^{-\kappa \tau}\right)^{2p}+\frac{K''\nu^2}{|\log \nu|} \| \hat w \|_{\out}^{2p-1}\right)\\
& \qquad  + \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} |\nabla \hat w^i|^2(\hat w^i)^{2p-2}\frac{dz}{\zeta^2}+C \frac{e^{-\kappa \tau}}{\nu}\| \hat w \|_{\out}^{2}\| \hat w^\perp\|_{\out}^{2p-2}\\
& \quad \leq   \int \left(1-\chi_{\frac{\zeta^*}{4}} \right) \zeta^{(2-\frac 14)2p} |\nabla \hat w^i|^2(\hat w^i)^{2p-2}\frac{dz}{\zeta^2}+\left(K'e^{-\kappa \tau}\right)^{2p},
\end{align*}

for any where we used Young inequality on the last line and took $\tau_0$ large enough. The collection of above inequalities yields (<ref>). We sum the identities (<ref>) and (<ref>) for $i=1,2$, concluding the proof of (<ref>).

§.§ Proof of Proposition <ref>

We give the proof of Proposition (<ref>) from which directly implies the conclusion of Theorem <ref>. Assume that the solution is initially trapped in the sense of Definition <ref>. We then define

\tau^*=\sup \left\{ \tau_1\geq \tau_0, \mbox{ such that the solution is trapped on } [\tau_0,\tau_1] \right\}.

We assume by contradiction that $\tau_1<\infty$ and will show that this is impossible by integrating in time the various modulation equations and Lyapunov functionals. Throughout the proof, $C$ denotes a universal constant that is independent of the bootstrap constants $\kappa,N,K,K',K''$ and the dependence on $K,K',K''$ in the various $\Oc$'s is explicitly mentioned. Recall Remark <ref> regarding the order in which the constants are chosen.

Step 1: Improved modulation for $\nu$ and $\beta$. We inject the identity (<ref>) and the bootstrap bound (<ref>) in (<ref>), then use the eigenvalue expansion (<ref>) and the compatibility condition (<ref>) to get

\begin{align}\label{bd:reintegrationinternu}
\frac{\nu_\tau}{\nu}&=\beta-\beta \frac{a_1}{4\nu^2}(\tilde \alpha_1-1-\tilde \alpha_0)+\Oc\left(\frac{K\mathcal D(\tau)}{|\ln \nu|^2}\right)+\Oc\left(\frac{1}{|\ln \nu|^3} \right)\\
\nonumber &=\beta-\beta \left(\left(-1+\frac{1}{2\ln \nu}+\frac{\ln 2-\gamma-1-\ln \beta}{4|\ln \nu|^2}\right)\left(-1-\frac{1}{4|\ln \nu|^2}\right)\right)+\Oc\left(\frac{K\mathcal D(\tau)}{|\ln \nu|^2}\right)+\Oc\left(\frac{1}{|\ln \nu|^3} \right) \nonumber \\
&=\beta \left(\frac{1}{2\ln \nu}+\frac{\ln 2 -\gamma-2-\ln \beta}{4|\ln \nu|^2}\right)+\Oc\left(\frac{K\mathcal D(\tau)}{|\ln \nu|^2} \right)+\Oc\left(\frac{1}{|\ln \nu|^3}\right).\nonumber
\end{align}

We then inject this identity, the bootstrap bound (<ref>) and the eigenvalue expansion (<ref>) into (<ref>) to obtain

\begin{eqnarray*}
a_{1,\tau}&=&2\beta a_1\tilde \alpha_1-a_1\frac{\beta}{2|\ln \nu|^2}+\frac{a_1\beta_\tau}{\beta}+\Oc\left(\frac{1}{|\ln \nu|^3}\right)+\Oc\left(\frac{K\mathcal D(\tau)}{|\ln \nu|^2}\right)\\
&=& \beta a_1 \left(\frac{1}{\ln \nu}+\frac{\ln 2 -\gamma-2-\ln \beta}{2|\ln \nu|^2}\right)+a_1\frac{\beta_\tau}{\beta}+\Oc\left(\frac{1}{|\ln \nu|^3}\right)+\Oc\left(\frac{K\mathcal D(\tau)}{|\ln \nu|^2}\right).
\end{eqnarray*}

We differentiate the compatibility condition $-1+\frac{1}{2\ln \nu}+\frac{(\ln 2-\gamma-1-\ln \beta)}{4|\ln \nu|^2}=\frac{a_1}{4\nu^2}$ and inject the two identities above to arrive at

\begin{align*}
&\left(\frac{\beta_\tau}{\beta}+\Oc\left(\frac{1}{|\ln \nu|^3}+\frac{K\mathcal D(\tau)}{|\ln \nu|^2}\right)\right)\frac{a_1}{4\nu^2}= \left(\frac{a_{1,\tau}}{a_1}-2\frac{\nu_\tau}{\nu} \right)\frac{a_1}{4\nu^2}\\
&=\frac{\nu_\tau}{\nu}\left(-\frac{1}{2|\ln \nu|^2}-\frac{\ln 2 -\gamma-1-\ln \beta}{2|\ln \nu|^3}\right)=\Oc\left(\frac{1}{|\ln \nu|^3}+\frac{K\mathcal D(\tau)}{|\ln \nu|^4} \right).
\end{align*}

Since $\frac{a_1}{4\nu^2}=-1+\Oc(|\ln \nu|^{-1})$ and $\beta \sim 1/2$ in the bootstrap, we obtain $\beta_\tau=O\left(\frac{1}{|\ln \nu|^3}+\frac{K\mathcal D(\tau)}{|\ln \nu|^2}\right)$, from which and (<ref>), we arrive at the system

\begin{equation} \label{bd:betatau}
\left\{ \begin{array}{l l} \frac{\nu_\tau}{\nu}=\beta \left(\frac{1}{2\ln \nu}+\frac{\ln 2 -\gamma-2-\ln \beta}{4|\ln \nu|^2}\right)+\Oc\left(\frac{K}{|\ln \nu|^3}\right),\\ \beta_\tau=\Oc\left(\frac{K}{|\ln \nu|^3}\right).\end{array}\right.
\end{equation}

Step 2: Reintegrating the modulation equations. We introduce the renormalised time $\tilde \tau =2\beta_0\tau_0+2\int_{\tau_0}^\tau \beta$ and write $\nu=\sqrt{2\beta^{-1}} e^{-\frac{2+\gamma}{2}}e^{-\sqrt{\tilde \tau/2}}\left(1+ \nu'\right)$. From (<ref>), we have $\tilde \tau=2\beta \tau+O(\tau^{-1/2})$ and

\frac{\nu_\tau}{\nu}=\frac{\nu'_\tau}{1+\nu'}-\frac{2\sqrt 2 \beta}{\sqrt{\tilde \tau}}+\Oc(|\beta_{\tau}|)=\frac{\nu'_\tau}{1+\nu'}-\frac{ \beta }{\sqrt{ 2 \tilde \tau}}+\Oc(\tau^{-3/2}).

From (<ref>) and (<ref>), one has $|\nu'|\lesssim (\ln |\ln \nu|)|\ln \nu|^{-1}$ and the linearisation provides

\begin{align*}
\frac{1}{2\ln \nu}+\frac{\ln 2 -\gamma-2-\ln \beta}{4|\ln \nu|^2}&=\frac{1}{2\left(\ln (\nu'+1)+\frac{\ln 2-\gamma-2-\ln \beta}{2}-\sqrt{\frac{\tilde \tau}{2}}\right)}  +\frac{\ln 2 -\gamma-2-\ln \beta}{2\tilde \tau +\Oc(\sqrt{\tau})}\\
&=-\frac{1}{\sqrt{2\tilde \tau}}-\frac{\nu'}{\tilde \tau}+\Oc(\tau^{-\frac 32}).
\end{align*}

Equation (<ref>) is then transformed into

\nu'_\tau =-\frac{\beta \nu'}{\tilde \tau}+\Oc(K\tau^{-\frac 32}) \quad \mbox{ so that } \quad \nu'_{\tilde \tau} =-\frac{ \nu'}{2\tilde \tau}+\Oc(K\tilde \tau^{-\frac 32})

which reintegrated in time by  using (<ref>) gives

\begin{equation} \label{bd:reintegrationnu}
|\nu'(\tau_1)|=|\nu'(\tilde \tau_1)|\leq \frac{\tilde \tau_0}{\tilde \tau_1}|\nu'(\tau_0)|+\frac{CK\ln \tilde \tau_1}{\sqrt{\tilde \tau_1}}\leq \frac{K'}{2}\frac{\ln |\ln \nu|}{|\ln \nu|}.
\end{equation}

As $\beta_\tau =\Oc(K|\ln \nu|^{-3})=\Oc(K\tau^{-3/2})$, we use (<ref>) to obtain

\begin{equation} \label{bd:reintegrationbeta}
\left|\beta (\tau_1)-\frac 12\right| =\left|\beta (\tau_0)-\frac 12 +\Oc(K\tau_0^{-1/2})\right|\leq CK\tau_0^{-\frac 12}\leq \frac{K'}{2|\ln \nu_0|}.
\end{equation}

Finally, we reintegrate the modulation equations (<ref>) for the other parameters $a_i$ for $i=2,...,n$. Using (<ref>) and the fact that $\beta \geq 1/4$ and $|\tilde \alpha_i|\leq 1/2$ yield

\begin{align*}
\frac{d}{d\tau}(a_i^2)&=4\beta (1 - n + \tilde{\alpha}_n)a_i^2+2a_i \left( \Oc\left( \frac{\Dc(\tau)}{|\ln \nu|} \|m_\e\|_{L^2_{\frac{\omega_\nu}{\zeta}}} \right) + \Oc\left(\frac{\nu^2}{|\ln \nu|^2}\right) \right) \leq -\frac{a_i^2}{2}+K\tau^{-2}e^{-2\sqrt{\beta \tau}}.
\end{align*}

Reintegrating in time, using (<ref>) and (<ref>) yields that for some universal constant $C>0$ the estimate

\begin{align}\label{bd:intermodaiigeq2}
a_i^2(\tau_1)& \leq e^{-\frac{\tau_1-\tau_0}{2}}a_i^2(\tau_0)+e^{-\frac{\tau_1}{2}}\int_{\tau_0}^{\tau_1}e^{\frac{\tau}{2}}K\tau^{-2}e^{-2\sqrt{\beta \tau}}d\tau \leq C \tau_1^{-2}e^{-2\sqrt{\beta \tau_1}}+CK\tau_1^{-2}e^{-2\sqrt{\beta \tau_1}} \\
&\leq CK\frac{\nu^4}{|\log \nu|^4}\leq \frac{K^2}{2}\frac{\nu^4}{|\log \nu|^4}.\nonumber
\end{align}

Step 3: The Lyapunov functionals. We inject the estimate $|\mathcal D(\tau)|\leq C |\ln \nu|^{-1}$ into (<ref>) and (<ref>) to obtain the estimate

|\mbox{Mod}_0|+|\mbox{Mod}_1|\leq C\frac{\nu^2}{|\ln \nu|^{2}}.

We inject this estimate and the bootstrap bound (<ref>) and the last estimate in (<ref>) into (<ref>) to get

\begin{align*}
\frac{1}{2}\frac{d}{d\tau} \big\|\bar m_\e \big\|^2_{L^2_{\frac{\bar \omega_\nu}{\zeta}}} &\leq -2\beta\left(N - C \right)\big\|\bar m_\e \big\|^2_{L^2_{\frac{\bar \omega_\nu}{\zeta}}}+K\frac{\nu^2}{|\log \nu|^3}+CK\frac{\nu^4}{|\log \nu|^2}+C\frac{\nu^4}{|\log \nu|^2} \\
& \quad \leq -\big\|\bar m_\e \big\|^2_{L^2_{\frac{\bar \omega_\nu}{\zeta}}}+CK\frac{e^{-4\sqrt{\beta \tau}}}{\tau}
\end{align*}

for $N$ large enough. We integrate in time this identity  and use the initial condition (<ref>),

\begin{align} \label{bd:reintegrationL2omega}
& \big\|\bar m_\e(\tau_1) \big\|^2_{L^2_{\frac{\bar \omega_\nu}{\zeta}}} \leq  e^{-2(\tau_1-\tau_0)}\big\|\bar m_\e(\tau_0) \big\|^2_{L^2_{\frac{\bar \omega_\nu}{\zeta}}}+CKe^{-2\tau_1}\int_{\tau_0}^{\tau_1} e^{2\tau}\frac{e^{-4\sqrt{\beta \tau}}}{\tau}d\tau  \leq CK\frac{e^{-4\sqrt{\beta \tau}}}{\tau}\leq \frac{K^2}{2}\frac{\nu^4}{|\log \nu|^2},
\end{align}

(where we used $|\beta_\tau|\lesssim K|\log \nu|^3$ to integrate by parts in time). Next, we directly have from (<ref>) the bound

\begin{equation} \label{bd:reintegrationmid}
\| m_\e (\tau_1) \|_{H^2(\zeta_* \leq \zeta \leq \zeta^*)}\leq \frac{K'}{2} \frac{\nu^2}{|\ln \nu|}.
\end{equation}

We now turn to the inner estimates. Let $M\gg 1$ be a large constant. Then for $\zeta_*$ small enough, we obtain from (<ref>) and the bootstrap bounds (<ref>):

\begin{align*}
&\frac{d}{d\tau} \left[- e^{M\tau}\tnu^2\int \tm_v^* \As_0 \tm_v^* \frac{\omega_0}{\tr} \right] +e^{M\tau} \|\As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}}   \leq  CK^{'2}e^{M\tau}\frac{\nu^4}{|\log \nu|^2}\leq CK^{'2}e^{M\tau}\frac{e^{-4\sqrt{\beta \tau}}}{\tau},
\end{align*}

from which and a reintegration in time  and the dissipation estimate from (<ref>), we derive

\begin{align*}
&-\tnu^2(\tau_1)\int \tm_v^*(\tau_1) \As_0 \tm_v^* (\tau_1)\frac{\omega_0}{\tr}+e^{-M\tau_1}\int_{\tau_0}^{\tau_1}e^{M\tau} \|\As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}}d\tau\\
& \quad \leq  -e^{-M(\tau_1-\tau_0)} \tnu^2(\tau_0)\int \tm_v^*(\tau_0) \As_0 \tm_v^* (\tau_0)\frac{\omega_0}{\tr}+e^{-M\tau_1}CK^{'2}\int_{\tau_0}^{\tau_1}e^{M\tau}\frac{e^{-4\sqrt{\beta \tau}}}{\tau}d\tau \\
&\quad \leq  \frac{CK^{'2}}{M} \frac{e^{-4\sqrt{\beta \tau}}}{\tau}\leq \frac{K^{''2}}{2M}\frac{\nu^4}{|\log \nu|^2}.
\end{align*}

We inject the bound (<ref>) into (<ref>) to obtain

\frac{d}{d\tau}\left(e^{M\tau}\| \As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}}^2\right) \leq C \left(Me^{M\tau}\| \As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}}^2  + \frac{K'\nu^4}{|\ln \nu|^2}\right).

Reintegrating in time and using the previous dissipation estimate for $\| \As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}}^2$ and (<ref>) yield

\begin{align}\label{bd:reintegrationint}
& \| \As_0 \tm_v^*(\tau_1)\|_{L^2_\frac{\omega_0}{\tr}}^2\leq e^{-M(\tau_1-\tau_0)}\| \As_0 \tm_v^*(\tau_0)\|_{L^2_\frac{\omega_0}{\tr}}^2 +CMe^{-M\tau_1}\int_{\tau_0}^{\tau_1}e^{M\tau}\| \As_0 \tm_v^*\|_{L^2_\frac{\omega_0}{\tr}}^2d\tau+CK^{'2}\int_{\tau_0}^{\tau_1} e^{M\tau}\frac{e^{-4\sqrt{\beta \tau}}}{\tau}d\tau \\
& \qquad \quad \leq  \frac{CK^{'2}e^{-4\sqrt{\beta \tau}}}{\tau}\leq \frac{K^{''2}}{2}\frac{\nu^4}{|\log \nu|^2}.\nonumber
\end{align}

We turn to the estimates for the nonradial part. Reintegrating in time (<ref>) directly gives the bound

\begin{equation} \label{bd:reintegrationL2omegaperp}
\| \e^\perp \|_{0}\leq Ce^{-\kappa \tau}\leq \frac{K}{2}e^{-\kappa \tau}.
\end{equation}

From the bootstrap bound (<ref>), we have

\begin{align*}
&\frac{d}{d\tau} \left[ \int \tilde{q}^\perp_* \Ms \tilde{q}_*^\perp  dy - \int \Ls_0 \tilde q^\perp_* \Ms \tilde{q}_*^\perp dy \right] \leq  -\delta_2'\left[ \frac{1}{\zeta_*^2}\int \tilde{q}^\perp_* \Ms \tilde{q}_*^\perp  dy - \int \Ls_0 \tilde q^\perp_* \Ms \tilde{q}_*^\perp dy  \right] + \frac{C(K^{2}+K^{'2})}{\nu^2}e^{-2\kappa \tau}.
\end{align*}

For $\kappa$ small enough depending on $\delta_2'$, we have from (<ref>),

\begin{align} \label{bd:reintegrationintperp}
&\left[ \int \tilde{q}^\perp_*(\tau_1) \Ms \tilde{q}_*^\perp (\tau_1)  dy - \int \Ls_0 \tilde q^\perp_*(\tau_1) \Ms \tilde{q}_*^\perp (\tau_1) dy \right] \leq C(K^2+K^{'2})\frac{e^{-2\kappa \tau}}{\nu^2}\leq \frac{K^{''2}}{2\nu^2}e^{-2\kappa \tau}.
\end{align}

Finally,(<ref>) and (<ref>) are directly reintegrated in time for $\zeta^*$ large enough and $\kappa$ small enough, using the initial bounds (<ref>) and (<ref>),

\begin{equation} \label{bd:reintegrationout}
\| \hat w^0\|_{\out}\leq \frac{K''\nu^2}{2|\ln \nu|}, \quad \| \hat w^\perp \|_{\out}\leq \frac{K''}{2}e^{-\kappa \tau}.
\end{equation}

Step 4: End of the proof of Proposition <ref>. In Step 1 and Step 2, all the bounds involved in the Definition <ref> have been improved by a factor $1/2$ at time $\tau_1$, from (<ref>), (<ref>), (<ref>), (<ref>), (<ref>), (<ref>), (<ref>), (<ref>), (<ref>). Hence, by a continuity argument, these bounds also hold true on some time interval $[\tau_1,\tau_1+\delta]$ for some small $\delta>0$. This contradicts the definition of $\tau_1$. Hence $\tau_1=\infty$ and the solution is trapped in the regime <ref> for $\tau \in [\tau_0, +\infty)$. Knowing the solution is global in time $\tau$, reintegrating (<ref>) yields

\beta(\tau)=\beta_\infty+\Oc(\tau^{-1/2}), \ \ \beta_\infty=\beta(\tau_0)+\int_{\tau_0}^\infty \beta_\tau d\tau.

Recall the renormalised time $\tilde \tau$ of Step 2, we obtain from the above identity that

\begin{equation} \label{reinteproxtautildetau}
\tau=\tilde \tau/2\beta_{\infty} +\Oc(\sqrt{\tilde \tau}),
\end{equation}

and since by definition $\mu_\tau/\mu=\beta$ that $\mu(\tilde \tau)=e^{-\tilde \tau/2}$. To go back to the original time variable, we integrate

\frac{d\tilde \tau}{dt}=\frac{d\tau}{dt}\frac{d\tilde \tau}{d\tau}=\frac{2}{\mu^2}\beta=2e^{\tilde \tau} \beta_\infty (1+\Oc(\tilde \tau^{-1/2})).

Solving the above equation, there exists $T>0$ such that:

\begin{equation} \label{reinte:exprs}
\tilde \tau=-\log \left(2\beta_\infty (T-t)\right)+\Oc\left(\frac{1}{|\log \left((T-t)\right)|^{\frac 12}}\right).
\end{equation}

Hence, we obtain the following expression for the parabolic scale $\mu$,

\begin{equation} \label{reinte:exprmu}
\mu= e^{-\frac{\tilde \tau}{2}} =\sqrt{2\beta_\infty}\sqrt{T-t}\left(1+\Oc\left(\frac{1}{|\log \left((T-t)\right)|^{\frac 12}}\right)\right).
\end{equation}

We get from $\beta=\beta_\infty+\Oc(\tau^{-1/2})=\beta_{\infty}+\Oc(|\ln(T-t)|^{-1/2})$ and (<ref>) that

\begin{align*}
\nu &=\sqrt{\frac{2}{\beta}}e^{-\frac{2+\gamma}{2}}e^{-\sqrt{\frac{\tilde \tau}{2}}}\left(1+\Oc\left(\frac{1}{|\ln \nu|^{\frac 12}}\right)\right)=\sqrt{\frac{2}{\beta_{\infty}}}e^{-\frac{2+\gamma}{2}}e^{-\sqrt{\frac{|\ln (T-t)|}{2}}}\left(1+\Oc\left(\frac{1}{|\ln (T-t)|^{\frac{1}{4}}}\right)\right)
\end{align*}

and hence, we get the desired blowup speed

\lambda =\mu \nu =2e^{-\frac{2+\gamma}{2}}\sqrt{T-t}e^{-\sqrt{\frac{|\ln (T-t)|}{2}}}\left(1+\Oc\left(\frac{1}{|\ln (T-t)|^{\frac{1}{4}}}\right)\right).

From (<ref>) (the right hand side being less than $1$) and (<ref>), we get the rough bound:

|x^*_t|=\frac{1}{\mu} \frac{|x^*_\tau |}{\mu}\lesssim \frac{1}{\sqrt{T-t}}.

This implies that $x^*(t)$ converges to some $X\in \mathbb R^2$ as $t\rightarrow T$. We now turn to the proof of the continuity of the blowup time and blowup point with respect to the initial datum. Fix $u_0\in \mathcal O$ with blowup time $T(u_0)$, blowup point $x^*_u$ with limit $X(u_0)$, and renormalised time $\tau_u$, and $\delta >0$. From (<ref>), (<ref>) and the above inequality, for any $\tau_1$ large enough, there exists $T(u_0)-\delta/2\leq T_1<T(u_0)$ such that $\tau_u(T_1)\geq \tau_1+1$ and $|x^*_u(T)-X(u_0)|\leq \delta/3$. By continuity we then obtain that for another solution $v$, $v$ and all $v$-related parameters converge to $u$ and $u$-related parameters on $[0,T_1]$ as $v_0\rightarrow u_0$ in $\mathcal E$. In particular, $\tau_v(T_1)\geq \tau_1$ and $|x^*_v(T_1)-X(u_0)|\leq \delta/2$. By (<ref>), (<ref>) and the above inequality, we get that for $\tau_1$ large enough, $|X(v)-x^*_v(T_1)|\leq \delta/2$ and $|T(v_0)-T_1|\leq \delta/2$. Hence by summing we obtain the desired continuity $|X(u_0)-X(v_0)|\leq \delta$ and $|T(u_0)-T(v_0)|\leq \delta$. This concludes the proof of Proposition <ref> as well as Theorem <ref>. $\square$

§ UNSTABLE BLOWUP DYNAMICS

In this section, we sketch the idea to establish the existence of unstable blowup solutions to system (<ref>) in the radial setting. The strategy of the proof of Theorem <ref> has to be modified the following way. Fix $\ell \in \mathbb{N}$ with $\ell \geq 2$ and $N \in \mathbb{N}$ with $N \gg 1$ and consider the approximate solution of the form

\begin{equation}\label{def:mW7}
m_W(\zeta, \tau) = Q_\nu(\zeta) + a_\ell(\tau) \big(\phi_{\ell, \nu}(\zeta) - \phi_{0, \nu}(\zeta)\big) + \sum_{n = 1, n\ne \ell}^{N} a_n(\tau)\phi_{n, \nu}(\zeta) = Q_\nu + P_\nu,
\end{equation}

where the approximate perturbation is $P_\nu = P_{\ell,\nu} + P_{+, \nu} + P_{N, \nu}$ with

$$ P_{\ell, \nu} = a_\ell(\tau) \big(\phi_{\ell, \nu}(\zeta) - \phi_{0, \nu}(\zeta)\big), \;\; P_{+,\nu} = \sum_{i = 1}^{\ell - 1} a_i(\tau)\phi_{i, \nu}(\zeta),\;\; P_{N,\nu} = \sum_{i = \ell + 1}^{N} a_i(\tau)\phi_{i, \nu}(\zeta)$$

and $\mathbf{a}(\tau) = (a_1, \cdots, a_N)(\tau)$ and $\nu(\tau)$ are unknown functions to be determined,
and $(\phi_{n, \nu})_{0\leq n\leq N}$ are the eigenfunctions  described in Proposition <ref>, i.e,

\begin{equation}\label{eq:eigeninETA}
\As^\zeta \phi_{n,\nu}(\zeta) = 2\beta\Big(1 - n + \frac{1}{2 \ln \nu} + \bar{\alpha}_n(\nu) \Big)\phi_{n, \nu}(\zeta)
\end{equation}

with $ |\bar{\alpha}(\nu)| \lesssim \frac{1}{|\ln \nu |^2}$ and $\phi_{n,\nu}$ is given by (<ref>). Here, the leading order term in the approximate perturbation is $P_{\ell, \nu}$ which drives the law of blowup law. The first higher order term $P_{+, \nu}$ contains $\ell - 1$ "unstable" directions for $\ell \geq 2$, that can be controlled by tuning the initial datum in a suitable way (see Definition <ref> below) via a classical topological argument. The second higher order term $P_{N,\nu}$ added to (<ref>) is to get a large constant in the spectral gap (<ref>) which is only used for the control of the solution and does not affect the leading dynamic of blowup

*The generated error and derivation of unstable blowup rates.  The following lemma is similar to Lemma <ref>, from which we can redesign the bootstrap regime <ref> adapted for this case.
Assume that $(\beta,\nu, \mathbf{a})$ are $\Cc^1$ maps $(\beta,\nu, \mathbf{a}): [\tau_0, \tau_1) \mapsto [1/2,2]\times (0,\nu^*] \times (0, a^*]^{\ell + 1}$, for  $0 < \nu^*, a^* \ll 1$ and $1 \ll \tau_0 <\tau_1 \leq +\infty$, with a priori bounds:

\begin{equation} \label{bd:aprioriinstable}
|\beta_\tau|\lesssim |\ln \nu|^{\frac 32}, \quad \left|\frac{\nu_\tau}{\nu} \right| \lesssim 1, \quad  |a_\ell| \lesssim \nu^2, \quad |a_n| \lesssim \frac{\nu^2}{|\ln \nu|} \;\; \textup{for}\;\; 1 \leq n \ne \ell \leq N.
\end{equation}

Then the error generated by (<ref>) to the flow (<ref>) can be decomposed as

\begin{align}\label{exp:mEtilee}
m_E &= -\pa_\tau m_W + \pa_\zeta^2 m_W - \frac{\pa_\zeta m_W}{\zeta} + \frac{\pa_\zeta(m_W^2)}{2\zeta} - \beta \zeta\pa_\zeta m_W = \sum_{j = 0}^\ell \textup{Mod}_j  \times  \phi_{j, \nu} + \tilde{m}_E + \frac{\pa_\zeta P_\nu^2}{2\zeta},
\end{align}

\begin{align*}
\textup{Mod}_0 &= \left(\frac{\nu_\tau}{\nu} - \beta \right)8\nu^2 +  a_{\ell, \tau} - 2\beta a_\ell \big(1 + \frac{1}{2\ln \nu} + \bar \alpha_0 \big),\\
\textup{Mod}_j &= - a_{j, \tau}  + 2 \beta a_j \big(1 - j + \frac{1}{2\ln \nu} + \bar \alpha_j \big) \quad \textup{for} \quad 1 \leq j \leq N,
\end{align*}

\begin{align} \label{bd:tildemeinstable2}
&\big \la \tm_E,\phi_{\ell,\nu} \big\ra_{L^2_\frac{\omega_\nu}{\zeta}} =  -a_\ell \| \phi_{\ell, \nu}\|^2_{L^2_\frac{\omega_\nu}{\zeta}} \left(\frac{\nu_\tau}{
\nu} \frac{1}{\ln \nu}+\frac{\beta_\tau}{\beta} \ell  \right)+O(|\ln \nu|^{\frac 12}),\\
\label{bd:tildemeinstable3} & \big \la \tm_E,\phi_{0,\nu} \big\ra_{L^2_\frac{\omega_\nu}{\zeta}} \sim  -a_\ell \frac{\nu_\tau}{8\nu}+O(|\ln \nu|^{-\frac 12}),
\end{align}

\begin{equation} \label{bd:tildemeinstable}
\sum_{n = 1, n \ne \ell}^{N}\Big|\big \la \tm_E,\phi_{n,\nu} \big\ra_{L^2_\frac{\omega_\nu}{\zeta}}\Big| \lesssim |\ln \nu | \nu^2, \quad  \|\tm_E\|_{L^2_\frac{\omega_\nu}{\zeta}} \lesssim \nu^2.
\end{equation}

Since the proof is exactly the same to that of Lemma <ref>, we only sketch it. From the a priori bounds, we focus on the leading order term of $\tilde{m}_E$ which is

\begin{equation} \label{id:instablemainterm}
\tm_E \sim -a_\ell \left( \frac{\nu_\tau}{\nu} \nu \pa_\nu (\phi_{\ell, \nu} - \phi_{0, \nu})+\frac{\beta_\tau}{\beta} \beta \pa_\beta (\phi_{\ell, \nu} - \phi_{0, \nu})\right).
\end{equation}

We introduce (the logarithmic cancellation as $r\to \infty$ is a consequence of (<ref>)):

\begin{equation} \label{id:asymptthetainstable}
\Theta_j = (2j - 2)T_j - r\pa_r T_j =_{r\to \infty} \left\{ \begin{array}{l l} -\hat d_j r^{2j - 2}+O(r^{2j-4}\ln r) \qquad \mbox{if }j\geq 1, \\ O(r^{-4}) \qquad \qquad \qquad \quad \mbox{if }j=0,\end{array} \right.
\end{equation}

with $\hat d_1 = -\frac{1}{2}$, $\hat d_{j} = -\frac{\hat d_{j-1}}{4j(j-1)}$. This and (<ref>) imply the identities:

\begin{align} \label{id:instableidentity1}
&\nu\pa_\nu (\phi_{\ell, \nu} - \phi_{0, \nu}) = \sum_{j=1}^\ell c_{\ell,j}\beta^j \nu^{2j-2}  \Theta_j(\zeta/\nu) +\nu\pa_\nu \tilde \phi_{\ell, \nu} -\nu\pa_\nu \tilde \phi_{0, \nu}, \\
\label{id:instableidentity2}&\beta\pa_\beta (\phi_{\ell, \nu} - \phi_{0, \nu}) = \sum_{j=1}^\ell j c_{\ell,j}\beta^j \nu^{2j-2}  T_j(\zeta/\nu) +\beta \pa_\beta \tilde \phi_{\ell, \nu} -\beta \pa_\beta \tilde \phi_{0, \nu}.
\end{align}

From Proposition 1 in [10] we have the following estimates for the error in (<ref>):

\begin{equation} \label{id:estimationtildephiinstable}
\| D^k\tilde \phi_{j,\nu}\|_{L^2_{\frac{\omega_\nu}{\zeta}}}+\| D^k\nu \pa_\nu \tilde \phi_{j,\nu}\|_{L^2_{\frac{\omega_\nu}{\zeta}}}+\| D^k\beta \pa_\beta \tilde \phi_{j,\nu}\|_{L^2_{\frac{\omega_\nu}{\zeta}}}\lesssim |\ln \nu|^{-\frac12}
\end{equation}

for $k=0,1,2$ and $0\leq j\leq N$. Thus, the a priori bounds (<ref>), the identities (<ref>) and (<ref>), the asymptotics (<ref>) and (<ref>) and the bounds (<ref>) prove the second bound in (<ref>). This bound in turns imply the first one in (<ref>) after applying Cauchy-Schhwarz and (<ref>). We have proved (<ref>) for the leading term (<ref>).

Next, we write $\la \cdot \ra=\la \cdot \ra_{L^2_{\frac{\omega_\nu}{\zeta}}}$. The identities (<ref>) and (<ref>), the decomposition (<ref>) and the bound (<ref>) imply that, where we change variables  $\zeta\mapsto r=\zeta/\nu$ for the main term:

\begin{align*}
\la \nu \pa_\nu (\phi_{\ell, \nu} - \phi_{0, \nu}),\phi_{0,\nu}\ra = \sum_{j=1}^\ell c_{\ell,j}\beta^j \nu^{2j} \int_0^{\infty} \Theta_j(r) T_0(r) \frac{e^{-\beta \nu^2 r^2/2}}{rU(r)} dr+O(|\ln \nu|^{-\frac 12}),\\
\la \beta \pa_\beta (\phi_{\ell, \nu} - \phi_{0, \nu}),\phi_{0,\nu}\ra =\sum_{j=1}^\ell j c_{\ell,j}\beta^j \nu^{2j} \int_0^{\infty} T_j(r) T_0(r) \frac{e^{-\beta \nu^2 r^2/2}}{rU(r)} dr +O(|\ln \nu|^{\frac 12}).
\end{align*}

We compute the numbers appearing above. By (<ref>) and (<ref>):

\begin{align*}
&\sum_{j=1}^\ell c_{\ell,j}\beta^j \nu^{2j} \int_0^{\infty} \Theta_j(r) T_0(r) \frac{e^{-\beta \nu^2 r^2/2}}{rU(r)} dr= -\sum_{j = 1}^{\ell}c_{\ell, j} \hat d_j 2^{j - 4} \int_{0}^{\infty} \xi^{j - 1}e^{-\xi} d\xi +O(\nu) \\
& =\frac 18 \sum_{j = 1}^\ell (-1)^{j + 1} \frac{\ell!}{(\ell - j)! j!}\ = \  \frac{1}{8}+O(\nu),
\end{align*}

using that $\hat d_j = (-1)^j \frac{2^{-2j + 1}}{j[(j -1)!]^2}$ and $c_{\ell, j} = 2^j \frac{\ell!}{(\ell - j)!}$ from (<ref>) and (<ref>). Hence

\begin{equation} \label{instableerror1}
\la \nu \pa_\nu (\phi_{\ell, \nu} - \phi_{0, \nu}),\phi_{0,\nu}\ra=\frac18+O(|\ln \nu|^{-\frac 12}).
\end{equation}

Similarly, using (<ref>) and the identity $\int_{0}^{\infty} \xi^{j - 1}e^{-\xi} d\xi=(j-1)!$:

\begin{align*}
& \sum_{j=1}^\ell jc_{\ell,j}\beta^j \nu^{2j} \int_0^{\infty} T_j(r) T_0(r) \frac{e^{-\beta \nu^2 r^2/2}}{rU(r)} dr = |\ln \nu |\sum_{j = 1}^{\ell} c_{\ell, j} \hat d_j 2^{j - 4} j!+O(1)\\
& = \frac 18 |\ln \nu| \sum_{j = 1}^\ell (-1)^{j } \frac{\ell!}{(\ell - j)! (j-1)!}+O(1) \ = \ O(1).
\end{align*}

\begin{equation} \label{instableerror2}
\la \beta \pa_\beta (\phi_{\ell, \nu} - \phi_{0, \nu}),\phi_{0,\nu}\ra =O(|\ln \nu|^{\frac 12}).
\end{equation}

Next, we recall that $\| \phi_{\ell,\nu}\|_{L^2_{\frac{\omega_\nu}{\zeta}}}=c_\ell \ln^2 \nu$ from (<ref>). Differentiating, we get:

2\la \nu \pa_\nu \phi_{\ell,\nu},\phi_{\ell,\nu}\ra=2c_{\ell}\ln \nu-\la  \phi_{\ell,\nu},\phi_{\ell,\nu}\frac{\nu \pa_\nu \omega_\nu}{\omega_\nu}\ra=2c_{\ell}\ln \nu+O(|\ln \nu|^{\frac 12})

where we used the cancellation $|\nu \pa_\nu \omega_\nu|\lesssim \omega_\nu \la \zeta/\nu\ra^{-2}$, (<ref>) and (<ref>). We recall the orthogonality $\la \phi_{\ell,\nu},\phi_{0,\nu} \ra=0$. Differentiating it, using the same cancellation, (<ref>) and (<ref>) we get:

2\la \nu \pa_\nu \phi_{0,\nu},\phi_{\ell,\nu}\ra=-\la  \phi_{0,\nu},\phi_{\ell,\nu}\frac{\nu \pa_\nu \omega_\nu}{\omega_\nu}\ra=O(|\ln \nu|^{\frac 12}).

Collecting the two estimates above we have proved:

\begin{equation} \label{instableerror3}
\la \nu \pa_\nu (\phi_{\ell, \nu} - \phi_{0, \nu}),\phi_{\ell,\nu}\ra=c_{\ell}\ln \nu+O(|\ln \nu|^{\frac 12}).
\end{equation}

Next, we write from (<ref>) and (<ref>), using that $\Ac_0 T_j=-T_{j-1}$:

j T_j =T_j+\frac 12 r \pa_r T_j+\frac 12 \Theta_j=T_j-(2\beta \nu^2)^{-1}\Ac T_{j}-(2\beta \nu^2)^{-1}T_{j-1}+\frac 12 \Theta_j .

Above, we recall that from (<ref>) and (<ref>), for $j<n$ one has $\nu^{2j-2}T_j(\zeta/\nu)=\sum_0^j \tilde c_{j,i}\phi_{i,\nu}+O_{L^2_{\omega_\nu/\zeta}}(|\ln \nu|^{1/2})$ for some constants $\tilde c_{j,i}$. Using this, the above identity, the identity (<ref>), the fact that $\Ac^\zeta \phi_{\ell,\nu}=\alpha_{\ell,\nu} \phi_{\ell,\nu}$ and (<ref>) we obtain:

\beta \pa_\beta \phi_{\ell,\nu}=\ell \phi_{\ell,\nu}+\sum_{j<n}\tilde c_j \phi_{j,\nu}+O_{L^2_{\omega_\nu/\zeta}}(|\ln \nu|^{1/2})

for some constants $\tilde c_j$. Therefore, from the orthogonality of the eigenfunctions:

\la  \beta \pa_\beta \phi_{\ell,\nu},\phi_{\ell,\nu}\ra=\ell \| \phi_{\ell,\nu}\|_{L^2_{\omega_\nu/\zeta}}^2+O(|\ln \nu|^{3/2}).

Since from (<ref>), $\pa_\beta \phi_{0,\nu}=\pa_\beta \tilde \phi_{0,\nu}$, from (<ref>), (<ref>) and Cauchy-Schwarz we get $\la  \beta \pa_\beta \phi_{0,\nu},\phi_{\ell,\nu}\ra=O(|\ln \nu|^{3/2})$. This and the above give

\begin{equation} \label{instableerror4}
\la  \beta \pa_\beta (\phi_{\ell,\nu}-\phi_{0,\nu}),\phi_{\ell,\nu}\ra=\ell \| \phi_{\ell,\nu}\|_{L^2_{\omega_\nu/\zeta}}^2+O(|\ln \nu|^{3/2}).
\end{equation}

Collecting (<ref>), (<ref>), (<ref>) and (<ref>), using the a priori bounds (<ref>), we have established (<ref>) and (<ref>) for the main order term (<ref>).

From Lemma <ref>, we project (<ref>) onto $\phi_{0, \nu}$ and $\phi_{\ell, \nu}$ to obtain the following system of ODEs

\begin{equation}\label{eq:companuaellsystem}
\left\{ \begin{array}{l l} &\left(\frac{\nu_\tau}{\nu} - \beta \right)8\nu^2 +  a_{\ell, \tau} - 2\beta a_\ell \big(1 + \frac{1}{2\ln \nu} \big) + a_\ell \frac{\nu_\tau}{\nu} \frac{1}{\ln \nu} = \Oc\left(\frac{\nu^2}{|\ln \nu|^{3/2}} \right),\\
& a_{\ell, \tau} - 2\beta a_\ell \big(1 - \ell + \frac{1}{2\ln \nu}\big) + a_\ell \frac{\nu_\tau}{\nu} \frac{1}{\ln \nu} +\ell a_\ell \frac{\beta_\tau}{\beta}= \Oc\left(\frac{\nu^2}{|\ln \nu|^{3/2}} \right).
\end{array} \right.
\end{equation}

We solve this system for $0<\nu\ll 1$ and $\beta\approx 1$ under the compatibility condition

\begin{equation}\label{eq:companuaell}
\frac{a_{\ell}}{4\nu^2} = -1+\frac{1}{2 \ln \nu},
\end{equation}

which is a constraint on $\beta$. Namely one obtains from (<ref>) that this condition is satisfied provided that $\beta_\tau =O(|\ln \nu|^{-3/2})$. Under (<ref>), (<ref>) gives:

\frac{\nu_\tau}{\nu} = \beta(1 - \ell)+\frac{\beta \ell}{2\ln \nu}+O(|\ln \nu|^{-\frac 32}).

Solving this yields that $\beta\to \beta_{\infty}$ and $\nu(\tau)\sim e^{\beta (1-\ell)\tau}\tau^{\frac{\ell}{2(1-\ell)}}\nu_{\infty}$ for some $\beta_{\infty},\nu_{\infty}>0$. Since $\mu_\tau=\beta \mu$, $\pa_t \tau=\mu^{-2}$ and $\lambda =\mu \nu$, we get that for some blow up time $T>0$:

\begin{equation}\label{eq:lawlambdaell}
\lambda(t) \sim C(u_0) (T-t)^\frac{\ell}{2} |\ln (T-t)|^{-\frac{\ell}{2(\ell - 1)}}.
\end{equation}

Note that $\int |\ln \nu|^{-\alpha}d\tau<\infty$ for all $\alpha>1$, which is not the case for the stable blow-up law (<ref>) where $\alpha>2$ is needed. This is of a simplification for the instable case: one can only perform the analysis with an accuracy of one order in $|\ln \nu|^{-1}$ less than for the stable case and still be able to close the estimates.

*Bootstrap regime. Lemma <ref> provides information about the size of the error and Remark <ref> formally gives us the law of $\nu$, from which we can redesign the bootstrap regime (<ref>) adapted to the case $\ell \geq 2$. In particular, we control the remainder $\e$ according to the following regime.

Let $\ell \in \mathbb{N}$ with $\ell \geq 2$ and $\tau_0\gg 1$. A solution $w$ is said to be trapped on $[\tau_0,\tau^*]$ if it satisfies the initial bootstrap conditions in the sense of Definition <ref> at time $\tau_0$ and the following conditions on $(\tau_0,\tau^*]$. There exists $\mu \in C^1([0,t^*],(0,\infty))$ and constants $K''\gg K'\gg K\gg 1$ such that the solution can be decomposed according to (<ref>), (<ref>) on $(\tau_0,\tau^*]$ with:

(i) (Compatibility condition for the renormalisation rate $\beta$)

$$\frac{a_{\ell}}{4\nu^2} = -1+\frac{1}{2 \ln \nu},$$

(ii) (Modulation parameters)

e^{\frac{(1 - \ell)\tau}{2}} \tau^{\frac{ \ell}{2(1 - \ell)}} \leq \nu(\tau) \leq 2 e^{\frac{(1 - \ell)\tau}{2}} \tau^{\frac{ \ell}{2(1 - \ell)}} \ \ \mbox{ and } \ \ \frac 12 < \beta <2,

|a_n|< \frac{\nu^2}{|\ln \nu|} \ \mbox{ for } 1 \leq n \ne \ell \leq N.

(iii) (Remainder)

\|m_\varepsilon(\tau) \|_{L^2_{\frac{\omega_\nu}{\zeta}}}<K \nu^2, \quad \| m_\e(\tau) \|_{H^2(\zeta_*\leq \zeta\leq \zeta^*)} < K' \nu^2,

$$\quad \| \tilde m_{v} \|_{\inn}  < K''\nu^2, \quad \| \hat w^{0} \|_{\out} < K''\nu^2.$$

The bootstrap definition <ref> is almost the same as the one defined in Definition <ref>, except for the bounds on $m_\e$ which are of size $\nu^2$ only, see remark <ref>. All the energy estimates as well as the derivation of the modulation equations given in Section <ref> can be adapted to the new definition <ref> without any difficulties to derive the conclusion of Theorem <ref>.

§ SOME USEFUL ESTIMATES

Hardy-Poincaré type inequality: Let us recall the following estimates in spaces with weights involving $\rho$, for functions $v$ and $u$ without radial components. The first one is a Poincaré-type inequality

\begin{equation} \label{bd:genpoincare}
\int_{\mathbb R^2} v^2 |z|^{2k}(1+|z|^{2})e^{-\frac{z^2}{2}}dz \lesssim \int_{\mathbb R^2} |\nabla v|^2 |z|^{2k} e^{-\frac{z^2}{2}}dz,
\end{equation}

for any $k\geq 0$. The second one is a Hardy type inequality: for $0<b<1$, there exists $C > 0$ independent of $b$ such that

\begin{equation} \label{bd:hardyL2rho}
\int_{\mathbb R^2} (1+|y|^2) u^2 e^{-\frac{b|y|^2}{2}}  \leq C \int_{\mathbb R^2} (1+|y|^4)|\nabla u|^2e^{-\frac{b|y|^2}{2}}.
\end{equation}

By the change of variables $z=\sqrt by$, the two above inequalities imply for any $1\leq \alpha \leq 3$:

\begin{equation}\label{est:HaPo1}
b^{\alpha - 1} \int_{\mathbb R^2} |q^\perp|^2  (1 + |y|^{2\alpha}) \rho dy \leq C_\alpha \int_{\mathbb R^2} \frac{|\nabla q^\perp|^2}{U}\rho dy.
\end{equation}

Estimates on the Poison field: For $u$ localized on a single spherical harmonics, the Laplace operator is  written as

\Delta u (x)=\Delta^{(k)} (u^{(k,i)})(r) \phi^{(k,i)}(\theta), \ \ \Delta^{(k)}:= \pa_{rr}+\frac{1}{r}\pa_r-\frac{k^2}{r^2}.

The fundamental solutions to $\Delta^{(k)}f=0$ are

\left\{ \ba{l l} \log(r) \ \ \text{and} \ \ 1\ \ \text{for} \ k=0, \\r^k \ \ \text{and} \ \ r^{-k} \ \text{for} \ k\geq 1. \ea \right.

and their Wronskian relation

W^{(0)}=\frac{d}{dr}\log (r)=r^{-1} \ \ \text{and} \ \ W^{(k)}=\frac{d}{dr}(r^k)r^{-k}-r^k \frac{d}{dr}(r^{-k})=2kr^{-1} \ \text{for} \ k\geq1.

The solution to the Laplace equation $-\Delta \Phi_u=u$ given by $\Phi_u=-(2\pi)^{-1}\log (|x|)*u$ is given on spherical harmonics by:

\Phi_u^{(0,0)}(r)=-\log (r) \int_0^r u^{(0,0)}(\tilde r)\tilde r d \tilde r -\int_r^{+\infty} u^{(0,0)}(\tilde r)\log(\tilde r)\tilde r d\tilde r ,

\nabla \Phi_u^{(0,0)}(x)=-\frac{x}{|x|^2} \int_0^{|x|} u^{(0,0)}(\tilde r)\tilde r d \tilde r,

\begin{equation} \label{id:Phiki}
\Phi_u^{(k,i)}(r)= \frac{r^k}{2k} \int_r^{+\infty} u^{(k,i)}(\tilde r)\tilde r^{1-k} d \tilde r +\frac{r^{-k}}{2k} \int_0^r u^{(k,i)}(\tilde r) \tilde r^{1+k} d\tilde r,
\end{equation}

\begin{equation} \label{id:nablaPhiki}
\pa_r \Phi_u^{(k,i)}(r)= \frac{r^{k-1}}{2} \int_r^{+\infty} u^{(k,i)}(\tilde r)\tilde r^{1-k} d \tilde r -\frac{r^{-k-1}}{2} \int_0^r u^{(k,i)}(\tilde r) \tilde r^{1+k} d\tilde r,
\end{equation}

The following lemma gives pointwise estimates of the Poison field.

If $u^{(0,0)}=0$, there holds the estimate for any $\alpha>0$,

\begin{equation} \label{bd:poisson1}
|\Phi_{u}|^2+ |y|^2 |\nabla \Phi_{u}|^2\lesssim |y|^2(1+|y|)^{-2\alpha} \big(1 + \mathbf{1}_{|y| \leq 1} |\log |y||\big)\int_{\mathbb R^2}  |u|^2(1+|y|)^{2\alpha }dy.
\end{equation}

See Lemma A.1 in [10].

For the control of the outer part, we need the following estimates for the Poison field in term of the outer norm.

For $|z|\geq 1$ one has the estimate,

\begin{equation} \label{estimationpoissonexte}
\frac{1}{\zeta}|\Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w}(z)| + |\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w}(z)|+\zeta |\nabla^2 \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w}(z)|\lesssim \frac{1}{\zeta^{\frac 12}} \| \hat w \|_{\out}.
\end{equation}

Notice that for $v(y)=\nu^2w(z)$, the Poisson field scales in $L^{\infty}$, that is: $\Phi_v(y)=\Phi_w(z)$. Hence (<ref>) holds also in $z$ variables. We apply it with $\alpha=\frac 12$, so that for $|z|\geq 1$:

\begin{align*}
&\frac{1}{|z|}|\Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w}|^2+ |z||\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w}|^2 \lesssim  \int_{\mathbb R^2}  |(1-\chi_{\frac{\zeta^*}{2}})\hat w|^2|z|dz\lesssim  \int_{\mathbb R^2}  (1-\chi_{\frac{\zeta^*}{2}}) \hat w^2|z|^{4-\frac 12} |z|^{-\frac 12}\frac{dz}{|z|^2}\\
&\quad \lesssim  \left(\int_{\mathbb R^2}  (1-\chi_{\frac{\zeta^*}{2}}) |\hat w|z|^{2-\frac 14}|^{2p}\frac{dz}{|z|^2}\right)^{\frac{1}{p}}  \left(\int_{\mathbb R^2}  (1-\chi_{\frac{\zeta^*}{2}}) |z|^{-\frac 12 p'}\frac{dz}{|z|^2}\right)^{\frac{1}{p'}} \lesssim  \| \hat w\|_{\out}^{2}
\end{align*}

where we used $(1-\chi_{\frac{\zeta^*}{2}})\leq 1$ and Hölder with $p'$ the conjugate exponent of $p$. Next, we decompose $\hat w=\hat w^{0}+\hat w^\perp$ between radial and non radial components. One has $\| \hat w^{\perp}\|_{\out}+\| \hat w^0\|_{\out}\lesssim \| \hat w \|_{\out}$. from the Sobolev embedding of $W^{3/p,2p}$ into the Hölder space $C^{1/p}$ in dimension $2$, we obtain by interpolation that for any $|z|\geq 1$:

\begin{align*}
&\| (1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp \|_{C^{\frac 1p}(B(z,\frac{|z|}{2}))}  \leq  C \| \hat w^\perp \|_{W^{\frac{3}{p},2p}(B(z,\frac{|z|}{2}))}\leq C  \| \hat w^\perp \|_{L^{2p}(B(z,\frac{|z|}{2}))}^{1-\frac 3p} \| \nabla \hat w^\perp \|_{L^{2p}(B(z,\frac{|z|}{2}))}^{\frac 3p} \\
& \quad \leq  \frac{C}{\zeta^{\frac 32}}  \| \zeta^{2-\frac 12}\hat w^\perp \|_{L^{2p}(B(z,\frac{|z|}{2}))}^{1-\frac 3p} \| \zeta^{2-\frac 12} \nabla \hat w^\perp \|_{L^{2p}(B(z,\frac{|z|}{2}))}^{\frac 3p}\leq \frac{C}{\zeta^{\frac 32}} \| \hat w \|_{\out},
\end{align*}

where we used the definition of the $\| \hat w \|_{\out}$ norm and the inequality $1/2\geq 1/4+1/p$ for $p$ large enough. Therefore, in the ball $B(z,|z|/2)$, $\Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp}$ solves $-\Delta \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp}=(1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp$ with the estimates:

\frac{1}{|z|^{\frac 12}}|\Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp}|+|z|^{\frac 12}|\nabla \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp}+|z|^{\frac 32}\|(1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp\|_{C^{1/p}(B(z,|z|/2)}\leq C\| \hat w \|_{\out}.

By standard regularity properties of the Dirichlet problem and a rescaling argument, we obtain that $\| \nabla^2 \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^\perp} \|_{B(z,\frac{|z|}{2}}\leq |z|^{-3/2}C\| \hat w \|_{\out}$. This proves (<ref>) for the nonradial part of $\hat w$. Next, for the radial part, we have for $i=1,2$ and $\zeta \geq 1$:

\begin{eqnarray*}
\nabla \partial_{z_i} \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^0}(z)  &= & -\nabla \left(\frac{z_i}{\zeta^2} \int_0^{\zeta} (1-\chi_{\frac{\zeta^*}{2}})\hat w^0(\tilde \zeta)\tilde \zeta d \tilde \zeta\right),\\
&=&-\nabla (z_i) \frac{1}{\zeta^2} \int_0^{\zeta} (1-\chi_{\frac{\zeta^*}{2}})\hat w^0(\tilde \zeta)\tilde \zeta d \tilde{\zeta} +\frac{2z_iz}{\zeta^4} \int_0^{\zeta} (1-\chi_{\frac{\zeta^*}{2}})\hat w^0(\tilde \zeta)\tilde \zeta d \tilde \zeta -\frac{2z_iz}{\zeta^2} (1-\chi_{\frac{\zeta^*}{2}})\hat w^0(\zeta).
\end{eqnarray*}

From Hölder, where $(2p)'$ is the Hölder conjugate of $2p$:

\begin{align*}
&\left| \int_0^{\zeta} (1-\chi_{\frac{\zeta^*}{2}})\hat w^0(\tilde \zeta)\tilde \zeta d \tilde \zeta \right|=\left| \int_0^{\zeta} (1-\chi_{\frac{\zeta^*}{2}})\tilde \zeta^{2-\frac 14 -\frac 1p}\hat w^0(\tilde \zeta)\tilde \zeta^{-1+\frac 14 +\frac 1p} d \tilde \zeta \right|  \lesssim  \left| \int_0^{\zeta} (1-\chi_{\frac{\zeta^*}{2}})\tilde \zeta^{2-\frac 14 -\frac{1}{2p}}\hat w^0(\tilde \zeta)\tilde \zeta^{-1+\frac 14 +\frac{1}{2p}} d \tilde \zeta \right|\\
& \quad \lesssim   \left( \int_0^{\zeta} (1-\chi_{\frac{\zeta^*}{2}})|\tilde \zeta^{2-\frac 14}\hat w^0(\tilde \zeta)|\frac{d\tilde \zeta}{\zeta} \right)^{\frac{1}{2p}} \left( \int_0^{\zeta} (1-\chi_{\frac{\zeta^*}{2}}) \tilde \zeta^{(-1+\frac{1}{4}+\frac{1}{2p})(2p)'} d\tilde \zeta\right)^{\frac{1}{(2p)'}} \lesssim  \| \hat w \|_{\out} \zeta^{\frac 14}.
\end{align*}

We recall that $|\hat w^{0}|\lesssim \zeta^{-3/2}$ for $\zeta \geq 1$ from (<ref>). This, and the two above identities imply that for $\zeta \geq 1$:

\left| \nabla \partial_{z_i} \Phi_{(1-\chi_{\frac{\zeta^*}{2}})\hat w^0}(z) \right|\lesssim \zeta^{-\frac 32} \| \hat w \|_{\out}.

This proves (<ref>) for the radial part of $\hat w$.

§ COERCIVITY OF $\AS_0$

In this section, we aim at deriving the coercive estimate of $\As_0$ which is the key to establish the monotonicity formula of the inner norm (<ref>). We first claim the following.

Assume that $f:[0,\infty)\rightarrow \mathbb R$ satisfies

\begin{equation} \label{coerciviteA0:hp1}
\int_0^{\infty} \left(\frac{|f|^2}{1+r^2}+|\pa_r f|^2\right)\frac{\omega_0(r)}{r}dr<\infty,
\end{equation}

$$\int_0^\infty f(r)T_0(r) \chi_{_M}(r) \frac{\omega_0}{r} dr=0.$$

Then, there exists a constant $\delta_0>0$ such that

\int_0^\infty f \As_0 f\frac{\omega_0}{r} dr \leq -\delta_0 \int_0^\infty  \left(\frac{|f|^2}{1+r^2}+|\pa_r f|^2\right)\frac{\omega_0}{r} dr.

Step 1 We first claim the Hardy inequality for any $f$ satisfying (<ref>):

\begin{equation} \label{coerciviteA0:hardy}
\int_0^\infty |\pa_r f|^2\frac{\omega_0}{r}dr\gtrsim \int_0^\infty \frac{f^2}{1+r^2}\frac{\omega_0}{r}dr.
\end{equation}

We first study the function near the origin. If $f$ satisfies (<ref>) then by standard one dimensional Sobolev embedding, $f$ is continuous on $[0,1]$. Hence the estimate (<ref>) implies $f(0)=0$. From the fundamental Theorem of Calculus (which is justified for $f$ via a standard approximation procedure):

\begin{align*}
\left|f(r)\right|&=\left| f(0)+\int_0^r \pa_r (f)d\tilde r \right|=\left|\int_0^r \pa_r (f)d\tilde r \right| \leq \left( \int_0^r \frac{|\pa_r (f)|^2}{\tilde r}d\tilde r \right)^{\frac 12} \left( \int_0^r \tilde rd\tilde r \right)^{\frac 12}\lesssim r \int_0^\infty |\pa_r f|^2\frac{\omega_0}{\tilde r}d\tilde r.
\end{align*}

This proves that:

\begin{equation} \label{coerciviteA0:inter1}
\int_0^\infty |\pa_r f|^2\frac{\omega_0}{r}dr \gtrsim \int_0^1  \frac{f^2}{1+r^2}\frac{\omega_0}{r}dr.
\end{equation}

Away from the origin, integrating by parts:

-\int_1^\infty \pa_r f f r^2 dr=\int_1^\infty f^2rdr+f^2(1).

By Cauchy-Schwarz and Hardy:

\left| \int_1^\infty \pa_r f f r^2 dr \right| \leq \frac12 \int_1^\infty |\pa_r f|^2r^3dr+\frac 12 \int_1^\infty f^2rdr.

The two above identities then give:

\begin{align*}
\int_1^\infty f^2rdr&\leq -\int_1^\infty \pa_r f f r^2 dr-f^2(1)\leq -\int_1^\infty \pa_r f f r^2 dr \leq \frac12 \int_1^\infty |\pa_r f|^2r^3dr+\frac 12 \int_1^\infty f^2rdr,
\end{align*}

from what one deduces, since for $r\geq 1$, one has $cr^4\leq \omega_0(r)\leq r^4/c$ for some $c>0$:

\begin{equation} \label{coerciviteA0:inter2}
\int_1^\infty f^2rdr \leq  \int_1^\infty |\pa_r f|^2r^3dr\lesssim \int_0^\infty |\pa_r f|^2\frac{\omega_0}{r}dr.
\end{equation}

The two estimates (<ref>) and (<ref>) imply the desired Hardy inequality (<ref>).

Step 2 Proof of the coercivity estimate. $\As_0$ has $T_0(r)$ in its kernel, with $T_0$ a strictly positive function on $(0,\infty)$ implies that the spectrum of this self-adjoint operator is nonnegative by standard Sturm-Liouville argument. Hence for any $f$ satisfying (<ref>):

\int_0^\infty f \As_0 f\frac{\omega_0}{r}dr\leq 0 .

Integrating by parts:

\begin{equation} \label{coerciviteA0:inter4}
\int_0^\infty f \As_0 f\frac{\omega_0}{r} dr=-\int_0^\infty |\pa_r f|^2\frac{\omega_0}{r}dr+\int_0^\infty f^2\frac{dr}{r}.
\end{equation}

Combining this and Step 1 gives that for some $c,C>0$, for any $f$ satisfying (<ref>):

\begin{equation} \label{coerciviteA0:inter3}
\int_0^\infty f \As_0 f\frac{\omega_0}{r} dr \leq -c \int_0^\infty  \left(\frac{|f|^2}{1+r^2}+|\pa_r f|^2\right)\frac{\omega_0}{r} dr+C\int_0^\infty \frac{f^2}{r}dr.
\end{equation}

We now assume by contradiction that there exists a sequence of functions $f_n$ with

\begin{equation} \label{coerciviteA0:hpfn}
\int_0^{\infty} \left(\frac{|f_n|^2}{1+r^2}+|\pa_r f_n|^2\right)\frac{\omega_0(r)}{r}dr=1,
\end{equation}

$$ \int_0^\infty f_n \As_0 f_n\frac{\omega_0}{r} dr\rightarrow 0, \ \ \int_0^\infty f_n(r)\chi(r) T_0(r)dr=0.$$

Up to a subsequence, $f_n$ converges weakly in $H^1_{loc}$ and strongly in $L^2_{loc}$ to some function $f_\infty$. By lower-semicontinuity of the above norm, and by strong continuity in $L^2_{loc}$ it satisfies:

\int_0^{\infty} \left(\frac{|f_\infty|^2}{1+r^2}+|\pa_r f_\infty|^2\right)\frac{\omega_0(r)}{r}dr\leq 1, \ \ \int_0^\infty f_{\infty}(r)\chi(r) T_0(r)dr=0.

From the bound $\int_0^{\infty} |f_\infty|^2\omega_0/(r+r^3)dr\leq 1$ and the strong continuity in $L^2_{loc}$ one has:

\int_0^\infty \frac{f_n^2}{r}dr\rightarrow \int_0^\infty \frac{f_\infty^2}{r}dr.

The subcoercivity (<ref>) and the first bound in (<ref>) imply $\int_0^\infty \frac{f_n^2}{r}dr\geq c$ for some $c>0$. From the above strong convergence this implies:

\int_0^\infty \frac{f_\infty^2}{r}dr \geq c >0.

From (<ref>), (<ref>), the aforementioned strong convergence and lower semi-continuity:

\begin{align*}
0\leq -\int_0^\infty f \As_0 f\frac{\omega_0}{r}&=\int_0^\infty |\pa_r f|^2\frac{\omega_0}{r}dr-\int_0^\infty f^2\frac{dr}{r}\leq \liminf\left( \int_0^\infty |\pa_r f_n|^2\frac{\omega_0}{r}dr-\int_0^\infty f_n^2\frac{dr}{r} \right)=0.
\end{align*}

Hence $\int_0^\infty f \As_0 f\frac{\omega_0}{r}=0$. Hence $f_\infty=\lambda T_0(r)$, for some $\lambda \neq 0$, which contradicts with $\int_0^\infty f_{\infty}\chi T_0=0$. This concludes the proof of Lemma <ref>.

We also have the following coercivity estimate for $\As_0$.
Assume that $f:[0,\infty)\rightarrow \mathbb R$ satisfies

\begin{equation*}
\int_0^{\infty} \left(|\pa^2_r f|^2 + \frac{|\pa_r f|^2}{1 + r^2} + \frac{|f|^2}{(1+r^4)(1  +\ln ^2\la r \ra)}\right)\frac{\omega_0(r)}{r}dr<\infty,
\end{equation*}

$$\int_0^\infty f(r)T_0(r) \chi_{_M}(r) \frac{\omega_0}{r} dr=0.$$

Then, there exists a constant $\delta_1>0$ such that

\int_0^\infty |\As_0 f|^2\frac{\omega_0}{r} dr \geq \delta_1 \int_0^\infty  \left(|\pa^2_r f|^2 + \frac{|\pa_r f|^2}{1 + r^2} + \frac{|f|^2}{(1+r^4)(1  +\ln ^2\la r\ra )}\right)\frac{\omega_0(r)}{r} dr.

Since the proof follows exactly the same lines to the one of Lemma <ref>, apart from the following Hardy inequality (in the critical case) which is used to obtain a sub-coercivity estimate,

\begin{equation}
\int_0^{+\infty} \frac{|\pa_r f|^2}{1 + r^2} \frac{\omega_0}{r}dr \geq C \int_0^{+\infty} \frac{|f|^2}{(1 + r^4)(1 + \ln^2\la r\ra )} \frac{\omega_0}{r}dr.
\end{equation}

so we omit the proof.

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