\section{Introduction and Main results}

Consider  the stable solution of the following equation
 \begin{equation}\label{8LE}
(-\Delta)^s u=|u|^{p-1}u\;\;\;\;\hbox{in}\;\;\;\;\;  \R^n,
\end{equation}
where $(-\Delta)^s$ is the fractional Laplacian operator for $2<s<3$.

 \vskip0.1in
 The motivation of  studying    such an equation is originated from the classical   Lane-Emden equation
 \begin{equation}\label{LZ=000123}
 -\Delta  u=|u|^{p-1}u \;\;\;\;\hbox{in}\;\;\;\;\;  \R^n
\end{equation}
and its parabolic counterpart, which  have played a crucial   role in the development of nonlinear PDEs in the last
decades. These  arise  in astrophysics and Riemannian geometry.  The pioneering  works on  Eq.\eqref{LZ=000123} were  contributed by   R. Fowler \cite{Fowler=1,Fowler=2}. Later, the  ground-breaking  result on equation \eqref{LZ=000123} is the  fundamental    Liouville-type theorems  established by  Gidas and Spruck \cite{Gidas-Sp=1981}, they claimed   that the Eq. \eqref{LZ=000123} has no positive solution whenever $p\in (1, 2^\ast-1)$, where $2^\ast=2n/(n-2)$ if $n\geq 3$ and
 $2^\ast=\infty$ if $n\leq 2.$  The critical case  $p=2^\ast-1$,     Eq.\eqref{LZ=000123}  has a unique positive solution  up to translation  and rescaling  which is radial and explicitly formulated, see Caffarelli-Gidas-Spruck \cite{Caffarelli1989}. Since then
many experts in partial differential equations devote to the above equations for various
parameters $s$ and $p$.

 \vskip0.1in

For the nonlocal case of $0 < s < 1,$  a counterpart of the classification results of Gidas and Spruck \cite{Gidas-Sp=1981}, and Caffarelli-
Gidas-Spruck  \cite{Caffarelli1989}  holds for the fractional Lane-Emden equation  \eqref{8LE}, see the works due to   Li \cite{YLi04} and Chen-Li-Ou \cite{Chen-Li-Ou}. In these cases, the Sobolev exponent is given by $P_S(n, s)=(n+2s)/(n-2s)$ if $n>2s$, and otherwise $P_S(n, s)=\infty$.

\vskip0.123in
Recently, for the nonlocal case of $0<s<1$, Davila, Dupaigne and Wei in \cite{Wei0=1} gave a complete classification of finite Morse index solution of \eqref{8LE}; for the nonlocal case of $1<s<2$, Fazly and Wei in \cite{Wei1=2} gave a complete classification of finite Morse index solution of \eqref{8LE}.
For the local cases $s=1$ and $s=2$, such kind of  classification is proved by Farina in \cite{Farina2007} and Davila, Dupaigne, Wang and Wei in \cite{Wei=2}, respectively. For the case $s=3$, the Joseph-Lundgren exponent (for the triharmonic Lane-Emden equation) is obtained
and classification is proved by in \cite{LWZ2016=3}.

 \vskip0.1in

However, when $2<s<3$,  the equation \eqref{8LE} has not been  considered  so far. In this paper we classify the stable solution of \eqref{8LE}.
\vskip0.23in

There are many ways of defining the fractional Laplacian $(-\Delta)^s$, where $s$ is any positive, noninteger number.
Caffarelli and Silvestre in \cite{Caffarelli2007} gave a characterization of the fractional Laplacian when $0<s<1$ as the
Dirichlet-to-Neumann map for a function $u_e$ satisfying a higher order elliptic equation in the upper half space
with one extra spatial dimension. This idea  was  later generalized by Yang in \cite{Yang2100} when  the $s$ is being any positive, noninteger number.
See also Chang-Gonzales \cite{Chang2011} and Case-Chang \cite{Case2015} for general manifolds.

\vskip0.123in
To introduce the fractional operator $(-\Delta)^s$ for $2<s<3$, just like the case of $1<s<2$, via  the
Fourier transform, we can define
\be\nonumber
\widehat{(-\Delta)^s}u(\xi)=|\xi|^{2s} \widehat{u}(\xi)
\ee
or equivalently define this operator inductively by $(-\Delta)^s=(-\Delta)^{s-2}\circ(-\Delta)^2$.

\vskip0.1in
\begin{definition}
 We say a solution $u$ of \eqref{8LE} is stable outside a compact set if there exists $R_0>0$ such that
 \be\label{8stable}
 \int_{\R^n}\int_{\R^n}\frac{(\varphi(x)-\varphi(y))^2}{|x-y|^{n+2s}}dx dy-p\int_{\R^n}|u|^{p-1}\varphi^2dx\geq0
 \ee
 for any $\varphi\in C_c^{\infty}(\R^n\backslash \overline{B_{R_0}})$.
 \end{definition}
Set
\be\nonumber
p_s(n)=
\begin{cases}
\infty\;\;&\hbox{if}\;\; n\leq 2s,\\
\frac{n+2s}{n-2s}\;\;&\hbox{if}\;\; n>2s.
\end{cases}
\ee
The first main result  of the present paper is the following

\bt\label{8Liouvillec}
Suppose that $n>2s$ and $2<s<\delta<3$. Let $u\in C^{2\delta}(\R^n)\cap L^1(\R^n,(1+|z|)^{n+2s}dz)$ be a solution of
\eqref{8LE} which  is stable outside a compact set. Assume

\begin{itemize}
\item [(1)]   $1<p<p_s(n)$
or
\item [(2)]  $p_s(n)< p$  and
\be\label{8gamma}
p\frac{\Gamma(\frac{n}{2}-\frac{s}{p-1})\Gamma(s+\frac{s}{p-1})}{\Gamma(\frac{s}{p-1})\Gamma(\frac{n-2s}{2}-\frac{s}{p-1})}
>\frac{\Gamma(\frac{n+2s}{4})^2}{\Gamma(\frac{n-2s}{4})^2},
\ee
then the solution $u\equiv0$.
\item [(3)] $p=p_s(n)$, then $u$ has finite energy, i.e.,
\be\nonumber
\|u\|^2_{\dot{H}^s(\R^n)}:= \int_{\R^n}\int_{\R^n}\frac{(u(x)-u(y))^2}{|x-y|^{n+2s}}dx dy=\int_{\R^n}|u|^{p+1}<+\infty.
\ee
If in addition $u$ is stable, then $u\equiv0$.
\end{itemize}
\et

\br
In the Theorem  \ref{8Liouvillec} the condition \eqref{8gamma} is optimal. In fact,  the radial singular solution $u=|x|^{-\frac{2s}{p-1}}$ is stable if  \be\nonumber
p\frac{\Gamma(\frac{n}{2}-\frac{s}{p-1})\Gamma(s+\frac{s}{p-1})}{\Gamma(\frac{s}{p-1})\Gamma(\frac{n-2s}{2}-\frac{s}{p-1})}
\leq\frac{\Gamma(\frac{n+2s}{4})^2}{\Gamma(\frac{n-2s}{4})^2}.
\ee See \cite{LWZ}.
\er

\br The hypothesis $(2)$  of Theorem \ref{8Liouvillec} is  equivalent to
\be
p<p_c(n):=\begin{cases}
+\infty\;\;\;\;\;\;\;\;&\hbox{if}\;\;\;\;\;\;\;\; n\leq n_0(s),\\
\frac{n+2s-2-2a_{n,s}\sqrt{n}}{n-2s-2-2a_{n,s}\sqrt{n}}&\hbox{if}\;\;\;\;\;\;\;\;n> n_0(s),\\
\end{cases}
\ee
where $n_0(s)$ is the largest root  of $n-2s-2-2a_{n,s}\sqrt{n}=0$, see \cite{LWZ}. More details and further sharp results about $a_{n,s}$ and $n_0(s)$ see \cite{LWZ=00}.

\er

\br
In this remark, we further analyze the  hypothesis $(2)$  in Theorem \ref{8Liouvillec}. Recall that when $s=1$
the condition \eqref{8gamma} gives a upper bounded of $p$ (originated from Joseph and Lundgren \cite{Joseph1972}), it is
\be
p<p_c(n):=\begin{cases}
\;\;\;\;\;\;\;\;\;\infty\;\;\;\;\;\;\;\;&\hbox{if}\;\;\;\;\;\;\;\; n\leq10,\\
\frac{(n-2)^2-4n+8\sqrt{n-1}}{(n-2)(n-10)}\;\;\;\;\;\;\;\;&\hbox{if}\;\;\;\;\;\;\;\; n\geq11.
\end{cases}
\ee
For the case $s=2$, \eqref{8gamma} induce the upper bound of $p$ which is given by the following formula (cf. Gazzola and Grunau \cite{Gazzola2006}):
\be
p<p_c(n)=\begin{cases}
\;\;\;\;\;\;\;\;\;\;\infty\;\;\;\;\;\;\;\;&\hbox{if}\;\;\;\;\;\;\;\; n\leq12,\\
\frac{n+2-\sqrt{n^2+4-n\sqrt{n^2-8n+32}}}{n-6-\sqrt{n^2+4-n\sqrt{n^2-8n+32}}}\;\;\;\;\;\;\;\;&\hbox{if}\;\;\;\;\;\;\;\; n\geq13.
\end{cases}
\ee

\vskip0.1in

In the triharmonic case, the corresponding exponent given by see (\cite{LWZ2016=3}) is the following

\be\nonumber
p<p_{c}(n)=
\begin{cases}
\;\;\;\infty\;\;&\hbox{if}\;\; n\leq 14,\\
\frac{n+4-2D(n)}{n-8-2D(n)}\;\;&\hbox{if}\;\; n\geq15,
\end{cases}
\ee
where
\be\nonumber
D(n):=\frac{1}{6}\Big(9n^2+96-\frac{1536+1152n^2}{D_0(n)}-\frac{3}{2}D_0(n)\Big)^{1/2};
\ee
\be\nonumber
D_0(n):=-(D_1(n)+36\sqrt{D_2(n)})^{1/3};
\ee
\be\nonumber\aligned
D_1(n):=-94976+20736n+103104n^2-10368n^3+1296n^5-3024n^4-108n^6;
\endaligned\ee
\be\nonumber\aligned
D_2(n):&=6131712-16644096n^2+6915840n^4-690432n^6-3039232n\\
&\quad+4818944n^3-1936384n^5+251136n^7-30864n^8-4320n^9\\
&\quad+1800n^{10}-216n^{11}+9n^{12}.
\endaligned\ee

\er

\s{Preliminary}

Throughout this paper we denote $b:=5-2s$ and define the operator
\be\nonumber
\Delta_b w:=\Delta w+\frac{b}{y}w_y=y^{-b}\mathbf{div}(y^b\nabla w)
\ee
for a function $w\in W^{3,2}(\R^{n+1};y^bdxdy)$.  We firstly quote the following result.

\bt\label{ththYang} (See \cite{Yang2100} ) Assume $2<s<3$.  Let  $u_e\in W^{3,2}(\R^{n+1};y^bdxdy)$ satisfy  the equation
\be\label{8eqs}
\Delta_b^3u_e=0
\ee
on the upper half space for $(x,y)\in \R^n\times\R_+$ (where $y$ is the spacial direction) and the boundary conditions:
\be\label{8eqs1}\aligned
&u_e(x,0)=f(x),\\
&\lim_{y\rightarrow0}y^b\frac{\pa u_e}{\pa y}=0,\\
&\frac{\pa^2 u_e}{\pa y^2}\mid_{y=0}=\frac{1}{2s}\Delta_{x}u_e\mid_{y=0},\\
&\lim_{y\rightarrow0}C_{n,s}y^b\frac{\pa}{\pa y}\Delta_b^2 u_e=(-\Delta)^s f(x),
\endaligned
\ee
where $f(x)$ is some function defined on $H^s(\R^n)$.  Then we have
\be\label{Yangmore}
\int_{\R^n}|\xi|^{2s}|\hat{u}(\xi)|^2d\xi=C_{n,s}\int_{\R^{n+1}_+} y^b|\nabla\Delta_b u_e(x,y)|^2dxdy.
\ee
\et

\vskip0.23in

Applying the above theorem to solutions of \eqref{8LE},  we conclude that the extended function $u_e(x,y)$,
where $x=(x_1,...,x_n)\in\R^n$ and $y\in\R^+$, satisfies

\be\label{8LEE}\begin{cases}
\Delta_b^3 u_e=0\;\;\hbox{in}\;\;\R^{n+1}_+,\\
\lim_{y\rightarrow0}y^b\frac{\pa u_e}{\pa y}=0\;\;\hbox{on}\;\;\pa\R^{n+1}_+,\\
\frac{\pa^2 u_e}{\pa y^2}\mid_{y=0}=\frac{1}{2s}\Delta_{x}u_e\mid_{y=0}\;\;\hbox{on}\;\;\pa\R^{n+1}_+,\\
\lim_{y\rightarrow0}y^b\frac{\pa}{\pa y}\Delta_b^2 u_e=-C_{n,s}|u_e|^{p-1}u_e\;\;\hbox{in}\;\;\R^{n+1}_+.
\end{cases}\ee
Moreover,
\be\nonumber
\int_{\R^n}|\xi|^{2s}|\hat{u}(\xi)|^2d\xi=C_{n,s}\int_{\R^{n+1}_+} y^b|\nabla\Delta_b u_e(x,y)|^2dxdy
\ee
and $u(x)=u_e(x,0)$.

\vskip0.11in

\vskip0.1in

Define
\be\label{8exu}\aligned
&E(\la,x,u_e)=\int_{\R^{n+1}_+\cap\pa B_1}\frac{1}{2}\theta_1^b|\nabla\Delta_b u_e^\la|^2-\frac{C_{n,s}}{p+1}\int_{\pa \R^{n+1}_+\cap B_1}|u_e^\la|^{p+1}\\
&+\sum_{0\leq i,j\leq 4,i+j\leq5}C_{i,j}^1\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^{i+j}\frac{d^i u^\la_e}{d\la^i}\frac{d^j u^\la_e}{d\la^j}\\
&+\sum_{0\leq t,s\leq 2,t+s\leq3}C_{t,s}^2\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^{t+s}\nabla_{S^n}\frac{d^t u^\la_e}{d\la^t}\nabla_{S^n}\frac{d^s u^\la_e}{d\la^s}\\
&+\sum_{0\leq l,k\leq 1,l+k\leq1}C_{l,k}^3\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^{l+k}\Delta_{S^n}\frac{d^l u^\la_e}{d\la^l}\Delta_{S^n}\frac{d^k u^\la_e}{d\la^k}\\
&+(\frac{s}{p-1}+1)\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\Delta_b u^\la_e)^2.
\endaligned\ee

\vskip0.1in

The following is the monotonicity formula which will paly an important role.
\bt\label{8monoid}
Let $u_e$ satisfy  the equation \eqref{8eqs} with the boundary conditions \eqref{8eqs1}, we have the following
\be\aligned
\frac{d E(\la,x,u_e)}{d\la}&=\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\Big(
3\la^5(\frac{d^3u_e^\la}{d\la^3})^2+A_1\la^3(\frac{d^2u_e^\la}{d\la^2})^2+A_2\la(\frac{du^\la_e}{d\la})^2\Big)\\
&+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\Big(
2\la^3|\nabla_{S^n}\frac{d^2u_e^\la}{d\la^2}|^2+B_1\la|\nabla_{S^n}\frac{du_e^\la}{d\la}|\Big)\\
&+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la(\Delta_{S^n}\frac{du_e^\la}{d\la})^2,
\endaligned
\ee
\et
where $\theta_1=\frac{y}{r}$ and
\be\nonumber\aligned
&A_1:=10\delta_1-2\delta_2-56+\alpha_0^2-2\al_0-2\beta_0-4,\\
&A_2:=-18\delta_1+6\delta_2-4\delta_3+2\delta_4+72-\al_0^2+\beta_0^2+2\al_0+2\beta_0,\\
&B_1:=8\al-4\beta-2\beta_0+4(n+b)-14,
\endaligned\ee
\be\nonumber\aligned
&\al:=n+b-2-\frac{4s}{p-1},\beta:=\frac{2s}{p-1}(3+\frac{2s}{p-1}-n-b),\quad\quad \\
&\al_0:=n+b-\frac{4s}{p-1},\;\;\;\beta_0:=\frac{2s}{p-1}(1+\frac{2s}{p-1}-n-b)\quad\quad
\endaligned\ee
and
\be\label{8delta}\aligned
\delta_1=&2(n+b)-\frac{8s}{p-1},\\
\delta_2=&(n+b)(n+b-2)-(n+b)\frac{12s}{p-1}+\frac{12s}{p-1}(1+\frac{2s}{p-1}),\\
\delta_3=&-\frac{8s}{p-1}(1+\frac{2s}{p-1})(2+\frac{2s}{p-1})+2(n+b)\frac{6s}{p-1}(1+\frac{2s}{p-1})\\
           &-(n+b)(n+b-2)(1+\frac{4s}{p-1}),\\
\delta_4=&(3+\frac{2s}{p-1})(2+\frac{2s}{p-1})(1+\frac{2s}{p-1})\frac{2s}{p-1}\\
&\quad-2(n+b)(1+\frac{2s}{p-1})(2+\frac{2s}{p-1})\frac{2s}{p-1}\\
          &\quad+(n+b)(n+b-2)(2+\frac{2s}{p-1})\frac{2s}{p-1}.\\
\endaligned\ee

We will give the proof of Theorem   \ref{8monoid} in the next section. Now we would like to
state a consequent result of  Theorem   \ref{8monoid}.
Recall that  $E(\la,x,u_e)$, defined in \eqref{8exu},  can be divided into two parts:  the integral  over  the ball $B_\la$
and the terms on the boundary $\pa B_\la$.  We note  that in our blow-down analysis, the coefficients (including positive or negative, big or small) of the boundary terms can be estimated  in a unified way, therefore we may change some coefficients of the boundary terms  in $E(\la,x,u_e)$. After such a change,  we denote the new functional  by $E^c(\la,x,u_e)$.

\vskip0.1in
Define
\be\label{728-pm}
p_m(n):=\begin{cases}
+\infty\;\;\;\;\;\;\;\;&\hbox{if}\;\;\;\;\;\;\;\; n<2s+6+\sqrt{73},\\
\frac{5n+10s-\sqrt{15(n-2s)^2+120(n-2s)+370}}{5n-10s-\sqrt{15(n-2s)^2+120(n-2s)+370}}&\hbox{if}\;\;\;\;\;\;\;\;n\geq2s+6+\sqrt{73}.\\
\end{cases}
\ee
 We have the following
\bt\label{8Monotonem} Assume that $\frac{n+2s}{n-2s}<p<p_m(n)$, then $E^c(\la,x,u_e)$ is a nondecreasing function of $\la>0$. Furthermore,
\be\nonumber
\frac{d E^c(\la,x,u_e)}{d\la}\geq C(n,s,p)\la^{2s\frac{p+1}{p-1}-6-n}
\int_{\R^{n+1}_+\cap\pa B_\la(x_0)}y^b\Big(\frac{2s}{p-1}u_e+\la\pa_r u_e\Big)^2,
\ee
where $C(n,s,p)$ is a constant independent of $\la$.
\et

By carefully comparing $\frac{n+2s}{n-2s}<p<p_m(n)$ with $p>\frac{n+2s}{n-2s}$ and \eqref{8gamma},  we get the following (see the last section of the current paper) monotonicity formula for our blow down analysis.

\bt\label{8Monotone} Assume that $p>\frac{n+2s}{n-2s}$ and \eqref{8gamma}, then $E^c(\la,x,u_e)$ is a nondecreasing function of $\la>0$. Furthermore,
\be\nonumber
\frac{d E^c(\la,x,u_e)}{d\la}\geq C(n,s,p)\la^{2s\frac{p+1}{p-1}-6-n}
\int_{\R^{n+1}_+\cap\pa B_\la(x_0)}y^b\Big(\frac{2s}{p-1}u_e+\la\pa_r u_e\Big)^2,
\ee
where $C(n,s,p)$ is a constant independent of $\la$.
\et

\section{Monotonicity formula and the proof of Theorem   \ref{8monoid}}

The derivation of the monotonicity for the \eqref{8LE} when $2<s<3$ is complicated in its process, we divide it into several subsections.
In subsection $3.1$, we  derive  $\frac{d}{d\la}\overline{E}(u_e,\la)$.
In subsection $3.2$, we calculate  $\frac{\pa^j}{\pa r^j}u_e^\la$  and  $\frac{\pa^i}{\pa \la^i}u_e^\la,\;\; i,j=1,2,3,4$.
In subsection $3.3$,  the operator   $\Delta_b^2$   and its representation will be given.
In subsection $3.4$, we  decompose $\frac{d}{d\la}\overline{E}(u_e^\la,1)$.
Finally, combine with the above four subsections, we can obtain the  monotonicity formula, hence get the proof of  Theorem \ref{8monoid}.

\vskip0.1in

Suppose that $x_0=0$ and denote by $B_\la$  the balls  centered at zero with radius $\la$. Set
\be\nonumber
\overline{E}(u_e,\la):=\la^{2s\frac{p+1}{p-1}-n}\Big(\int_{\R^{n+1}_+\cap B_\la}\frac{1}{2}y^b|\nabla\Delta_b u_e|^2
-\frac{C_{n,s}}{p+1}\int_{\pa\R^{n+1}_+\cap B_\la}|u_e|^{p+1}\Big).\\
\ee

\subsection{The derivation of $\frac{d}{d\la}\overline{E}(u_e,\la)$}
Define
\be\label{8uvw}\aligned
&v_e:=\Delta_b u_e, u_e^\la(X):=\la^{\frac{2s}{p-1}}u_e(\la X),\;\; w_e(X):=\Delta_b v_e\\
&v_e^\la(X):=\la^{\frac{2s}{p-1}+2}v_e(\la X),\;\;  w_e^\la(X):=\la^{\frac{2s}{p-1}+4}w_e(\la X),
\endaligned\ee
where $X=(x,y)\in \R^{n+1}_+$.
Therefore,
\be\label{8lamda}
\Delta_b u_e^\la(X)=v_e^\la(X), \Delta_b v_e^\la(X)=w_e^\la(X).
\ee
Hence
\be\nonumber\aligned
&\Delta_b w_e^\la=0,\\
&\lim_{y\rightarrow0}y^b\frac{\pa u_e}{\pa y}=0,\\
&\frac{\pa^2 u_e}{\pa y^2}\mid_{y=0}=\frac{1}{2s}\Delta_x u_e\mid_{y=0},\\
&\lim_{y\rightarrow0}C_{n,s}y^b\frac{\pa}{\pa y}w_e^\la=-C_{n,s}|u_e|^{p-1}u_e.
\endaligned
\ee
In addition, differentiating \eqref{8lamda} with respect to $\la$ we have
\be\nonumber
\Delta_b \frac{du_e^{\la}}{d\la}=\frac{dv_e^{\la}}{d\la},\;\;\; \Delta_b \frac{dv_e^{\la}}{d\la}=\frac{dw_e^{\la}}{d\la}.
\ee
Note that
\be\nonumber
\overline{E}(u_e,\la)=\overline{E}(u_e^\la,1)=\int_{\R^{n+1}_+\cap B_1}\frac{1}{2}y^b|\nabla v_e^\la|^2
-\frac{C_{n,s}}{p+1}\int_{\pa\R^{n+1}_+\cap B_1}|u_e^\la|^{p+1}.
\ee
Taking derivative of the energy $\overline{E}(u_e^\la,1)$ with respect to $\la$ and integrating by part we have:

\be\label{8overE}\aligned
&\frac{d\overline{E}(u_e^\la,1)}{d\la}=\int_{\R^{n+1}_+\cap B_1}y^b\nabla v_e^\la\nabla\frac{dv_e^\la}{d\la}
-C_{n,s}\int_{\pa\R^{n+1}_+\cap B_1}|u_e^\la|^{p-1}u_e^\la\frac{du_e^\la}{d\la}\\
=&\int_{\pa(\R^{n+1}_+\cap B_1)}y^b\frac{\pa v_e^\la}{\pa n}\frac{dv_e^\la}{d\la}-
\int_{\R^{n+1}_+\cap B_1}(y^b\Delta v_e^\la+by^{b-1}\frac{\pa v_e^\la}{\pa y})\frac{dv_e^\la}{d\la}\\
&-C_{n,s}\int_{\pa\R^{n+1}_+\cap B_1}|u_e^\la|^{p-1}u_e^\la\frac{du_e^\la}{d\la}\\
=&-\int_{\pa \R^{n+1}\cap B_1}y^b\frac{\pa v_e^\la}{\pa y}\frac{dv_e^\la}{d\la}
+\int_{\R^{n+1}_+\cap\pa B_1}y^b\frac{\pa v_e^\la}{\pa r}\\
&-\int_{\R^{n+1}_+\cap B_1}y^b\Delta v_e^\la\frac{dv_e^\la}{d\la}+by^{b-1}\frac{\pa v_e^\la}{\pa y}\frac{v_e^\la}{d\la}
-\int_{\pa\R^{n+1}_+\cap B_1}y^b\frac{w_e^\la}{\pa y}\frac{u_e^\la}{d\la}.
\endaligned\ee
Now note that from the definition of $v_e^\la$ and by differentiating  it  with respect to $\la$,  we get the following identity for $X\in\R^{n+1}_+$,
\be\nonumber
r\frac{\pa v_e^\la}{\pa r}=\la \pa_{\la} v_e^\la-(\frac{2s}{p-1}+2)v_e^\la.
\ee
Hence,
\be\nonumber
\aligned
\int_{\R^{n+1}_+\cap B_1}y^b\frac{\pa v_e^\la}{\pa r}&\frac{dv_e^\la}{d\la}
=\int_{\R^{n+1}_+\cap B_1}y^b\Big(\la\frac{dv_e^\la}{d\la}\frac{dv_e^\la}{d\la}-(\frac{2s}{p-1}+2)v_e^\la\frac{dv_e^\la}{d\la}\Big)\\
&=\la\int_{\R^{n+1}_+\cap \pa B_1}y^b(\frac{dv_e^\la}{d\la})^2-(\frac{s}{p-1}+1)\frac{d}{d\la}\int_{\R^{n+1}_+\cap \pa B_1}y^b(v_e^\la)^2.\\
\endaligned
\ee
Note that
\be\nonumber
\aligned
-\int_{\R^{n+1}_+\cap  B_1}y^b\Delta v_e^\la\frac{dv_e^\la}{d\la}=&
\int_{\pa\R^{n+1}_+\cap  B_1}y^b\frac{\pa v_e^\la}{\pa y}\frac{v_e^\la}{d\la}-\int_{\R^{n+1}_+\cap \pa B_1}y^b\frac{\pa v_e^\la}{\pa r}\frac{v_e^\la}{d\la}\\
&+\int_{\R^{n+1}_+\cap  B_1}\nabla v_e^\la\nabla(y^b\frac{dv_e^\la}{d\la}).
\endaligned
\ee
Integration by part we have
\be\nonumber\aligned
&\int_{\R^{n+1}_+\cap  B_1}y^b\nabla v_e^\la \nabla\frac{dv_e^\la}{d\la}\\
&=-\int_{\pa\R^{n+1}_+\cap  B_1}y^b\frac{\pa v_e^\la}{\pa y}\frac{dv_e^\la}{d\la}+\int_{\R^{n+1}_+\cap  \pa B_1}y^b\frac{\pa v_e^\la}{\pa r}\frac{dv_e^\la}{d\la}\\
&\quad -\int_{\R^{n+1}_+\cap  B_1}\nabla\cdot(y^b\nabla v^\la_e)\frac{dv_e^\la}{d\la}\\
&=-\int_{\pa\R^{n+1}_+\cap  B_1}y^b\frac{\pa v_e^\la}{\pa y}\frac{dv_e^\la}{d\la}+\int_{\R^{n+1}_+\cap  \pa B_1}y^b\frac{\pa v_e^\la}{\pa r}\frac{dv_e^\la}{d\la}\\
&\quad -\int_{\R^{n+1}_+\cap  B_1}y^b\Delta_b v_e^\la\Delta_b\frac{du_e^\la}{d\la}.\\
\endaligned
\ee
Now
\be\nonumber\aligned
-\int_{\R^{n+1}_+\cap  B_1}&y^b\Delta_b v_e^\la\Delta_b\frac{du_e^\la}{d\la}
=-\int_{\R^{n+1}_+\cap  B_1}y^b\Delta_b v_e^\la(\Delta\frac{du_e\la}{d\la}+\frac{b}{y}\frac{\pa}{\pa y}\frac{du_e^\la}{d\la})\\
=&-\int_{\pa(\R^{n+1}_+\cap  B_1)}y^b\Delta_b v_e^\la\frac{\pa}{\pa n}\frac{du_e^\la}{d\la}
+\int_{\R^{n+1}_+\cap  B_1}\nabla(y^b\Delta_b v_e^\la)\nabla\frac{du_e^\la}{d\la}\\
&-\int_{\R^{n+1}_+\cap  B_1}by^{b-1}\Delta_b v_e^\la\frac{\pa}{\pa y} \frac{du_e^\la}{d\la}\\
=&-\int_{\pa(\R^{n+1}_+\cap  B_1)}y^b\Delta_b v_e^\la \frac{\pa}{\pa n}\frac{d u_e^\la}{d\la}
+\int_{\R^{n+1}_+\cap  B_1}y^b\nabla\Delta_b v_e^\la\nabla \frac{du_e^\la}{d\la}\\
=&-\int_{\pa(\R^{n+1}_+\cap  B_1)}y^b\Delta_b v_e^\la \frac{\pa}{\pa n}\frac{d u_e^\la}{d\la}
+\int_{\pa(\R^{n+1}_+\cap  B_1)}y^b\frac{\pa \Delta_b v_e^\la}{\pa n}\frac{d u_e^\la}{d\la}\\
&-\int_{\R^{n+1}_+\cap  B_1}y^b\Delta_b^2 v_e^\la\frac{d u_e^\la}{d\la}\\
=&-\int_{\pa(\R^{n+1}_+\cap  B_1)}y^b\Delta_b v_e^\la \frac{\pa}{\pa n}\frac{d u_e^\la}{d\la}
+\int_{\pa(\R^{n+1}_+\cap  B_1)}y^b\frac{\pa \Delta_b v_e^\la}{\pa n}\frac{d u_e^\la}{d\la}.\\
\endaligned
\ee
Here we have used that $\Delta_b^2 v_e^\la=\Delta_b^3 u_e^\la=0$.
Therefore, combine with the above arguments we get that
\be\label{8overE1}\aligned
\int_{\R^{n+1}_+\cap  B_1}&y^b\nabla v_e^\la \nabla\frac{dv_e^\la}{d\la}=-\int_{\pa\R^{n+1}_+\cap  B_1}y^b\frac{\pa v_e^\la}{\pa y}\frac{dv_e^\la}{d\la}
+\int_{\R^{n+1}_+\cap \pa B_1}y^b \frac{\pa v_e^\la}{\pa r}\frac{dv_e^\la}{d\la}\\
&+\int_{\pa\R^{n+1}_+\cap  B_1}y^b\Delta_b v_e^\la\frac{\pa}{\pa y}\frac{du_e^\la}{d\la}-\int_{\R^{n+1}_
+\cap \pa B_1}y^b \Delta v_e^\la \frac{\pa}{\pa r}\frac{du_e^\la}{d\la}\\
&-\int_{\pa\R^{n+1}_+\cap  B_1}y^b\frac{\pa}{\pa y}\Delta_b v_e^\la\frac{du_e^\la}{d\la}+\int_{\R^{n+1}_+\cap \pa B_1}y^b\frac{\pa}{\pa r}\Delta_b v_e^\la\frac{du_e^\la}{d\la}\\
=&\int_{\R^{n+1}_+\cap \pa B_1}y^b \frac{\pa v_e^\la}{\pa r}\frac{dv_e^\la}{d\la}-\int_{\R^{n+1}_
+\cap \pa B_1}y^b \Delta v_e^\la \frac{\pa}{\pa r}\frac{du_e^\la}{d\la}\\
&-C(n,s)\int_{\pa\R^{n+1}_+\cap  B_1}|u_e^\la|^{p-1}\frac{du_e^\la}{d\la}+\int_{\R^{n+1}_+\cap \pa B_1}y^b\frac{\pa}{\pa r}\Delta_b v_e^\la\frac{du_e^\la}{d\la}.\\
\endaligned
\ee
Here,  we have used that $\frac{\pa \Delta u_e^\la(x,0)}{\pa y}=0$, $\frac{\pa}{\pa y}\frac{du_e^\la}{d\la}=0$ on $\pa \R^{n+1}_+$
 and $\lim_{y\rightarrow0}y^b\frac{\pa}{\pa y}\Delta_b v_e^\la=-C_{n,s}|u_e^\la|^{p-1}u_e^\la$.
By \eqref{8overE} and \eqref{8overE1} we obtain that
\be\label{8overee}\aligned
\frac{d}{d\la}\overline{E}(u_e^\la,1)=&\int_{\R^{n+1}_+\cap \pa B_1}y^b\frac{\pa v_e^\la}{\pa r}\frac{dv_e^\la}{d\la}
+\int_{\R^{n+1}_+\cap \pa B_1}y^b\frac{\pa w_e^\la}{\pa r}\frac{du_e^\la}{d\la}\\
&-\int_{\R^{n+1}_+\cap \pa B_1}y^bw_e^\la\frac{\pa}{\pa r}\frac{du_e^\la}{d\la}.
\endaligned\ee
Recall  \eqref{8uvw} and
differentiate it with respect to $\la$, we have
\be\nonumber\aligned
\frac{du_e^\la(X)}{d\la}&=\frac{1}{\la}\big(\frac{2s}{p-1}u_e^\la(X)+r\pa_r u_e^\la(X)\big),\\
\frac{dv_e^\la(X)}{d\la}&=\frac{1}{\la}\big((\frac{2s}{p-1}+2)v_e^\la(X)+r\pa_r v_e^\la(X)\big),\\
\frac{dw_e^\la(X)}{d\la}&=\frac{1}{\la}\big((\frac{2s}{p-1}+4)w_e^\la(X)+r\pa_r w_e^\la(X)\big).\\
\endaligned
\ee
Differentiate the above equations with respect to $\la$ again we get
\be\nonumber
\la\frac{d^2u_e^\la(X)}{d\la^2}+\frac{du_e^\la(x)}{d\la}=\frac{2s}{p-1}\frac{du_e^\la(X)}{d\la}+r\pa_r\frac{du_e^\la}{d\la}.
\ee
Hence, for $X\in\R^{n+1}_+\cap B_1$, we have
\be\nonumber\aligned
\pa_r(u_e^\la(X))&=\la\frac{du_e^\la}{d\la}-\frac{2s}{p-1}u_e,\\
\pa_r(\frac{du_e^\la(X)}{d\la})&=\la\frac{d^2u_e^\la(X)}{d\la^2}+(1-\frac{2s}{p-1})\frac{du_e^\la}{d\la},\\
\pa_r(v_e^\la(X))&=\la\frac{dv_e^\la}{d\la}-(\frac{2s}{p-1}+2)v_e^\la,\\
\pa_r(w_e^\la(X))&=\la\frac{d w_e^\la}{d\la}-(\frac{2s}{p-1}+4)w_e^\la.\\
\endaligned
\ee
Plugging these equations into \eqref{8overee}, we get that
\be\label{8overlineE}\aligned
\frac{d}{d\la}\overline{E}(u_e^\la,1)=&\int_{\R^{n+1}_+\cap \pa B_1}y^b\big(\la\frac{dv_e^\la}{d\la}\frac{dv_e^\la}{d\la}-(\frac{2s}{p-1}+2)v_e^\la\frac{dv_e^\la}{d\la}\big)\\
&+y^b\big(\la\frac{d w_e^\la}{d\la}\frac{u_e^\la}{d\la}-(\frac{2s}{p-1}+4)w_e^\la\frac{du_e^\la}{d\la}\big)\\
&-y^b\big(\la w_e^\la\frac{d^2 u_e^\la}{d\la^2}+(1-\frac{2s}{p-1}w_e^\la\frac{du_e^\la}{d\la}\big)\\
 =&\underbrace{\int_{\R^{n+1}_+\cap \pa B_1}y^b\big[\la\frac{dv_e^\la}{d\la}\frac{dv_e^\la}{d\la}-(\frac{2s}{p-1}+2)v_e^\la\frac{dv_e^\la}{d\la}\big]}\\
&+\underbrace{y^b\big[\la \frac{d w_e^\la}{d\la}\frac{du_e^\la}{d\la}-\la w_e^\la\frac{d^2u_e^\la}{d\la^2}\big]-5y^bw_e^\la\frac{du_e^\la}{d\la}}\\
:=&\overline{E}_{d1}(u_e^\la,1)+\overline{E}_{d2}(u_e^\la,2).
\endaligned
\ee

\subsection{The calculations  of $\frac{\pa^j}{\pa r^j}u_e^\la$  and  $\frac{\pa^i}{\pa \la^i}u_e^\la,\; i,j=1,2,3,4$}

Note
\be\label{8ue0}
\la\frac{du_e^\la}{d\la}=\frac{2s}{p-1}u_e^\la+r\frac{\pa}{\pa r}u_e^\la.
\ee
Differentiating \eqref{8ue0} once, twice and  thrice with respect to $\la$ respectively,  we have
\be\label{8ue1}
\la\frac{d^2u_e^\la}{d\la^2}+\frac{du_e^\la}{d\la}=\frac{2s}{p-1}\frac{du_e^\la}{d\la}+r\frac{\pa}{\pa r}\frac{du_e^\la}{d\la},
\ee
\be\label{8ue2}
\la\frac{d^3u_e^\la}{d\la^3}+2\frac{d^2u_e^\la}{d\la^2}=\frac{2s}{p-1}\frac{d^2u_e^\la}{d\la^2}+r\frac{\pa}{\pa r}\frac{d^2u_e^\la}{d\la^2},
\ee
\be\label{8ue3}
\la\frac{d^4u_e^\la}{d\la^4}+3\frac{d^3u_e^\la}{d\la^3}=\frac{2s}{p-1}\frac{d^3u_e^\la}{d\la^3}+r\frac{\pa}{\pa r}\frac{d^3u_e^\la}{d\la^3}.
\ee
Similarly, differentiating \eqref{8ue0} once, twice and  thrice with respect to $r$ respectively we have
\be\label{8ue4}
\la\frac{\pa}{\pa r}\frac{du_e^\la}{d\la}=(\frac{2s}{p-1}+1)\frac{\pa}{\pa r}u_e^\la+r\frac{\pa^2}{\pa r^2}u_e^\la,
\ee
\be\label{8ue5}
\la\frac{\pa^2}{\pa r^2}\frac{du_e^\la}{d\la}=(\frac{2s}{p-1}+2)\frac{\pa^2}{\pa r^2}u_e^\la+r\frac{\pa^3}{\pa r^3}u_e^\la,
\ee
\be\label{8ue6}
\la\frac{\pa^3}{\pa r^3}\frac{du_e^\la}{d\la}=(\frac{2s}{p-1}+3)\frac{\pa^3}{\pa r^3}u_e^\la+r\frac{\pa^4}{\pa r^4}u_e^\la.
\ee
From \eqref{8ue0}, on $\R^{n+1}_+\cap \pa B_1$, we have
\be\nonumber
\frac{\pa u_e^\la}{\pa r}=\la \frac{du_e^\la}{d\la}-\frac{2s}{p-1}u_e^\la.
\ee
Next from \eqref{8ue1}, on $\R^{n+1}_+\cap \pa B_1$,  we derive that
\be\nonumber
\frac{\pa}{\pa r}\frac{d u_e^\la}{d\la}=\la\frac{d^2u_e^\la}{d\la^2}+(1-\frac{2s}{p-1})\frac{du_e^\la}{d\la}.
\ee
From \eqref{8ue4}, combine with the two equations above, on $\R^{n+1}_+\cap \pa B_1$, we get
\be\label{8r31}\aligned
\frac{\pa^2}{\pa r^2} u_e^\la&=\la\frac{\pa}{\pa r}\frac{du_e^\la}{d\la}-(1+\frac{2s}{p-1})\frac{\pa}{\pa r} u_e^\la\\
&=\la^2\frac{d^2 u_e^\la}{d\la^2}-\la\frac{4s}{p-1}\frac{du_e^\la}{d\la}+(1+\frac{2s}{p-1})\frac{2s}{p-1}u_e^\la.
\endaligned\ee
Differentiating \eqref{8ue1} with respect to $r$, and combine with \eqref{8ue1} and \eqref{8ue2}, we get that
\be\label{8r32}\aligned
\frac{\pa^2}{\pa r^2}\frac{du_e^\la}{d\la}&=\la\frac{\pa}{\pa r}\frac{d^2 u_e^\la}{d\la^2}-\frac{2s}{p-1}\frac{\pa}{\pa r}\frac{du_e^\la}{d\la}\\
&=\la^2\frac{d^3 u_e^\la}{d\la^3}+(2-\frac{4s}{p-1})\la\frac{d^2 u_e^\la}{d\la^2}-(1-\frac{2s}{p-1})\frac{2s}{p-1}\frac{du_e^\la}{d\la}.
\endaligned\ee
From \eqref{8ue5}, on $\R^{n+1}_+\cap \pa B_1$, combine with \eqref{8r31} and \eqref{8r32}, we have
\be\label{8ue7}\aligned
\frac{\pa^3}{\pa r^3}u_e^\la=&\la\frac{\pa^2}{\pa r^2}\frac{du_e^\la}{d\la}-(2+\frac{2s}{p-1})\frac{\pa^2}{\pa r^2}u_e^\la\\
=&\la^3\frac{d^3 u_e^\la}{d\la^3}-\la^2\frac{6s}{p-1}\frac{d^2 u_e^\la}{d\la^2}+\la(\frac{6s}{p-1}+\frac{12s^2}{(p-1)^2})\frac{du_e^\la}{d\la}\\
&-(2+\frac{2s}{p-1})(1+\frac{2s}{p-1})\frac{2s}{p-1}u_e^\la.
\endaligned\ee
Now differentiating \eqref{8ue1} once with respect to $r$, we get
\be\nonumber
\la\frac{\pa^2}{\pa r^2}\frac{d^2 u_e^\la}{d\la^2}=(\frac{2s}{p-1}+1)\frac{\pa^2}{\pa r^2}\frac{du_e^\la}{d\la}+r\frac{\pa^3}{\pa r^3}\frac{du_e^\la}{d\la},
\ee
then on $\R^{n+1}_+\cap \pa B_1$, we have
\be\label{8ue8}
\frac{\pa^3}{\pa r^3}\frac{du_e^\la}{d\la}=\la\frac{\pa^2}{\pa r^2}\frac{d^2 u_e^\la}{d\la^2}-(\frac{2s}{p-1}+1)\frac{\pa^2}{\pa r^2}\frac{du_e^\la}{d\la}.
\ee
Now differentiating \eqref{8ue2} twice with respect to $r$, we get
\be\nonumber
\la\frac{\pa}{\pa r}\frac{d^3 u_e^\la}{d\la^3}=(\frac{2s}{p-1}-1)\frac{\pa}{\pa r}\frac{d^2 u_e^\la}{d\la^2}+r\frac{\pa^2}{\pa r^2}\frac{d^2 u_e^\la}{d\la^2},
\ee
hence on $\R^{n+1}_+\cap \pa B_1$, combine with \eqref{8ue2} and \eqref{8ue3} there holds
\be\label{8ue9}
\aligned
\frac{\pa^2}{\pa r^2}&\frac{d^2 u_e^\la}{d\la^2}=\la\frac{\pa}{\pa r}\frac{d^3 u_e^\la}{d\la^3}+(1-\frac{2s}{p-1})\frac{\pa}{\pa r}\frac{d^2 u_e^\la}{d\la^2}\\
=&\la^2\frac{d^4 u_e^\la}{d\la^4}+\la(4-\frac{4s}{p-1})\frac{d^3 u_e^\la}{d\la^3}+(1-\frac{2s}{p-1})(2-\frac{2s}{p-1})\frac{d^2 u_e^\la}{d\la^2}.
\endaligned\ee
Now differentiating \eqref{8ue1} with respect to $r$, we have
\be\nonumber
\la\frac{\pa}{\pa r}\frac{d^2 u_e^\la}{d\la^2}=\frac{2s}{p-1}\frac{\pa}{\pa r}\frac{du_e^\la}{d\la}+r\frac{\pa^2}{\pa r^2}\frac{du_e^\la}{d\la}.
\ee
This combine with \eqref{8ue1} and \eqref{8ue2}, on $\R^{n+1}_+\cap \pa B_1$, we have
\be\label{8ue10}\aligned
\frac{\pa^2}{\pa r^2}\frac{du_e^\la}{d\la}&=\la\frac{\pa}{\pa r}\frac{d^2 u_e^\la}{d\la^2}-\frac{2s}{p-1}\frac{\pa}{\pa r}\frac{d u_e^\la}{d\la}\\
&=\la^2\frac{d^3 u_e^\la}{d\la^3}+\la(2-\frac{4s}{p-1})\frac{d^2 u_e^\la}{d\la^2}-\frac{2s}{p-1}(1-\frac{2s}{p-1})\frac{du_e^\la}{d\la}.
\endaligned\ee
Now from \eqref{8ue8}, combine with \eqref{8ue9} and \eqref{8ue10}, we get
\be\label{8ue11}\aligned
\frac{\pa^3}{\pa r^3}\frac{du_e^\la}{d\la}=&\la^3\frac{d^4 u_e^\la}{d\la^4}+\la^2(3-\frac{6s}{p-1})\frac{d^3 u_e^\la}{d\la^3}-\la(1-\frac{2s}{p-1})\frac{6s}{p-1}\frac{d^2 u_e^\la}{d\la^2}\\
&+(1-\frac{2s}{p-1})(1+\frac{2s}{p-1})\frac{2s}{p-1}\frac{du_e^\la}{d\la}.
\endaligned\ee
From \eqref{8ue6}, on $\R^{n+1}_+\cap \pa B_1$, combine with \eqref{8ue11} then
\be\nonumber\aligned
\frac{\pa^4}{\pa r^4}u_e^\la=&\la \frac{\pa^3}{\pa r^3}\frac{du_e^\la}{d\la}-(3+\frac{2s}{p-1})\frac{\pa^3}{\pa r^3}u_e^\la\\
=&\la^4\frac{d^4 u_e^\la}{d\la^4}-\la^3\frac{8s}{p-1}\frac{d^3 u_e^\la}{d\la^3}+\la^2(2+\frac{4s}{p-1})\frac{6s}{p-1}\frac{d^2 u_e^\la}{d\la^2}\\
&-\la (1+\frac{2s}{p-1})(1+\frac{s}{p-1})\frac{16s}{p-1}\frac{du_e^\la}{d\la}\\
&+(3+\frac{2s}{p-1})(2+\frac{2s}{p-1})(1+\frac{2s}{p-1})\frac{2s}{p-1}u_e^\la.\\
\endaligned\ee
In summary, we have that
\be\nonumber\aligned
\frac{\pa^3}{\pa r^3}u_e^\la
=&\la^3\frac{d^3 u_e^\la}{d\la^3}-\la^2\frac{6s}{p-1}\frac{d^2 u_e^\la}{d\la^2}+\la(\frac{6s}{p-1}+\frac{12s^2}{(p-1)^2})\frac{du_e^\la}{d\la}\\
&-(2+\frac{2s}{p-1})(1+\frac{2s}{p-1})\frac{2s}{p-1}u_e^\la
\endaligned\ee
and
\be\nonumber
\frac{\pa ^2}{\pa r^2}u_e^\la=\la^2\frac{d^2 u_e^\la}{d\la^2}-\la\frac{4s}{p-1}\frac{du_e^\la}{d\la}+(1+\frac{2s}{p-1})\frac{2s}{p-1}u_e^\la
\ee
\be\nonumber
\frac{\pa u_e^\la}{\pa r}=\la \frac{du_e^\la}{d\la}-\frac{2s}{p-1}u_e^\la.
\ee
\subsection{On the operator $\Delta_b^2$  and its representation }

Note that
\be\nonumber\aligned
\Delta_b u=&y^{-b}\nabla\cdot(y^b\nabla u)
=&u_{rr}+\frac{n+b}{r}u_r+\frac{1}{r^2}\theta_1^{-b}\mathbf{ div} _{S^n}(\theta_1^b\nabla_{S^n}u),
\endaligned\ee
where $\theta_1=\frac{y}{r},r=\sqrt{|x|^2+y^2}$. Set $v=\Delta_b u$  and $\Delta_b^2 u:=w$.
Then
\be\nonumber\aligned
w=&\Delta_b v=v_{rr}+\frac{n+b}{r}v_r+\frac{1}{r^2}\theta_1^{-b} \mathbf{ div} _{S^n}(\theta_1^b\nabla_{S^n}v)\\
=&\pa_{rrrr}u+\frac{2(n+b)}{r}\pa_{rrr}u+\frac{(n+b)(n+b-2)}{r^2}\pa_{rr}u-\frac{(n+b)(n+b-2)}{r^3}\pa_r u\\
&+r^{-4}\theta_1^{-b}{\bf div}_{S^n}(\theta_1^b\nabla(\theta_1^{-b}\mathbf{ div}_{S^n}(\theta_1^b\nabla_{S^n}u))\\
&+2r^{-2}\theta_1^{-b}{\bf div}_{S^n}(\theta_1^b\nabla_{S^n}(u_{rr}+\frac{n+b-2}{r}u_r))\\
&-2(n+b-3)r^{-4}\theta_1^{-b}\mathbf{ div}_{S^n}(\theta_1^b\nabla_{S^n}u).\\
\endaligned\ee
On $\R^{n+1}_+\cap \pa B_1$, we have
\be\nonumber\aligned
w=&\underbrace{\pa_{rrrr}u+2(n+b)\pa_{rrr}u+(n+b)(n+b-2)\pa_{rr}u-(n+b)(n+b-2)\pa_r u}\\
&\underbrace{+\theta_1^{-b}\mathbf{ div}_{S^n}(\theta_1^b\nabla(\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u))}\\
&\underbrace{+2\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}(u_{rr}+\frac{n+b-2}{r}u_r))}\\
&\underbrace{-2(n+b-3)\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u)}\\
 :=&I(u)+J(u)+K(u)+L(u).\\
\endaligned\ee
By these notations, we can rewrite the term $\overline{E}_{d2}(u_e^\la,1)$ appear in \eqref{8overlineE} as following
\be\label{8ijk}\aligned
&\overline{E}_{d2}(u_e^\la,1)\\
&=\int_{\R^{n+1}_+\cap \pa B_1}\theta_1^b(\la\frac{d w_e^\la}{d\la}\frac{d u_e^\la}{d\la}-\la w_e^\la\frac{d^2 u_e^\la}{d\la^2})-5\theta_1^b w_e^\la\frac{d u_e^\la}{d\la}\\
&=\underbrace{\int_{\R^{n+1}_+\cap\pa B_1}\la\theta_1^b\frac{d}{d\la}I(u_e^\la)\frac{d u_e^\la}{d\la}-\la\theta_1^b I(u_e^\la)\frac{d^2 u_e^\la}{d\la^2}-5\theta_1^bI(u_e^\la)\frac{du_e^\la}{d\la}}\\
&\quad \underbrace{+\int_{\R^{n+1}_+\cap\pa B_1}\la\theta_1^b\frac{d}{d\la}J(u_e^\la)\frac{d u_e^\la}{d\la}-\la\theta_1^b J(u_e^\la)\frac{d^2 u_e^\la}{d\la^2}-5\theta_1^bJ(u_e^\la)\frac{du_e^\la}{d\la}}\\
&\quad \underbrace{+\int_{\R^{n+1}_+\cap\pa B_1}\la\theta_1^b\frac{d}{d\la}K(u_e^\la)\frac{d u_e^\la}{d\la}-\la\theta_1^b K(u_e^\la)\frac{d^2 u_e^\la}{d\la^2}-5\theta_1^bK(u_e^\la)\frac{du_e^\la}{d\la}}\\
&\quad \underbrace{+\int_{\R^{n+1}_+\cap \pa B_1}\la\theta_1^b\frac{d}{d\la}L(u_e^\la)\frac{du_e^\la}{d\la}-\la\theta_1^bL(u_e^\la)\frac{d^2u_e^\la}{d\la^2}-5\theta_1^bL(u_e^\la)\frac{du_e^\la}{d\la}},
\endaligned\ee
we define as
\be\nonumber\aligned
\overline{E}_{d2}(u_e^\la,1):&=\mathcal{I}+\mathcal{J}+\mathcal{K}+\mathcal{L}\\
:&=I_1+I_2+I_3+J_1+J_2+J_3+K_1+K_2+K_3+L_1+L_2+L_3.
\endaligned\ee
where $I_1,I_2,I_3,J_1,J_2,J_3,K_1,K_2,K_3,L_1,L_2,L_3 $ are corresponding successively to the $12$ terms in \eqref{8ijk}.
By the conclusions of subsection $2.2$, we have
\be\label{8iuiu}\aligned
I(u_e^\la)&=\pa_{rrrr} u_e^\la+2(n+b)\pa_{rrr} u_e^\la\\
&\quad +(n+b)(n+b-2)\pa_{rr} u_e^\la-(n+b)(n+b-2)\pa_r u_e^\la\\
&=\la^4\frac{d^4 u_e^\la}{d\la^4}+\la^3\big(2(n+b)-\frac{8s}{p-1})\big)\frac{d^3 u_e^\la}{d\la^3}\\
&\quad+\la^2\big[\frac{12s}{p-1}(1+\frac{2s}{p-1})-(n+b)\frac{12s}{p-1}+(n+b)(n+b-2)\big]\frac{d^2 u_e^\la}{d\la^2}\\
&\quad+\la\big[-\frac{8s}{p-1}(1+\frac{2s}{p-1})(2+\frac{2s}{p-1})+2(n+b)\frac{6s}{p-1}(1+\frac{2s}{p-1})\\
&\quad+(n+b)(n+b-2)(-\frac{4s}{p-1}-1)\big]\frac{d u_e^\la}{d\la}\\
&\quad+\big[(1+\frac{2s}{p-1})(2+\frac{2s}{p-1})(3+\frac{2s}{p-1})\frac{2s}{p-1}\\
&\quad-(n+b)(1+\frac{2s}{p-1})(2+\frac{2s}{p-1})\frac{4s}{p-1}\\
&\quad+(n+b)(n+b-2)(\frac{2s}{p-1}+2)\frac{2s}{p-1}\big]u_e^\la.\\
\endaligned\ee
For convenience, we denote that
\be\label{8iue}
I(u_e^\la)=\la^4\frac{d^4 u_e^\la}{d\la^4}+\la^3 \delta_1\frac{d^3 u_e^\la}{d\la^3}+\la^2\delta_2\frac{d^2 u_e^\la}{d\la^2}+\la\delta_3\frac{du_e^\la}{d\la}+\delta_4 u_e^\la,
\ee
where $\delta_i$ are the corresponding coefficients of $\la^i\frac{d^i u_e^\la}{d\la^i}$ appeared in \eqref{8iuiu} for $i=1,2,3,4$.
Now taking the derivative of \eqref{8iue} with respect to $\la$, we get
\be\label{8diue}\aligned
\frac{d}{d\la}I(u_e^\la)=&\la^4\frac{d^5 u_e^\la}{d\la^5}+\la^3 (\delta_1+4)\frac{d^4 u_e^\la}{d\la^4}+\la^2(3\delta_1+\delta_2)\frac{d^3 u_e^\la}{d\la^3}\\
&+\la(2\delta_2+\delta_3)\frac{d^2u_e^\la}{d\la^2}+(\delta_3+\delta_4)\frac{du_e^\la}{d\la}
\endaligned\ee
and
\be\label{8urr}\aligned
\pa_{rr}& u_e^\la+(n+b-2)\pa_r u_e^\la \\ =&\la^2\frac{d^2u_e^\la}{d\la^2}+\la(n+b-2-\frac{4s}{p-1})\frac{du_e^\la}{d\la}+\frac{2s}{p-1}(3+\frac{2s}{p-1}-n-b)u_e^\la\\
:=&\la^2\frac{d^2 u_e^\la}{d\la^2}+\la\alpha\frac{du_e^\la}{d\la}+\beta u_e^\la.
\endaligned\ee
Hence,
\be\label{8urrla}\aligned
\frac{d}{d\la}[\pa_{rr}& u_e^\la+(n+b-2)\pa_r u_e^\la]=\la^2\frac{d^3 u_e^\la}{d\la^3}+\la(\alpha+2)\frac{d^2 u_e^\la}{d\la^2}+(\alpha+\beta)\frac{du_e^\la}{d\la},
\endaligned\ee
here $\alpha=n+b-2-\frac{4s}{p-1}$ and $\beta=\frac{2s}{p-1}(3+\frac{2s}{p-1}-n-b)$.

\subsection{The computations  of $I_1,I_2,I_3$ and $\mathcal{I}$}

\be\label{8i1}\aligned
I_1:=&\int_{\R^{n+1}_+\cap\pa B_1}\la \theta_1^b\frac{d}{d\la}I(u_e^\la)\frac{du_e^\la}{d\la}\\
=&\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big(\la^5\frac{d^5 u_e^\la}{d\la^5}+\la^4(4+\delta_1)\frac{d^4 u_e^\la}{d\la^4}+\la^3(3\delta_1+\delta_2)\frac{d^3 u_e^\la}{d\la^3}\\
&+\la^2(2\delta_2+\delta_3)\frac{d^2 u_e^\la}{d\la^2}+\la(\delta_3+\delta_4)\frac{du_e^\la}{d\la}\big)\frac{du_e^\la}{d\la}\\
=&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[\la^5\frac{d^4 u_e^\la}{d\la^4}\frac{d u_e^\la}{d\la}-\la^5\frac{d^3 u_e^\la}{d\la^3}\frac{d^2 u_e^\la}{d\la^2}+(\delta_1-1)\la^4\frac{d^3 u_e^\la}{d\la^3}\frac{d u_e^\la}{d\la}\\
&+(4-\delta_1+\delta_2)\la^3\frac{d^2 u_e^\la}{d\la^2}\frac{d u_e^\la}{d\la}+\frac{3\delta_1-\delta_2+\delta_3-12}{2}\la^2(\frac{d u_e^\la}{d\la})^2\big]\\
&+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b \big[(12-3\delta_1+\delta_2+\delta_4)\la(\frac{d u_e^\la}{d\la})^2\\
&+(\delta_1-4-\delta_2)\la^3(\frac{d^2 u_e^\la}{d\la^2})^2+\la^5(\frac{d^3 u_e^\la}{d\la^3})^2\big]\\
&+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(6-\delta_1)\la^4\frac{d^3 u_e^\la}{d\la^3}\frac{d^2 u_e^\la}{d\la^2},
\endaligned
\ee
where $\delta_i (i=1,2,3,4) $ are defined in \eqref{8iuiu} and \eqref{8iue}. Denote   $f=u_e^\la,f':=\frac{du_e^\la}{d\la}$,   we have  used  the following differential identities:
\be\nonumber\aligned
\la^5 f'''''f'=&\big[\la^5f''''f'-\la^5f'''f''-5\la^4f'''f'+20\la^3f''f'-30\la^2f'f'\big]'\\
&+60\la(f')^2-20\la^3(f'')^2+\la^5(f''')^2+10\la^4f'''f'',
\endaligned\ee
\be\nonumber\aligned
\la^4f''''f'=\big[\la^4f'''f'-4\la^3f''f'+6\la^2f'f'\big]'-12\la(f')^2+4\la^3(f'')^2-\la^4f'''f'',
\endaligned\ee
\be\nonumber\aligned
\la^3f'''f'=\big[\la^3f''f'-\frac{3\la^2}{2}f'f'\big]'+3\la(f')^2-\la^3(f'')^2,
\endaligned\ee
and
\be\nonumber\aligned
\la^2 f''f'=\big[\frac{\la^2}{2}f'f'\big]'-\la(f')^2.
\endaligned\ee

\be\label{8i2}\aligned
&I_2:=-\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b I(u_e^\la)\frac{d^2 u_e^\la}{d\la^2}\\
&=-\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b \big(\la^4\frac{d^4 u_e^\la}{d\la^4}+\la^3 \delta_1\frac{d^3 u_e^\la}{d\la^3}+\la^2\delta_2\frac{d^2 u_e^\la}{d\la^2}+\la\delta_3\frac{du_e^\la}{d\la}+\delta_4 u_e^\la\big)\frac{d^2 u_e^\la}{d\la^2}\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[-\la^5\frac{d^3 u_e^\la}{d\la^3}\frac{d^2 u_e^\la}{d\la^2}-\delta_4\la\frac{d u_e^\la}{d\la}u_e^\la\big]\\
&\quad+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[\la^5(\frac{d^3 u_e^\la}{d\la^3})^2-\delta_2\la^3(\frac{d^2 u_e^\la}{d\la^2})^2+\delta_4\la(\frac{d u_e^\la}{d\la})^2\big]\\
&\quad +\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[(5-\delta_1)\la^4\frac{d^3 u_e^\la}{d\la^3}\frac{d^2 u_e^\la}{d\la^2}-\delta_3\la^2\frac{d^2 u_e^\la}{d\la^2}\frac{d u_e^\la}{d\la}+\delta_4\frac{d u_e^\la}{d\la}u_e^\la\big].
\endaligned\ee
Here we have used that
\be\nonumber\aligned
-\la^5f''''f''=\big[-\la^5f'''f''\big]'+5\la^4f'''f''+\la^5(f''')^2
\endaligned\ee
and
\be\nonumber\aligned
-\la f''f=[-\la f'f]'+f'f+\la(f')^2.
\endaligned\ee

\be\label{8iu3}\aligned
I_3:=&-5\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b I(u_e^\la)\frac{d u_e^\la}{d\la}\\
=&-5\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[\la^4\frac{d^4 u_e^\la}{d\la^4}+\la^3 \delta_1\frac{d^3 u_e^\la}{d\la^3}+\la^2\delta_2\frac{d^2 u_e^\la}{d\la^2}+\la\delta_3\frac{du_e^\la}{d\la}+\delta_4 u_e^\la\big]\frac{d u_e^\la}{d\la}\\
=&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[-5\frac{d^3 u_e^\la}{d\la^3}\frac{d u_e^\la}{d\la}+(20-5\delta_1)\la^3\frac{d^2 u_e^\la}{d\la^2}\frac{d u_e^\la}{d\la}\big]\\
&+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[(5\delta_1-20)\la^3(\frac{d^2 u_e^\la}{d\la^2})^2-5\delta_3\la(\frac{d u_e^\la}{d\la})^2\big]\\
&+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[5\la^4\frac{d^3 u_e^\la}{d\la^3}\frac{d^2 u_e^\la}{d\la^2}+(15\delta_1-60-5\delta_2)\frac{d^2 u_e^\la}{d\la^2}\frac{d u_e^\la}{d\la}-5\delta_4\frac{d u_e^\la}{d\la}u_e^\la\big].
\endaligned\ee
Here we have use that
\be\nonumber\aligned
-\la^4 f''''f'=\big[-5\la^4f'''f'+20\la^3f''f'\big]'-20\la^3(f'')^2-60\la^2f''f'+5\la^4f'''f''
\endaligned\ee
and
\be\nonumber\aligned
-\la^3 f'''f'=\big[-\la^3f''f'\big]'+3\la^2 f''f'+\la^3 (f'')^2.
\endaligned\ee
Now we add up $I_1,I_2, I_3$ and further integrate by part, we can get the term $\mathcal{I}$.
\be\label{8I}\aligned
\mathcal{I}:=& I_1+I_2+I_3\\
=&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[\la^5\frac{d^4 u_e^\la}{d\la^4}\frac{d u_e^\la}{d\la}-2\la^5\frac{d^3 u_e^\la}{d\la^3}\frac{d^2 u_e^\la}{d\la^2}\\
&+(\delta_1-6)\la^4\frac{d^3 u_e^\la}{d\la^3}\frac{d u_e^\la}{d\la}+(24-6\delta_1+\delta_2)\la^3\frac{d^2 u_e^\la}{d\la^2}\frac{d u_e^\la}{d\la}\\
&+(9\delta_1-3\delta_2-36)\la^2\frac{d u_e^\la}{d\la}\frac{d u_e^\la}{d\la}\\
&+(8-\delta_1)\la^4(\frac{d^2 u_e^\la}{d\la^2})^2-\delta_4\la\frac{d u_e^\la}{d\la}u_e^\la-2\delta_4(u_e^\la)^2\big]\\
&+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\Big(
2\la^5(\frac{d^3u_e^\la}{d\la^3})^2+(10\delta_1-2\delta_2-56)\la^3(\frac{d^2u_e^\la}{d\la^2})^2\\
&+(-18\delta_1+\delta_2-4\delta_3+2\delta_4+72)\la(\frac{du_e^\la}{d\la})^2
\Big).
\endaligned\ee
Since $u_e^\la(X)=\la^{\frac{2s}{p-1}}u_e(\la X)$, we have the following
\be\nonumber\aligned
\la^4&\frac{d^4 u_e^\la}{d\la^4}=\la^{\frac{2s}{p-1}}\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1)(\frac{2s}{p-1}-2)(\frac{2s}{p-1}-3)u_e(\la X)\\
&+\frac{8s}{p-1}(\frac{2s}{p-1}-1)(\frac{2s}{p-1}-2)r\la\pa_r u_e(\la X)\\
&+\frac{12s}{p-1}(\frac{2s}{p-1}-1)r^2\la^2\pa_{rr}u_e(\la X)\\
&+\frac{8s}{p-1}r^3\la^3\pa_{rrr}u_e(\la X)+r^4\la^4\pa_{rrrr}u_e(\la X)\big],\\
\endaligned\ee
and
\be\nonumber\aligned
\la^3\frac{d^3 u_e^\la}{d\la^3}&=\la^{\frac{2s}{p-1}}\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1)(\frac{2s}{p-1}-2)u_e(\la X)\\
&+\frac{6s}{p-1}(\frac{2s}{p-1}-1)r\la\pa_r u_e(\la X)\\
&+\frac{6s}{p-1}r^2\la^2 \pa_{rr} u_e(\la X)+r^3\la^3\pa_{rrr} u_e(\la X)\big],
\endaligned\ee
\be\nonumber\aligned
&\la^2\frac{d^2 u_e^\la}{d\la^2}\\
&=\la^{\frac{2s}{p-1}}\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1) u_e(\la X)+\frac{4s}{p-1}r\la\pa_r u_e(\la X)+r^2\la^2\pa_{rr}u_e(\la X)\big]
\endaligned\ee
and
\be\nonumber\aligned
\la\frac{d u_e^\la}{d\la}=\la^{\frac{2s}{p-1}}\big[\frac{2s}{p-1}u_e(\la X)+r\la\pa_r u_e(\la X)\big].
\endaligned\ee
Hence,  by scaling we have the following derivatives:
\be\nonumber\aligned
\frac{d}{d\la}&\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^5\frac{d^4 u_e^\la}{d\la^4}\frac{d u_e^\la}{d\la}\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1)(\frac{2s}{p-1}-2)(\frac{2s}{p-1}-3)u_e\\
&\quad+\frac{8s}{p-1}(\frac{2s}{p-1}-1)(\frac{2s}{p-1}-2)\la\pa_r u_e+\frac{12s}{p-1}(\frac{2s}{p-1}-1)\la^2\pa_{rr}u_e\\
&\quad+\frac{8s}{p-1}\la^3\pa_{rrr}u_e+\la^4\pa_{rrrr}u_e\big]\big[\frac{2s}{p-1}u_e+r\la\pa_r u_e\big];
\endaligned\ee

\be\label{8Iscaling}\aligned
\frac{d}{d\la}&\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^5\frac{d^3 u_e^\la}{d\la^3}\frac{d^2 u_e^\la}{d\la^2}\quad\quad\quad\quad\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1)(\frac{2s}{p-1}-2)u_e\\
&+\frac{6s}{p-1}(\frac{2s}{p-1}-1)\la\pa_r u_e
+\frac{6s}{p-1}\la^2 \pa_{rr} u_e+\la^3\pa_{rrr} u_e\big]\\
&\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1) u_e+\frac{4s}{p-1}\la\pa_r u_e+\la^2\pa_{rr}u_e\big];
\endaligned\ee

\be\nonumber\aligned
\frac{d}{d\la}&\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^4\frac{d^3 u_e^\la}{d\la^3}\frac{d u_e^\la}{d\la}\quad\quad\quad\quad\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1)(\frac{2s}{p-1}-2)u_e\\
&+\frac{6s}{p-1}(\frac{2s}{p-1}-1)\la\pa_r u_e
+\frac{6s}{p-1}\la^2 \pa_{rr} u_e+\la^3\pa_{rrr} u_e\big]\\
&\big[\frac{2s}{p-1}u_e+\la\pa_r u_e\big];
\endaligned\ee

\be\nonumber\aligned
\frac{d}{d\la}&\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^3\frac{d^2 u_e^\la}{d\la^2}\frac{d u_e^\la}{d\la}\quad\quad\quad\quad\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b
\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1) u_e\\
&+\frac{4s}{p-1}\la\pa_r u_e
+\la^2\pa_{rr}u_e][\frac{2s}{p-1}u_e+\la\pa_r u_e\big]
\endaligned\ee
and
\be\nonumber\aligned
\frac{d}{d\la}&\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^2\frac{d u_e^\la}{d\la}\frac{d u_e^\la}{d\la}\quad\quad\quad\quad\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b
\big[\frac{2s}{p-1}u_e+\la\pa_r u_e\big]^2.
\endaligned\ee
Further,
\be\nonumber\aligned
\frac{d}{d\la}&\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^4\frac{d^2 u_e^\la}{d\la^2}\frac{d^2 u_e^\la}{d\la^2}\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b
\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1) u_e\\
&+\frac{4s}{p-1}\la\pa_r u_e+\la^2\pa_{rr}u_e\big]^2,
\endaligned\ee

\be\nonumber\aligned
\frac{d}{d\la}&\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la\frac{d u_e^\la}{d\la}u_e^\la\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b
\big[\frac{2s}{p-1}u_e+\la\pa_r u_e\big]u_e,
\endaligned\ee
and
\be\nonumber\aligned
\frac{d}{d\la}&\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b u_e^\la=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b
u_e^2.
\endaligned\ee

\subsection{The computations of $J_i,K_i,L_i (i=1,2,3)$ and $\mathcal{J},\mathcal{K},\mathcal{L}$}
Firstly,
\be\label{8j1}\aligned
J_1:=&\int_{\R^{n+1}_+\cap\pa B_1}\la\theta_1^b \frac{d}{d\la}J(u_e^\la)\frac{du_e^\la}{d\la}=\int_{\R^{n+1}_+\cap\pa B_1}\la\theta_1^b J(\frac{du_e^\la}{d\la})\frac{du_e^\la}{d\la}\\
=&\la\int_{\R^{n+1}_+\cap\pa B_1} \mathbf{div}_{S^n}\big(\theta_1^b\nabla_{S^n}(\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}\frac{du_e^\la}{d\la}))\big)\frac{du_e^\la}{d\la}\\
=&-\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\nabla_{S^n}\big(\theta_1^{-b} \mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}\frac{du_e^\la}{d\la})\big)\nabla_{S^n}\frac{du_e^\la}{d\la}\\
=&\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{-b}\big[\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}\frac{du_e^\la}{d\la})\big]^2\\
=&\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\Delta_{S^n}\frac{du_e^\la}{d\la})^2,
 \endaligned\ee
here we have used integrate by part formula on the unit  sphere $S^n$.
\be\label{8j2}\aligned
J_2:=&-\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b J(u_e^\la)\frac{d^2 u_e^\la}{d\la^2}\\
=&-\la\int_{\R^{n+1}_+\cap\pa B_1}\mathbf{div}_{S^n}\big(\theta_1^b\nabla_{S^n}(\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la))\big)\frac{d^2 u_e^\la}{d\la^2}\\
=&\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\nabla_{S^n}(\theta_1^{-b}\mathbf{ div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\nabla_{S^n}\frac{d^2u_e^\la}{d\la^2}\\
=&-\la\int_{\R^{n+1}_+\cap\pa B_1} \theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la) \frac{d^2}{d\la^2}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\\
=&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}-\la \theta_1^{-b}\big[\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\big]\frac{d}{d\la}\big[\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\big]\\
 &\quad+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{-b}\mathbf{ div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\cdot \frac{d}{d\la}\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\\
 &\quad+\la\int_{\R^{n+1}_+\cap\pa B_1} \theta_1^{-b}\big[\frac{d}{d\la}\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\big]^2\\
=&\frac{d}{d\la}\Big(\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\Big(
-\frac{1}{2}\la\frac{d}{d\la}(\Delta_{S^n}u_e^\la)^2+\frac{1}{2}(\Delta_{S^n}u_e^\la)^2
\Big)\Big)\\
&+\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{b}\la(\Delta_{S^n}\frac{du_e^\la}{d\la})^2,
 \endaligned\ee
here we denote that $g= \mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la),g'=\frac{d}{d\la}\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)$
and we have used that
\be\nonumber
-\la g g'=\big[-g g'\big]'+g g'+\la (g')^2=\big[-g g'+\frac{1}{2}g^2\big]'+\la(g')^2.
\ee
Further,
\be\label{8j3}\aligned
J_3:=&-5\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b J(u_e^\la)\frac{du_e^\la}{d\la}\\
=&-5\int_{\R^{n+1}_+\cap\pa B_1}\mathbf{div}_{S^n}\big(\theta_1^b \nabla_{S^n}(\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u))\big)\frac{du_e^\la}{d\la}\\
=&5\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b \nabla_{S^n}\big(\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\big)\nabla_{S^n}\frac{d u_e^\la}{d\la}\\
=&-5\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{-b}\mathbf{ div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\frac{d}{d\la}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\\
=&-\frac{5}{2}\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\Delta_{S^n}u_e^\la)^2.
 \endaligned\ee
Therefore, combine with \eqref{8j1}, \eqref{8j2} and \eqref{8j3}, we get that
\be\label{8j}\aligned
\mathcal{J}:=&J_1+J_2+J_3\\
=&2\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{-b}\big[\frac{d}{d\la}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\big]^2\\
 &\quad-4\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u)\frac{d}{d\la}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u)\\
 &\quad+\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}-\la \theta_1^b\big[\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\big]\frac{d}{d\la}\big[\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\big]\\
=&2\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{-b}\big[\frac{d}{d\la}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\big]^2\\
 &\quad-2\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\\
 &\quad+\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}-\la \theta_1^{-b}\big[\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\big]\frac{d}{d\la}\big[\mathbf{div}_{S^n}(\theta_1^b \nabla_{S^n}u_e^\la)\big]\\
=&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{b}\Big(-2(\Delta_{S^n}u_e^\la)^2-\frac{1}{2}\la\frac{d}{d\la}(\Delta_{S^n}u_e^\la)^2\Big)\\
 &\quad+2\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^{b}(\Delta_{S^n}u_e^\la)^2.
\endaligned\ee
Note that
\be\label{8Jscaling}\aligned
&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b \big[\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\big]^2\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}\big(\la^2\Delta_b u_e-\la^2\pa_{rr}u_e-(n+b)\la\pa_r u_e\big)^2,
\endaligned\ee
and
\be\nonumber\aligned
&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la \frac{d}{d\la}\big[\theta_1^{-b}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\big]^2\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-4}\frac{d}{d\la}\big(\la^2\Delta_b u_e-\la^2\pa_{rr}u_e-(n+b)\la\pa_r u_e\big)^2.
\endaligned\ee
Next we compute $K_1,K_2,K_3$ and $\mathcal{K}$.
\be\label{8k1}\aligned
&K_1\\
:=&\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\frac{d}{d\la}K(u_e^\la)\frac{du_e^\la}{d\la}\\
=&2\la\int_{\R^{n+1}_+\cap\pa B_1}\mathbf{div}_{S^n}\big(\theta_1^b\nabla_{S^n}(\frac{d}{d\la}(\pa_{rr}+(n+b-2)\pa_r)u_e^\la)\big)\frac{du_e^\la}{d\la}\\
=&2\la\int_{\R^{n+1}_+\cap\pa B_1}\mathbf{div}_{S^n}\big(\theta_1^b\nabla_{S^n}(\la^3\frac{d^3 u_e^\la}{d\la^3}+\la^2(\alpha+2)\frac{d^2 u_e^\la}{d\la^2}+\la(\alpha+\beta)\frac{d u_e^\la}{d\la})\big)\frac{du_e^\la}{d\la}\\
=&-2\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\nabla_{S^n}\big(\la^3\frac{d^3 u_e^\la}{d\la^3}+\la^2(\alpha+2)\frac{d^2 u_e^\la}{d\la^2}+\la(\alpha+\beta)\frac{d u_e^\la}{d\la}\big)\nabla_{S^n}\frac{du_e^\la}{d\la}\\
=&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}-\la^3\theta_1^b\frac{d}{d\la}\big(\frac{d}{d\la}\nabla_{S^n} u_e^\la\big)^2
+(2-2\alpha)\la^2\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b (\frac{d}{d\la}\nabla_{S^n} u_e^\la)\\
\cdot &(\frac{d^2}{d\la^2}\nabla_{S^n} u_e^\la)
+2\la^3\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\frac{d^2}{d\la^2}\nabla_{S^n} u_e^\la)^2\\
&-(2\alpha+2\beta)\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\frac{d}{d\la}\nabla_{S^n} u_e^\la)^2.
\endaligned\ee
Here we denote that $h=\nabla_{S^n} u_e^\la,h'=\frac{d}{d\la}\nabla_{S^n} u_e^\la$, and have used that
\be\nonumber
-\la^3h'h'''=\big[-\frac{\la^3}{2}\frac{d}{d\la}(h')^2\big]'+3\la^2 h'h''+\la^3(h'')^2.
\ee
Next,
\be\label{8k2}\aligned
K_2:=&-\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b K(u_e^\la)\frac{d^2 u_e^\la}{d\la^2}\\
=&-2\la\int_{\R^{n+1}_+\cap\pa B_1}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}\big(\la^2\frac{d^2 u_e^\la}{d\la^2}+\la\alpha\frac{d u_e^\la}{d\la}+\beta u_e^\la)\big)\frac{d^2 u_e^\la}{d\la^2}\\
=&2\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\nabla_{S^n}\big(\la^2\frac{d^2 u_e^\la}{d\la^2}+\la\alpha\frac{d u_e^\la}{d\la}+\beta u_e^\la\big)\nabla_{S^n}\frac{d^2 u_e^\la}{d\la^2}\\
=&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[2\beta\la\nabla_{S^n}u_e^\la \frac{d}{d\la}\nabla_{S^n}u_e^\la-\beta(\nabla_{S^n}u_e^\la)^2\big]\\
&+2\la^3\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\frac{d^2}{d\la^2}\nabla_{S^n}u_e^\la)^2-2\la\beta\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\frac{d}{d\la}\nabla_{S^n}u_e^\la)^2\\
&+2\la^2\alpha\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\frac{d}{d\la}\nabla_{S^n}u_e^\la\frac{d^2}{d\la^2}\nabla_{S^n}u_e^\la.
\endaligned\ee
Here we have used that
\be\nonumber
2\la h h''=[2\la h h'-h^2]'-2\la(h')^2.
\ee
Further,
\be\label{8k3}\aligned
K_3:=&-5\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b K(u_e^\la)\frac{du_e^\la}{d\la}\\
=&-10\int_{\R^{n+1}_+\cap\pa B_1}\mathbf{div}_{S^n}\big(\theta_1^b\nabla_{S^n}(\la^2\frac{d^2 u_e^\la}{d\la^2}+\la\alpha\frac{du_e^\la}{d\la}+\beta u_e^\la)\big)\frac{du_e^\la}{d\la}\\
=&10\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\nabla_{S^n}\big(\la^2\frac{d^2 u_e^\la}{d\la^2}+\la\alpha\frac{du_e^\la}{d\la}+\beta u_e^\la\big)\nabla_{S^n}\frac{du_e^\la}{d\la}\\
=&\frac{d}{d\la}\big[5\beta\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b \nabla_{S^n} u_e^\la \nabla_{S^n} u_e^\la\big]+10\la\alpha\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big(\frac{d}{d\la}\nabla_{S^n} u_e^\la\big)^2\\
&+10\la^2\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b \frac{d}{d\la}\nabla_{S^n} u_e^\la \frac{d^2}{d\la^2}\nabla_{S^n} u_e^\la.
\endaligned\ee
Now combine with \eqref{8k1}, \eqref{8k2} and \eqref{8k3}, we get that
\be\label{8k}\aligned
\mathcal{K}:=&K_1+K_2+K_3\\
=&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\Big[-\la^3\frac{d}{d\la}(\frac{d}{d\la}\nabla_{S^n} u_e^\la)^2\\
&+2\beta\la\nabla_{S^n} u_e^\la\frac{d}{d\la}\nabla_{S^n} u_e^\la+4\beta(\nabla_{S^n} u_e^\la)^2+6\la^2(\nabla_{S^n}\frac{du_e^\la}{d\la})^2\Big]\\
&+4\la^3\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\frac{d^2}{d\la^2}\nabla_{S^n} u_e^\la)^2\\
&+(8\alpha-4\beta-12)\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\frac{d}{d\la}\nabla_{S^n} u_e^\la)^2.\\
\endaligned\ee
Notice that by scaling we have
\be\label{8Kscaling}\aligned
&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\nabla_{S^n} u_e^\la)^2\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b\big[\la^2|\nabla u_e|^2-\la^2|\pa_r u_e|^2\big].
\endaligned\ee
\be\nonumber\aligned
&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la\frac{d}{d\la}(\nabla_{S^n} u_e^\la)^2\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-4}y^b\frac{d}{d\la}\big[\la^2|\nabla u_e|^2-\la^2|\pa_r u_e|^2\big]
\endaligned\ee
and
\be\nonumber\aligned
&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\la^3\frac{d}{d\la}(\frac{d}{d\la}\nabla_{S^n} u_e^\la)^2\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-4}y^b\frac{d}{d\la}\big[\frac{2s}{p-1}\la\nabla u_e+\la^2\nabla \pa_r u_e\big]^2.
\endaligned\ee
Finally, we compute $\mathcal{L}$.
\be\nonumber\aligned
&L_1:=\int_{\R^{n+1}_+\cap\pa B_1}\la\theta_1^b\frac{d}{d\la}L(u_e^\la)\frac{du_e^\la}{d\la}\\
&=-2(n+b-3)\la\int_{\R^{n+1}_+\cap\pa B_1}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}\frac{du_e^\la}{d\la})\frac{du_e^\la}{d\la}\\
&=2(n+b-3)\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b(\nabla_{S^n}\frac{du_e^\la}{d\la})^2;
\endaligned\ee

\be\nonumber\aligned
&L_2:=\int_{\R^{n+1}_+\cap\pa B_1}-\la\theta_1^bL(u_e^\la)\frac{d^2u_e^\la}{d\la^2}\\
&=2(n+b-3)\la\int_{\R^{n+1}_+\cap\pa B_1}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\frac{d^2 u_e^\la}{d\la^2}\\
&=-2(n+b-3)\int_{\R^{n+1}_+\cap\pa B_1}\la\theta_1^b\nabla_{S^n}u_e^\la\frac{d^2}{d\la^2}\nabla_{S^n}u_e^\la\\
&=-(n+b-3)\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\frac{d}{d\la}\big[2\la\nabla_{S^n}u_e^\la\nabla_{S^n}\frac{du_e^\la}{d\la}-(\nabla_{S^n}u_e^\la)^2\big]\\
&+2(n+b-3)\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b|\nabla_{S^n}\frac{du_e^\la}{d\la}|^2;\\
\endaligned\ee

\be\nonumber\aligned
&L_3:=\int_{\R^{n+1}_+\cap\pa B_1}-5\theta_1^bL(u_e^\la)\frac{du_e^\la}{d\la}\\
&=10(n+b-3)\int_{\R^{n+1}_+\cap\pa B_1}\mathbf{div}_{S^n}(\theta_1^b\nabla_{S^n}u_e^\la)\frac{du_e^\la}{d\la}\\
&=-10(n+b-3)\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\nabla_{S^n}u_e^\la\nabla_{S^n}\frac{du_e^\la}{d\la}\\
&=-5(n+b-3)\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b[\nabla_{S^n}u_e^\la]^2.\\
\endaligned\ee
Hence,
\be\nonumber\aligned
&\mathcal{L}:=L_1+L_2+L_3\\
&=-(n+b-3)\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[\la\frac{d}{d\la}(\nabla_{S^n}u_e^\la)^2\big]\\
&\quad-4(n+b-3)\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b[\nabla_{S^n}u_e^\la]^2\\
&\quad+4(n+b-4)\la\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b|\nabla_{S^n}\frac{du_e^\la}{d\la}|^2\\
&=-(n+b-3)\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-4}y^b\frac{d}{d\la}\big[\la^2|\nabla u_e|^2-\la^2|\pa_r u_e|^2\big]\\
&\quad-4(n+b-3)\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b\big[\la^2|\nabla u_e|^2-\la^2|\pa_r u_e|^2\big].
\endaligned\ee
By rescaling, we have
\be\label{8Lscaling}\aligned
&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b\big[\la\frac{d}{d\la}(\nabla_{S^n}u_e^\la)^2\big]\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-4}y^b\frac{d}{d\la}\big[\la^2|\nabla u_e|^2-\la^2|\pa_r u_e|^2\big];\\
&\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_1}\theta_1^b[\nabla_{S^n}u_e^\la]^2\\
&=\frac{d}{d\la}\int_{\R^{n+1}_+\cap\pa B_\la}\la^{2s\frac{p+1}{p-1}-n-5}y^b\big[\la^2|\nabla u_e|^2-\la^2|\pa_r u_e|^2\big].
\endaligned\ee

\subsection{The term $\overline{E}_{d_1}$}
Notice that on the boundary $\pa B_1$,
\be\nonumber\aligned
&v_e^\la=\Delta_b u_e^\la\\
&=\la^2\frac{d^2u_e^\la}{d\la^2}+(n+b-\frac{4s}{p-1})\la\frac{du_e^\la}{d\la}+\frac{2s}{p-1}(1+\frac{2s}{p-1}-n-b)u_e^\la+\Delta_\theta u_e^\la\\
&:=\la^2\frac{d^2u_e^\la}{d\la^2}+\alpha_0\la\frac{du_e^\la}{d\la}+\beta_0 u_e^\la+\Delta_\theta u_e^\la.
\endaligned\ee
Integrate by part, it follows that

\be\label{8ed1ed1}\aligned
&\int_{\pa B_1}y^b\la(\frac{dv_e^\la}{d\la})^2=\int_{\pa B_1}\Big(\la^5(\frac{d^3u_e^\la}{d\la^3})^2+(\alpha_0^2-2\alpha_0-2\beta_0-4)\la^3(\frac{d^2u_e^\la}{d\la^2})^2\\
&\quad\quad\quad\quad\quad\quad\quad\quad+(-\alpha_0^2+\beta_0^2+2\alpha_0+2\beta_0)\la(\frac{du_e^\la}{d\la})^2\Big)\\
&+\int_{\pa B_1}\Big(-2\la^3(\nabla_\theta \frac{d^2u_e^\la}{d\la^2})^2+(10-2\beta_0)\la(\nabla_\theta \frac{du_e^\la}{d\la})^2\Big)
+\int_{\pa B_1}\la(\Delta_\theta \frac{du_e^\la}{d\la})^2\\
&+\frac{d}{d\la}\Big(\int_{\pa B_1}\sum_{0\leq i,j\leq 2,i+j\leq2}{c^1_{i,j}}\la^{i+j}\frac{d^iu_e^\la}{d\la^i}\frac{d^ju_e^\la}{d\la^j}
+\sum_{0\leq s,t\leq 2,s+t\leq2}{c^2_{s,t}}\la^{s+t}\frac{d^su_e^\la}{d\la^s}\frac{d^tu_e^\la}{d\la^t}\Big),
\endaligned\ee
where $c^i_{i,j},c^2_{s,t}$ depending on   $a,b$ hence on  $p,n$.

\vskip0.2in

\noindent{\bf Proof of Theorem \ref{8monoid}}. Notice that the equation \eqref{8overlineE}, combine with the estimates on $\mathcal{I},\mathcal{J},\mathcal{K},\mathcal{L}$ and \eqref{8ed1ed1}, we obtain Theorem \ref{8monoid}.\hfill $\Box$

\section{Energy estimates and Blow down analysis }

 In this section, we do some energy estimates for the solutions of \eqref{8LE}, which are important when we perform a
 blow-down analysis in the next section.

 \subsection{Energy estimates}

\bl Let $u$ be a solution of \eqref{8LE} and $u_e$ satisfy \eqref{8LEE}, then there exists a positive constant $C$ such that
\be\label{8estimate1}\aligned
&\int_{\pa\R^{n+1}_+}|u_e|^{p+1}\eta^6+\int_{\R^{n+1}_+}y^b|\nabla\Delta_b u_e|^2\eta^6\\
&\leq C\big[\int_{\R^{n+1}_+}y^b|\Delta_b u_e|^2\eta^4|\nabla\eta|^2+\int_{\R^{n+1}_+}y^b|\nabla u_e|^2 \frac{|\Delta_b\eta^6|^2}{\eta^6}
+\int_{\R^{n+1}_+}y^bu_e^2\frac{|\nabla\Delta_b \eta^6|^2}{\eta^6}\\
&\quad +\int_{\R^{n+1}_+}y^b|\nabla u_e|^2\eta^2|\nabla \eta|^4
+\int_{\R^{n+1}_+}y^b|\nabla^2 u_e|^2\eta^4|\nabla\eta|^2\\
&\quad +\int_{\R^{n+1}_+}y^b|\nabla u_e|^2\eta^4|\nabla^2\eta|^2\big].
\endaligned\ee
\el
\bp
Multiply the equation \eqref{8LEE} with $y^b u_e\eta^6$, where $\eta$ is a test function, we get that
\be\aligned
0=&\int_{\R^{n+1}_+}y^b u_e\eta^6\Delta_b^3 u_e=\int_{\R^{n+1}_+} u_e\eta^6 \mathbf{div}(y^b\nabla\Delta_b^2u_e)\\
=&-\int_{\pa\R^{n+1}_+} u_e\eta^6 \frac{\pa}{\pa y}\Delta_b^2 u_e-\int_{\R^{n+1}_+}y^b\nabla(u_e\eta^6)\nabla\Delta_b^2u_e\\
=&C_{n,s}\int_{\pa\R^{n+1}_+}|u_e|^{p+1}\eta^6-\int_{\pa\R^{n+1}_+}y^b\frac{\pa (u_e\eta^6)}{\pa y}\Delta_b^2 u_e+\int_{\R^{n+1}_+}y^b\Delta_b(u_e\eta^6)\Delta_b^2u_e\\
=&C_{n,s}\int_{\pa\R^{n+1}_+}|u_e|^{p+1}\eta^6+\int_{\R^{n+1}_+}y^b\Delta_b(u_e\eta^6)\Delta_b^2u_e\\
=&C_{n,s}\int_{\pa\R^{n+1}_+}|u_e|^{p+1}\eta^6-\int_{\pa\R^{n+1}_+}\Delta_b(u_e\eta^6)y^b\frac{\pa \Delta_b u_e}{\pa y}\\
&-\int_{\R^{n+1}_+}y^b\nabla(\Delta_b(u_e\eta^6))\nabla\Delta_b u_e\\
=&C_{n,s}\int_{\pa\R^{n+1}_+}|u_e|^{p+1}\eta^6-\int_{\R^{n+1}_+}y^b\nabla(\Delta_b(u_e\eta^6))\nabla\Delta_b u_e.\\
\endaligned\ee
Hence, we have
\be\label{8eee}
C_{n,s}\int_{\pa\R^{n+1}_+}|u_e|^{p+1}\eta^6=\int_{\R^{n+1}_+}y^b\nabla(\Delta_b(u_e\eta^6))\nabla\Delta_b u_e.
\ee
Since $\Delta_b(\xi\eta)=\eta\Delta_b\xi+\xi\Delta_b\eta+2\nabla\xi\nabla\eta$, we have
\be\nonumber\aligned
\Delta_b(u_e\eta^6)=\eta^6\Delta_b u_e+u_e\Delta_b\eta^6+12\eta^5\nabla u_e\nabla\eta,
\endaligned\ee
therefore,
\be\label{8zhankai}\aligned
\nabla\Delta_b(u_e\eta^6)\nabla\Delta_b u_e=&6\eta^5\Delta_b u_e\nabla\eta\nabla\Delta_b u_e+(\eta)^6(\nabla\Delta_b u_e)^2+\Delta_b\eta^6\nabla u_e\nabla\Delta_b u_e\\
&+u_e\nabla\Delta_b \eta^6\nabla\Delta_b u_e+60\eta^4(\nabla\eta\nabla\Delta_b u_e)(\nabla u_e\nabla\eta)\\
&+12\eta^5\sum_{i,j}\pa_{ij}u_e\pa_i\eta\pa_j \Delta_b u_e+12\eta^5\sum_{i,j}\pa_i u_e\pa_{ij}\eta\pa_j\Delta_b u_e.
\endaligned\ee
here $\pa_j(j=1, ..., n, n+1)$ denote the derivatives  with respect to  $x_1,...,x_n,y$ respectively.
A similar way can be applied to deal with the following term $|\nabla\Delta_b (u_e\eta^3)|^2$.
On the other hand, by the stability condition, we have
\be\label{8stable1}\aligned
p\int_{\R^n}|u|^{p+1}\eta^6\leq\int_{\R^n}\int_{\R^n}\frac{\Big(u(x)\eta^3(x)-u(y)\eta^3(y)\Big)^2}{|x-y|^{n+2s}}=
\frac{1}{C_{n,s}}\int_{\R^{n+1}_+}y^b|\nabla\Delta_b(u_e\eta^3)|^2.
\endaligned\ee
(Here we notice that $u_e(x,0)=u(x)$, see Theorem \ref{ththYang}, \eqref{Yangmore})\\
Combine with \eqref{8eee}, \eqref{8zhankai} and \eqref{8stable1}, we have
\be\nonumber\aligned
&\int_{\R^{n+1}_+}y^b|\nabla\Delta_b u_e|^2\eta^6\\
&\leq  C\varepsilon\int_{\R^{n+1}_+}y^b(\nabla\Delta_b u_e)^2\eta^6+C(\varepsilon)\big[\int_{\R^{n+1}_+}y^b|\Delta_b u_e|^2\eta^4|\nabla\eta|^2\\
&\quad +\int_{\R^{n+1}_+}y^b|\nabla u_e|^2 (\frac{|\Delta_b\eta^6|^2}{\eta^6}+\eta^4|\nabla^2\eta|^2)\\
&\quad +\int_{\R^{n+1}_+}y^bu_e^2\frac{|\nabla\Delta_b \eta^6|^2}{\eta^6}
+\int_{\R^{n+1}_+}y^b|\nabla u_e|^2\eta^2|\nabla \eta|^4
+\int_{\R^{n+1}_+}y^b|\nabla^2 u_e|^2\eta^4|\nabla\eta|^2\big],\\
\endaligned\ee
we can select $\varepsilon$ so small   that $C\varepsilon\leq\frac{1}{2}$. Combine with \eqref{8eee} and \eqref{8zhankai}, we obtain our conclusion.
\ep

\bc\label{8hhhh}
 Let $u$ be a solution of \eqref{8LE} and $u_e$ satisfy \eqref{8LEE}, then

\be\nonumber\aligned
&\int_{\pa\R^{n+1}_+\cap B_{R/2}}|u_e|^{p+1}+\int_{\R^{n+1}_+\cap B_{R/2}}y^b(\nabla\Delta_b u_e)^2\\
&\leq C\big[R^{-6}\int_{\R^{n+1}_+\cap B_R}y^b u_e^2+R^{-4}\int_{\R^{n+1}_+\cap B_R}y^b|\nabla u_e|^2\\
&+R^{-2}\int_{\R^{n+1}_+\cap B_R}y^b(|\Delta_b u_e|^2+|\nabla^2 u_e|^2)\big].
\endaligned\ee
\ec
\bp
We let $\eta=\xi^m$ where $m>1$ in the estimate \eqref{8estimate1}. We have
\be\nonumber\aligned
&\int_{\pa\R^{n+1}_+}|u_e|^{p+1}\xi^{6m}+\int_{\R^{n+1}_+}y^b|\nabla\Delta_b u_e|^2\xi^{6m}\\
&\leq C[\int_{\R^{n+1}_+}y^b(|\Delta_b u_e|^2+|\nabla^2 u_e|^2)\xi^{6m-2}|\nabla\xi|^2\\
&\quad +\int_{\R^{n+1}_+}y^b|\nabla u_e|^2\xi^{6m-4}(|\nabla^2\xi|^2+|\nabla\xi|^4)+\int_{\R^{n+1}_+}y^b u_e^2\xi^{6m-6}|\nabla^3\xi|^2].
\endaligned\ee
Let $\xi=1$ in $B_{R/2}$ and $\xi=0$ in $B_R^C$, satisfying $|\nabla \xi |\leq\frac{C}{R}$, then we have the desired estimates.
\ep

\vskip0.1in
\bl\label{8lemma3}
Suppose that $u$ is a solution of \eqref{8LE} which is stable outside some ball $B_{R_0}\subset\R^n$. For $\eta\in C_c^\infty(\R^n\backslash\overline{B_{R_0}})$ and $x\in\R^n$, define
\be
\rho(x)=\int_{\R^n}\frac{(\eta(x)-\eta(y))^2}{|x-y|^{n+2s}}dy.
\ee
Then
\be
\int_{\R^n}|u|^{p+1}\eta^2dx+\int_{\R^n}\int_{\R^n}\frac{|u(x)\eta(x)-u(y)\eta(y)|^2}{|x-y|^{n+2s}}dxdy\leq C\int_{\R^n} u^2\rho dx.
\ee
\el

\bl\label{8lemma4}
Let $m>n/2$ and $x\in\R^n$. Set
\be\label{8420}
\rho(x):=\int_{\R^n}\frac{(\eta(x)-\eta(y))^2}{|x-y|^{n+2s}}dy\;\;\hbox{where}\;\;\eta(x)=(1+|x|^2)^{-m/2}.
\ee
Then there is a constant $C=C(n,s,m)>0$ such that
\be C^{-1}(1+|x|^2)^{-n/2-s}\leq \rho(x)\leq C(1+|x|^2)^{-n/2-s}.
\ee
\el

\bc\label{8corr1}
Suppose that $m>n/2$, $\eta$ is given by \eqref{8420} and $R>R_0>1$. Define
\be\label{8rhor}
\rho_R(x)=\int_{\R^n}\frac{(\eta_R(x)-\eta_R(y))^2}{|x-y|^{n+2s}}dy,\;\;\hbox{where}\;\;\eta_R(x)=\eta(x/R)\psi(x/R)
\ee
for a standard test function $\psi$ that $\psi\in C^{\infty}(\R^n)$, $0\leq\psi\leq1,\psi=0$ on $B_1$ and $\psi=1$ on $\R^n\setminus B_2$.
Then there exists a constant $C>0$ such that
\be\nonumber
\rho_R(x)\leq C\eta^2(x/R)|x|^{-(n+2s)}+R^{-2s}\rho(x/R).
\ee
\ec

\bl\label{8lemma5}
Suppose that $u$ is a   solution of \eqref{8LE} which  is stable outside a ball $B_{R_0}$. Consider $\rho_R$  which  is defined in \eqref{8rhor} for
$n/2<m<n/2+s(p+1)/2$. Then there exists a constant $C>0$ such that
\be\nonumber
\int_{\R^n}u^2\rho_R\leq C(\int_{B_{3R_0}}u^2\rho_R+R^{n-2s\frac{p+1}{p-1}})
\ee
for any $R>3R_0$.
\el
The proofs of Lemma \ref{8lemma3}, Corollary \ref{8corr1}, Lemma \ref{8lemma4} and Lemma \ref{8lemma5} are similar to that of  Lemmas $2.1, 2.2, 2.4$ in \cite{Wei0=1} and we omit the details here.

\bl\label{8estimate111}
Suppose that $p\neq\frac{n+2s}{n-2s}$. Let $u$ be a solution of \eqref{8LE} which is stable outside a ball $B_{R_0}$ and $u_e$ satisfy \eqref{8LEE}. Then
there exists a constant $C>0$ such that
\be\nonumber
\int_{B_R}y^b u_e^2\leq CR^{n+6-2s\frac{p+1}{p-1}},\quad \int_{B_R}y^b |\nabla u_e|^2\leq CR^{n+4-2s\frac{p+1}{p-1}},
\ee
\be\nonumber
\int_{B_R}y^b (|\nabla^2 u_e|^2+|\Delta_b u_e|^2)\leq CR^{n+2-2s\frac{p+1}{p-1}}.
\ee
\el
\bp
Recall that the Possion formula for the fractional equation for the case $0<s<1$  (see \cite{Caffarelli2007}), we can generalize the
 expression formula to the general case with non-integer positive real number. Therefore,
\be\nonumber
{u}_e(x,y)=C_{n,s}\int_{\R^n}u(z)\frac{y^{2s}}{(|x-z|^2+y^2)^{\frac{n+2s}{2}}}dz.
\ee
Then we have
\be\label{8pa0}
|{u}_e(x,y)|^2\leq C\int_{\R^n}u^2(z)\frac{y^{2s}}{(|x-z|^2+y^2)^{\frac{n+2s}{2}}}dz,
\ee
and
\be\nonumber
\pa_y{u}_e(x,y)=C_{n,s}\int_{\R^n}u(z)\big[\frac{2sy^{2s-1}}{(|x-z|^2+y^2)^{\frac{n+2s}{2}}}
-\frac{(n+2s)y^{2s+1}}{(|x-z|^2+y^2)^{\frac{n+2s+2}{2}}}\big]dz,
\ee
also
\be\nonumber
\pa_{x_j}{u}_e(x,y)=-C_{n,s}\int_{\R^n}u(z)\frac{(n+2s)(x_j-z_j)y^{2s}}{(|x-z|^2+y^2)^{\frac{n+2s+2}{2}}}dz,
\ee
for $j=1,2,...,n$.
Hence by H\"older's inequality we have
\be\label{8pa1}
|\nabla u_e(x,y)|^2\leq C\int_{\R^n}\frac{u^2(z)y^{2s-2}}{(|x-z|^2+y^2)^{\frac{n+2s}{2}}}dz.
\ee
By a straightforward calculation we have
\be\nonumber\aligned
\pa_{x_jx_j}{u}_e(x,y)=&C_{n,s}\int_{\R^n}u(z)\big[\frac{(n+2s)(n+2s+2)(x_j-z_j)^2y^{2s}}{(|x-z|^2+y^2)^{\frac{n+2s+4}{2}}}\\
&-\frac{(n+2s)y^{2s}}{(|x-z|^2+y^2)^{\frac{n+2s+2}{2}}}\big]dz,
\endaligned\ee

\be\nonumber\aligned
\pa_{x_jy}{u}_e(x,y)=&C_{n,s}\int_{\R^n}u(z)\big[\frac{(n+2s)(n+2s+2)(x_j-z_j)^2y^{2s+1}}{(|x-z|^2+y^2)^{\frac{n+2s+4}{2}}}\\
&-\frac{2s(n+2s)(x_j-z_j)^2y^{2s-1}}{(|x-z|^2+y^2)^{\frac{n+2s+2}{2}}}\big],
\endaligned\ee
and
\be\nonumber\aligned
\pa_{yy}{u}_e(x,y)=&C_{n,s}\int_{\R^n}u(z)\big[\frac{2s(2s-1)y^{2s-2}}{(|x-z|^2+y^2)^{\frac{n+2s}{2}}}\\
&-\frac{(n+2s)(4s+1)y^{2s}}{(|x-z|^2+y^2)^{\frac{n+2s+2}{2}}}
+\frac{(n+2s)(n+2s+2)y^{2s+2}}{(|x-z|^2+y^2)^{\frac{n+2s+4}{2}}}\big].
\endaligned\ee
Therefore, we have
\be\nonumber
|\nabla^2 u_e(x,y)|+|\Delta_b u_e(x,y)|\leq C\int_{\R^n}|u(z)|\frac{y^{2s-2}}{(|x-z|^2+y^2)^{\frac{n+2s}{2}}}dz.
\ee
Hence,
\be\label{8pa2}
|\nabla^2 u_e(x,y)|^2+|\Delta_b u_e(x,y)|^2\leq C\int_{\R^n}u^2(z)\frac{y^{2s-4}}{(|x-z|^2+y^2)^{\frac{n+2s}{2}}}dz.
\ee
Now we turn to estimate the following integration, which provides a unify way to deal with our desired estimates.

Define
\be\label{8ak}\aligned
A_k:=&\int_{|x|\leq R,z\in\R^n}u^2(z)\big[\int_0^R\frac{y^{2k+1}}{(|x-z|^2+y^2)^{\frac{n+2s}{2}}}dy\big]dzdx\\
=&\int_{|x|\leq R,z\in\R^n}u^2(z)\big[\int_0^{R^2}\frac{\alpha^k}{(|x-z|^2+\alpha)^{\frac{n+2s}{2}}}d\alpha]dzdx\\
\leq&\frac{1}{2}\int_{|x|\leq R,z\in\R^n}u^2(z)[\int_0^{R^2}\frac{d\alpha}{(|x-z|^2+\alpha)^{\frac{n+2s}{2}}-k}\big]dzdx\\
=&\frac{1}{2}(\frac{n+2s}{2}-k)\int_{|x|\leq R,z\in\R^n}u^2(z)\big[(|x-z|^2)^{k-\frac{n+2s}{2}+1}\\
&-(|x-z|^2+R^2)^{k-\frac{n+2s}{2}+1}\big],\\
\endaligned\ee
where $k=0,1,2$.
Split the integral to $|x-z|\leq 2R$ and $|x-z|>2R$, for the case of $|x-z|\leq 2R$, we see that

\be\nonumber\aligned
&\int_{|x|\leq R,|x-z|\leq 2R}u^2(z)\Big[(|x-z|^2)^{k-\frac{n+2s}{2}+1}-(|x-z|^2+R^2)^{k-\frac{n+2s}{2}+1}\Big]\\
&\leq\int_{|x|\leq R,|x-z|\leq 2R}u^2(z)\Big[(|x-z|^2)^{k-\frac{n+2s}{2}+1}\Big]\\
&\leq C R^{2k-2s+2}\int_{|z|\leq 3R}u^2(z)dz\\
&\leq R^{2k-2s+2}\Big(\int_{B_{3R}}|u|^{p+1}\eta_R^2\Big)^{2/(p+1)}\Big(\int_{B_{3R}}\eta_R^{-4/(p-1)} \Big)^{(p-1)/(p+1)}\\
&\leq C R^{2k-2s+2}\Big(\int_{B_{3R}}u^2(z)\rho_R(z)\Big)^{2/(p+1)}\\
&\leq C R^{n+2k+2-2s\frac{p+1}{p-1}}.
\endaligned
\ee
Here we have used Lemma \ref{8lemma3} and \ref{8lemma5}. For the case of $|x-z|>2R$, by the mean value theorem, we have
\be\nonumber\aligned
&\int_{|x|\leq R,|x-z|> 2R}u^2(z)\Big[(|x-z|^2)^{k-\frac{n+2s}{2}+1}-(|x-z|^2+R^2)^{k-\frac{n+2s}{2}+1}\Big]\\
&\leq R^2\int_{|x|\leq R,|x-z|> 2R}u^2(z)\Big[(|x-z|^2)^{k-\frac{n+2s}{2}}\Big]\\
&\leq C R^{n+2}\int_{|z|\geq R} u_e^2(z)|z|^{2k-n-2s}dz\\
&\leq C R^{n+2}\Big[\int_{|z|\geq R} (u^{p+1}_e(z))\Big]^{2/(p+1)}\Big(\int_{|z|\geq R}|z|^{(2k-n-2s)\frac{p+1}{p-1}}\Big)^{(p-1)/(p+1)}\\
&\leq C R^{n+2k+2-2s\frac{p+1}{p-1}},
\endaligned
\ee
here we have used Lemma \ref{8lemma3}.
Hence, we obtain that
\be\label{8akak}
A_k\leq C R^{n+2k+2-2s\frac{p+1}{p-1}},
\ee
where $C=C(n,s,p)$ independent of $R$.
Now, combine with \eqref{8pa0}, \eqref{8pa1} and \eqref{8pa2}, recall that $b=5-2s$, we have

\be\nonumber\aligned
&\int_{B_R}y^b u_e^2dxdy\leq A_2, \quad \int_{B_R}y^b|\nabla u_e|^2dxdy\leq A_1,\\
 &\int_{B_R}y^b\big(|\nabla^2 u_e|^2+|\Delta_b u_e|^2\big)dxdy\leq A_0.
\endaligned
\ee
Apply \eqref{8akak}, we finish our proof.
\ep

Combine Corollary \ref{8hhhh} and Lemma \ref{8lemma5}, we have the following lemma.
\bl\label{8lemmaE}
Let $u$ be a solution of \eqref{8LE} which is stable outside a ball $B_{R_0}$ and $u_e$ satisfy \eqref{8LEE}. Then there exists a positive constant $C$
such that
\be\nonumber\aligned
&\int_{\pa\R^{n+1}_+\cap B_R}|u_e|^{p+1}+R^{-6}\int_{\R^{n+1}_+\cap B_R}y^b |u_e|^2
+R^{-4}\int_{\R^{n+1}_+\cap B_R}y^b |\nabla u_e|^2\\
&+R^{-2}\int_{\R^{n+1}_+\cap B_R}y^b\big(|\Delta_b u_e|^2+|\nabla^2 u_e|^2\big)
+\int_{\R^{n+1}_+\cap B_R}y^b|\nabla\Delta_b u_e|^2
\leq C R^{n-2s\frac{p+1}{p-1}}.
\endaligned\ee
\el

\newpage
\subsection{ Blow down analysis and  the proof of Theorem \ref{8Liouvillec}}

  {\bf The proof of Theorem \ref{8Liouvillec}.}
 Suppose that $u$ is a solution of \eqref{8LE} which is stable outside the ball of radius $R_0$ and suppose that $u_e$ satisfies \eqref{8LEE}.
 In the subcritical case, i.e., $1<p<p_s(n)$, Lemma \ref{8lemma3} implies that $u\in \dot{H}^s(\R^n)\cap L^{p+1}(\R^n)$.
Multiplying \eqref{8LE} with $u$ and integrate, we obtain that
\be\label{841}
\int_{\R^n}|u|^{p+1}=\|u\|^2_{\dot{H}^s(\R^n)}.
\ee
Multiplying \eqref{8LE} with $u^\la(x)=u(\la x)$ yields
\be\nonumber
\int_{\R^n}|u|^{p-1} u^\la=\int_{\R^n} (-\Delta)^{s/2}u (-\Delta)^{s/2}u^\la=\la^s\int_{\R^n}w w_\la,
\ee
where $w=(-\Delta)^{s/2}u$. Following the ideas provided in \cite{Xavier2015,Xavier2014} and using the change of variable $z=\sqrt{\la}x$, we can get
the following Pohozaev identity
 \be\nonumber\aligned
-\frac{n}{p+1}\int_{\R^n}|u|^{p+1}&=\frac{2s-n}{2}\int_{\R^n}|w|^{2}
+\frac{d}{d\la}\int_{\R^n}w^{\sqrt{\la}}w^{1/\sqrt{\la}}dz \Big|_{\la=1}\\
&=\frac{2s-n}{2}\|u\|^2_{\dot{H}^s(\R^n)}.
\endaligned\ee
Hence, we have the following Pohozaev identity
 \be\nonumber\aligned
\frac{n}{p+1}\int_{\R^n}|u|^{p+1}=\frac{n-2s}{2}\|u\|^2_{\dot{H}^s(\R^n)}.
\endaligned\ee

For $p<p_s(n)$, this equality above together with \eqref{841} proves that $u\equiv0$.
For $p=p_s(n)$, this equality above means that the energy is finite. Further,  since $u\in \dot{H}^s(\R^n)$, apply the stability inequality with
test function $\psi=u\eta^2(\frac{x}{R})$, and let $R\rightarrow+\infty$ (where $\eta$ is cutoff function), then we get that
 \be\nonumber\aligned
p\int_{\R^n}|u|^{p+1}\leq\|u\|^2_{\dot{H}^s(\R^n)}.
\endaligned\ee
This together with \eqref{841} gives that $u\equiv0$.

\vskip0.1in

Now we consider the supercritical case, i.e., $p>\frac{n+2s}{n-2s}$, we perform the proof via a few steps.

\vskip0.1in

\noindent{\bf Step 1.}  $\lim_{\la\rightarrow\infty} E(u_e,0,\la)<\infty$.\\

From Theorem \ref{8Monotone} we know that $E$ is nondecreasing w.r.t. $\la$, so we only need to show that $E(u_e,0,\la)$ is bounded. Note that
\be\nonumber
E(u_e,0,\la)\leq\frac{1}{\la}\int_{\la}^{2\la} E(u_e,0,t)dt\leq\frac{1}{\la^2}\int_\la^{2\la}\int_{t}^{t+\la}E(u_e,0,\gamma)d\gamma dt.
\ee
From Lemma \ref{8lemmaE}, we have that
\be\nonumber\aligned
\frac{1}{\la^2}&\int_\la^{2\la}\int_{t}^{t+\la}\gamma^{2s\frac{p+1}{p-1}-n}\big[\int_{\R^{n+1}_+\cap B_\gamma}\frac{1}{2}y^b|\nabla\Delta_b u_e|^2dydx\\
&\quad\quad\quad\quad\quad\quad -\frac{C_{n,s}}{p+1}\int_{\pa\R^{n+1}_+\cap B_\gamma}|u_e|^{p+1}dx\big]d\gamma dt\leq C,
\endaligned\ee
where $C>0$ is independent of $\gamma$.
\be\label{8type1}\aligned
\frac{1}{\la^2}&\int_\la^{2\la}\int_{t}^{t+\la}\int_{\R^{n+1}_+\cap\pa B_\gamma}\gamma^{2s\frac{p+1}{p-1}-n-5}y^b\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1)(\frac{2s}{p-1}-2)u_e\\
&+\frac{6s}{p-1}(\frac{2s}{p-1}-1)\gamma\pa_r u_e
+\frac{6s}{p-1}\gamma^2 \pa_{rr} u_e+\gamma^3\pa_{rrr} u_e\big]\\
&\quad\quad\big[\frac{2s}{p-1}(\frac{2s}{p-1}-1) u_e+\frac{4s}{p-1}\gamma\pa_r u_e+\gamma^2\pa_{rr}u_e\big]\\
&\leq C\frac{1}{\la^2}\int_\la^{2\la}\int_{t}^{t+\la}t^{2s\frac{p+1}{p-1}-n-5}\int_{\R^{n+1}_+\cap\pa B_\gamma}\\
&\quad y^b\big[u_e^2+\gamma^2(\pa_r u_e)^2+\gamma^4(\pa_{rr}u_e)^2+\gamma^6(\pa_{rrr}u_e)^2\big]\\
&\leq C\frac{1}{\la^2}\int_\la^{2\la}t^{2s\frac{p+1}{p-1}-n-5}\int_{\R^{n+1}_+\cap B_{3\la}}\\
&\quad y^b\big[u_e^2+\gamma^2(\pa_r u_e)^2+\gamma^4(\pa_{rr}u_e)^2+\gamma^6(\pa_{rrr}u_e)^2\big]\\
&\leq C \la^{n-2s\frac{p+1}{p-1}+6}\frac{1}{\la^2}\int_\la^{2\la}t^{2s\frac{p+1}{p-1}-n-5}dt\\
&\leq C
\endaligned\ee
and

\be\label{8type2}\aligned
&\Big|\frac{1}{\la^2}\int_\la^{2\la}\int_{t}^{t+\la}\int_{\R^{n+1}_+\cap\pa B_\gamma}\gamma^{2s\frac{p+1}{p-1}-n-4}y^b\frac{d}{d\gamma}(\gamma^2\Delta_b u_e-\gamma^2\pa_{rr}u_e-(n+b)\gamma\pa_r u_e)^2\Big|\\
&\leq\frac{1}{\la^2}\int_\la^{2\la}t^{2s\frac{p+1}{p-1}-n-5}\int_{t}^{t+\la}\int_{\R^{n+1}_+\cap\pa B_\gamma}y^b\big[2\gamma^2\Delta_b u_e-2\gamma^2\pa_{rr}u_e-(n+b)\gamma\pa_r u_e\big]\\
&\quad \big[\gamma^2\Delta_b u_e-\gamma^2\pa_{rr}u_e-(n+b)\gamma\pa_r u_e\big]\\
&\leq\frac{1}{\la^2}\int_\la^{2\la}t^{2s\frac{p+1}{p-1}-n-5}\int_{t}^{t+\la}\int_{\R^{n+1}_+\cap B_{3\la}}y^b[2\gamma^2\Delta_b u_e-2\gamma^2\pa_{rr}u_e-(n+b)\gamma\pa_r u_e]\\
&\quad\big[\gamma^2\Delta_b u_e-\gamma^2\pa_{rr}u_e-(n+b)\gamma\pa_r u_e\big]\\
&\leq C\frac{1}{\la^2}\int_\la^{2\la}t^{2s\frac{p+1}{p-1}-n-5}\int_{t}^{t+\la}\int_{\R^{n+1}_+\cap B_{3\la}}y^b
\big[\gamma^4(\Delta_b u_e)^2+\gamma^4(\pa_{rr}u_e)^2+\gamma^2\pa_r u_e\big]\\
&\leq C \la^{n-2s\frac{p+1}{p-1}+6}\frac{1}{\la^2}\int_\la^{2\la}t^{2s\frac{p+1}{p-1}-n-5}dt\\
&\leq C.
\endaligned\ee
\vskip0.1in
Integrate by part,  by  the scaling identity of section 3, for example \eqref{8Iscaling}, \eqref{8Jscaling}, \eqref{8Kscaling} and \eqref{8Lscaling},
we can treat the remaining    terms  by a  similar way as the estimates \eqref{8type1} and \eqref{8type2}.

\vskip0.12in

\noindent{\bf Step 2.}  There exists a sequence $\la_i\rightarrow\infty$ such that $(u_e^{\la_i})$ converges weakly
to a function $u_e^\infty$  in $H^3_{loc}(\R^n;y^bdxdy)$,  this is a direct consequence of Lemma \ref{8lemmaE}.

\vskip0.12in

\noindent{\bf Step 3.}  {\bf The function $u_e^\infty$ is homogeneous.}
Due to the scaling invariance of $E $  (i.e., $E(u_e,0,R\la)=E(u_e^{\la},0,R)$ )
and the monotonicity formula, for any given $R_2>R_1>0$, we see that
\be\nonumber\aligned
0=&\lim_{i\rightarrow\infty}\big(E(u_e,0,R_2\la_i)-E(u_e,0,R_1\la_i)\big)\\
=&\lim_{i\rightarrow\infty}\big(E(u_e^{\la_i},0,R_2)-E(u_e^{\la_i},0,R_1)\big)\\
\geq&\liminf_{i\rightarrow\infty}\int_{(B_{R_2}\setminus B_{R_1})\cap \R^{n+1}_+}
y^b r^{2s\frac{p+1}{p-1}-n-6}\big(\frac{2s}{p-1}{u_e^{\la_i}}+r\frac{\pa u_e^{\la_i}}{\pa r}\big)^2dydx\\
 \geq&\int_{(B_{R_2}\setminus B_{R_1})\cap \R^{n+1}_+}
y^b r^{2s\frac{p+1}{p-1}-n-6}\big(\frac{2s}{p-1}{u_e^{\infty}}+r\frac{\pa u_e^{\infty}}{\pa r}\big)^2dydx.\\
\endaligned\ee
In the last inequality we have used the weak convergence of  the sequence $(u_e^{\la_i})$ to the function $u_e^{\infty}$ in $H^3_{loc}(\R^n;y^bdxdy)$. This implies that
\be\nonumber
\frac{2s}{p-1}\frac{u_e^{\infty}}{r}+\frac{\pa u_e^{\infty}}{\pa r}=0\;\;\hbox{a.e.}\;\;\hbox{in}\;\;\R^{n+1}_+.
\ee
Therefore,  $u_e^\infty$ is homogeneous.

\vskip0.12in

\noindent{\bf Step 4}. {  \bf $u_e^\infty=0$.}
This is a direct consequence of Theorem $3.1$ in \cite{Wei1=2}.

\vskip0.12in

\noindent{\bf Step 5}. $(u_e^{\la_i})$ converges strongly to zero in $H^3(B_R\setminus B_\varepsilon;y^bdxdy)$ and $(u_e^{\la_i})$ converges strongly in
$L^{p+1}(\pa\R^{n+1}_+\cap (B_R\setminus B_\varepsilon))$ for all $R>\varepsilon>0$. These are consequent results  of Lemma \ref{8lemmaE} and Theorem $1.5$ in \cite{Fabes1982}.

\vskip0.12in

\noindent{\bf Step 6}. $u_e=0$. Note that
\be\nonumber\aligned
\overline{E}(u_e,\la)=&\overline{E}(u_e^\la,1)\\
=&\frac{1}{2}\int_{\R^{n+1}_+\cap B_1}y^b|\nabla\Delta_b u_e^\la|^2dxdy-\frac{C_{n,s}}{p+1}\int_{\pa\R^{n+1}_+\cap B_1}|u_e^\la|^{p+1}dx\\
=&\frac{1}{2}\int_{\R^{n+1}_+\cap B_\varepsilon}y^b|\nabla\Delta_b u_e^\la|^2dxdy-\frac{C_{n,s}}{p+1}\int_{\pa\R^{n+1}_+\cap B_\varepsilon}|u_e^\la|^{p+1}dx\\
&+\frac{1}{2}\int_{\R^{n+1}_+\cap( B_1\setminus B_\varepsilon)}y^b|\nabla\Delta_b u_e^\la|^2dxdy-\frac{C_{n,s}}{p+1}\int_{\pa\R^{n+1}_+\cap( B_1\setminus B_\varepsilon)}|u_e^\la|^{p+1}dx\\
=&\varepsilon^{n-2s\frac{p+1}{p-1}}\overline{E}(u_e,\la\varepsilon)+\frac{1}{2}\int_{\R^{n+1}_+\cap( B_1\setminus B_\varepsilon)}y^b|\nabla\Delta_b u_e^\la|^2dxdy\\
&-\frac{C_{n,s}}{p+1}\int_{\pa\R^{n+1}_+\cap( B_1\setminus B_\varepsilon)}|u_e^\la|^{p+1}dx.\\
\endaligned\ee
Letting $\la\rightarrow+\infty$ and then $\varepsilon\rightarrow0$, we deduce that $\lim_{\la\rightarrow+\infty}\overline{E}(u_e,\la)=0$.
Using the monotonicity of $E$,
\be\aligned
E(u_e,\la)&\leq\frac{1}{\la}\int_{\la}^{2\la} E(t)dt\leq \sup_{[\la,2\la]} \overline{E}+C\frac{1}{\la}\int_\la^{2\la}[E-\overline{E}]\\
&\leq\sup_{[\la,2\la]} \overline{E}+C\frac{1}{\la} \int_\la^{2\la}\la^{2s\frac{p+1}{p-1}-n-5}\int_{\R^{n+1}_+\cap\pa B_\la}y^b
 \big[(u_e)^2+\la^2|\nabla u_e|^2\\
 &\quad +\la^4(|\Delta_b u_e|^2+|\nabla^2 u_e|^2)+\la^6|\nabla\Delta_b u_e|^2\big]\\
&=\sup_{[\la,2\la]} \overline{E}+C\frac{1}{\la}\la^{2s\frac{p+1}{p-1}-n-5}\int_{\R^{n+1}_+\cap( B_{2\la}\backslash B_\la)}y^b
\big[(u_e)^2+\la^2|\nabla u_e|^2\\
 &\quad +\la^4(|\Delta_b u_e|^2+|\nabla^2 u_e|^2)+\la^6|\nabla\Delta_b u_e|^2\big]\\
 &=\sup_{[\la,2\la]} \overline{E}+C\frac{1}{\la}\int_{\R^{n+1}_+\cap( B_{2}\backslash B_1)}y^b
\big[(u_e^\la)^2+|\nabla u_e^\la|^2\\
 &\quad +|\Delta_b u_e^\la|^2+|\nabla^2 u_e^\la|^2+|\nabla\Delta_b u_e^\la|^2\big]\\
\endaligned\ee
and so $\lim_{\la\rightarrow\infty} E(u_e,\la)=0$. Since $u$ is smooth, we also have $E(u_e,0)=0$. Since $E$ is monotone, $E\equiv0$
and so $\overline{u}$ must be homogenous, a contradiction unless $u_e\equiv0$.

\section{Algebraic analysis: The proof of Theorem \ref{8Monotonem}}

Let $k:=\frac{2s}{p-1}$ and $m:=n-2s$. By  a direct calculation,  we obtain  that
\be\label{8A1A2}\aligned
A_1&= -10k^2+10mk-m^2+12m+25,\\
A_2&=3k^4-6mk^3+(3m^2-12m-30)k^2+(12m^2+30m)k+9m^2+36m+27,\\
B_1&=-6k^2+6mk+12m+30.
\endaligned\ee

Notice that our supercritical condition $p>\frac{n+2s}{n-2s}$ is equivalent to $0<k<\frac{n-2s}{2}=\frac{m}{2}$.
Next, we have the following lemma which yields the sign of $A_2$ and $B_1$.

\bl\label{8A2-0} If $p>\frac{n+2s}{n-2s}$, then $A_2>0$ and $B_1>0$.
\el

\bp From  \eqref{8A1A2}, we derive that
\be\label{8A2}
A_2=3(k+1)(k+3)(k-(m+1))(k-(m+3)),
\ee
and the roots of $B_1=0$ are

\be\nonumber
\frac{1}{2}m-\frac{1}{2}\sqrt{m^2+8m+20},\quad \frac{1}{2}m+\frac{1}{2}\sqrt{m^2+8m+20}.
\ee
Recall that $p>\frac{n+2s}{n-2s}$ is equivalent to $0<k<\frac{m}{2}$, we get the conclusion.
\ep

 To show monotonicity formula,  we  proceed   to prove the following inequality. That is,  there exist real numbers $c_{i,j}$ and positive real number
 $\epsilon$ such that

\be\aligned
\label{8keyinequality}
&3\la^5(\frac{d^3u^\la}{d\la^3})^2+A_{1}\la^3\Big(\frac{d^2u^\la}{d\la^2}\Big)^2+A_{2}\la(\frac{du^\la}{d\la})^2\\
&\geq \epsilon\la(\frac{du^\la}{d\la})^2+\frac{d}{d\la}
\Big(\sum_{0\leq i,j\leq2}c_{i,j}\la^{i+j}\frac{d^iu^\la}{d\la^i}\frac{d^ju^\la}{d\la^j}
\Big).
\endaligned\ee
 To deal with the rest of the dimensions, we employ the second idea: we find nonnegative constants $d_1, d_2$ and constants $c_1, c_2$ such that we have the following Jordan form decomposition:
\be\label{8quad}\aligned
&3\la^5(f''')^2+A_1\la^3(f'')^2+A_2\la(f')^2=3\la(\la^2f'''+c_1\la f'')^2+d_1\la(\la f''+c_2 f')^2\\
&\;\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+d_2\la(f')^2
+\frac{d}{d\la}(\sum_{i,j}e_{i,j}\la^{i+j}f^{(i)}f^{(j)}),
\endaligned\ee
where the  unknown constants are to be determined.

\bl
\label{860011}
Let $ p>\frac{n+2s}{n-2s}$ and $A_1$ satisfy
\be
\label{8A11}
 A_1+12 >0,
 \ee
 then there exist  nonnegative numbers $d_1,d_2$,    and real numbers $c_1,c_2,e_{i,j}$ such that
the  differential inequality (\ref{8quad}) holds.

 \el

 \bp Since $$4\la^4f'''f''=\frac{d}{d\la}(2\la^4(f'')^2)-8\la^3(f'')^2$$ and
 $$2\la^2f''f'=\frac{d}{d\la}(\la^2(f')^2)-2\la(f')^2,$$ by comparing the coefficients of $\lambda^3(f'')^2$ and $\lambda (f')^2,$  we have that

\be\nonumber
d_1=A_1-3c_1^2+12c_1, \quad d_2=A_2-(c_2^2-2c_2)(A_1-3c_1^2+12c_1).
\ee
In particular, $$\max_{c_1}{d_1(c_1)}=A_1+12   \hbox{  and the critical point is }   c_1=2. $$
Since $A_2>0$,  we select that $c_1=2,c_2=0$.  Hence,  in this case,
by a direct calculation we see that $d_1=A_1+12>0$.  Then we get  the conclusion.
\ep

We conclude from Lemma \ref{860011} that if $A_1+12>0$ then  (\ref{8keyinequality}) holds.  This implies that when $m<6+\sqrt{73}, p>\frac{n+2s}{n-2s}$ or $m\geq 6+\sqrt{73}$ and
\be
\label{pee}
\frac{n+2s}{n-2s}<p<\frac{5m+20s-\sqrt{15m^2+120m+370}}{5m-\sqrt{15m^2+120m+370}},
\ee
 then (\ref{8keyinequality}) holds.
\vskip0.2in

Let
\be\label{727=1}
p_m(n):=\begin{cases}
+\infty\;\;\;\;\;\;\;\;&\hbox{if}\;  n<2s+6+\sqrt{73},\\
\frac{5n+10s-\sqrt{15(n-2s)^2+120(n-2s)+370}}{5n-10s-\sqrt{15(n-2s)^2+120(n-2s)+370}}&\hbox{if}\; n\geq2s+6+\sqrt{73}.\\
\end{cases}
\ee
Combining  all the lemmas of this section,  we obtain the Theorem \ref{8Monotonem}.

\vskip0.2in

\vskip0.16in

Now we proceed to prove Theorem \ref{8Monotone}. From  Corollary 1.1 of  \cite{LWZ}, we know that
 if $n>2s,s>0,p>\frac{n+2s}{n-2s}$,  then there exists  $n_0(s)$, where $\frac{1}{\sqrt{n}}<a_{n,s}<\frac{1}{2}\frac{n-2s}{\sqrt{n}}+\frac{1}{\sqrt{n}}$, such that   the inequality  \eqref{8gamma} always holds whenever
 $n\leq n_0(s)$; while when  $n> n_0(s)$, then  the inequality   \eqref{8gamma}  is true  if and only if
$$p<p_2:=\frac{n+2s-2-2a_{n,s}\sqrt{n}}{n-2s-2-2a_{n,s}\sqrt{n}},$$
where $n_0(s)$ is in fact the largest $n$ satisfying    $n-2s-2-2a_{n,s}\sqrt{n}\leq0$.
In particular, $    \frac{n+2s}{n-2s}<\frac{n+2s-4}{n-2s-4}<p_2<+\infty.$  Therefore, we introduce
\be\label{8pcn88}
p_c(n):=\begin{cases}
+\infty\;\;\;\;\;\;\;\;&\hbox{if}\;\;\;\;\;\;\;\; n\leq n_0(s),\\
\frac{n+2s-2-2a_{n,s}\sqrt{n}}{n-2s-2-2a_{n,s}\sqrt{n}}&\hbox{if}\;\;\;\;\;\;\;\;n> n_0(s).\\
\end{cases}
\ee

From \cite{LWZ=00}, we use the sharp estimate $n_0(s)<2s+8.998$ for $2<s<3$, then
\be\label{8n0s}
n_0(s)\leq 2s+8.998<2s+6+\sqrt{73}\simeq2s+14.544.
\ee
On the other hand, via the sharp estimate $a_{n,s}<1$ from \cite{LWZ=00}
\be\label{8compare}\aligned
\frac{5n+10s-\sqrt{15(n-2s)^2+120(n-2s)+370}}{5n-10s-\sqrt{15(n-2s)^2+120(n-2s)+370}}>\frac{n+2s-2-2a_{n,s}\sqrt{n}}{n-2s-2-2a_{n,s}\sqrt{n}}
\endaligned\ee
provided that   $s\in(2,3)$ and that
\be\label{728-mm}\aligned
225m^4-720m^3-17244m^2-29088m+7236>0,   \hbox{ where }  m=n-2s.
\endaligned\ee
The  \eqref{728-mm} holds  whenever  $m>11.12$, that is $n>2s+11.12$.
This combine with \eqref{8n0s} we obtain that $p_c(n)<p_m(n)$.
Therefore we get   Theorem \ref{8Monotone}. \hfill $\Box$

\newpage


\subsection{#2}{#4}\par\hfill{\arx{#1}}\par\hfill\emph{#3}}
\nc{\arXiv}[3]{\subsection{#2} {#3}, \arx{#1}\par\hfill}
\nc{\DOIpaper}[5]{\subsection{#2}{#4}\par\hfill{\url{http://dx.doi.org/#1}}\par\hfill\emph{#3}}
\nc{\AMSPaper}[5]{\subsection{#3}{#5}\par\hfill{\url{#1}}\par\hfill\emph{#4}\par\hfill{#2}}
\nc{\nAMSPaper}[4]{\subsection{#2} {#3}, {#4}, \url{#1}}
\nc{\AMS}[3]{\subsection{#1} {#2}, \url{#3}}
\nc{\SPMBul}{\textbf{$\mathcal{SPM}$ Bulletin}}
\nc{\BulEnd}{\par\bigskip\noindent
Boaz Tsaban\\
\emph{E-mail}: tsaban@math.biu.ac.il\\
\emph{URL}: http://www.cs.biu.ac.il/\~{}tsaban}

\nc{\arx}[1]{\url{http://arxiv.org/abs/#1}}

\nc{\probissue}{\emph{Problem of the issue}}

\title[$\mathcal{SPM}$ Bulletin \textbf{\issuenumber} (\issuemonth{} \issueyear)]
$\mathcal{SPM}$ Bulletin\\[0.5cm]
Issue \issuenumber{} (\issuemonth{} \issueyear{})}

\begin{document}
\maketitle

\section{Editor's note}

In addition to the beautiful research announcements, this issue includes an announcement of a festive conference dedicated to Selection Principles, celebrating the
60th birthday of Marion Scheepers. This will be an opportunity for all active researchers in the field to exchange new methods and results in the field, and for researchers looking for new, promising research directions to consider this topic in a thorough manner.

\medskip

With best regards,

\by{Boaz Tsaban}{tsaban@math.biu.ac.il}

\hfill \texttt{http://www.cs.biu.ac.il/\~{}tsaban}

\section{New conference: Frontiers of Selection Principles}

Dear Friends and Colleagues,

\medskip

We are glad to invite you to the conference Frontiers of Selection Principles, celebrating the 60th birthday of Marion Scheepers. This festive conference will take place at the Dewajtis Campus, Bielanski forest, Warsaw, 21.8--1.9, 2017: 21.8--25.8 tutorial, 26.8 excursion, 27.8--1.9 conference. (This is immediately after the Logic Colloquium and immediately before the Bedlewo meeting on set theoretic topology and analysis.)

\emph{Selection principles} connect topology, set theory, and functional analysis and make it possible to transport and apply methods from each of these fields to the other ones. It is now one of the most active streams of research within set theory and general topology. This conference will be fully dedicated to selection principles and their applications. It will begin with a one week tutorial, aimed mainly at students and researchers with no prior knowledge of selection principles. This tutorial will provide an overview of the filed, and a detailed introduction to its main methods. Participants with no prior knowledge attending the tutorial will thus fully benefit from the lectures of the second week, and will be able to consider possible directions for research in the field.

The second conference week will consist of invited lectures, by leading experts, on results in the frontiers of selection principles. We recommend that students and participants with expertise in other disciplines attend both weeks of the conference.

The expected accommodation cost is approximately 20EUR per night (full board: 30EUR).

Registration fee: Early registration: 50EUR. Normal registration: 80EUR. Registration fees will help reducing the costs for some students or early career researchers that would otherwise not be able to attend this meeting. Discount possibilities will be provided in future emails.

\medskip

\noindent Tentative list of conference speakers:

\begin{center}
\begin{tabular}{lll}
Leandro Aurichi	&	Liljana Babinkostova	&	Taras Banakh\\
Angello Bella*	&	Daniel Bernal Santos	&	Maddalena Bonanzinga*\\
Lev Bukovsky	&	Steven Clontz*		&	Samuel Da-Silva*\\
Rodrigo Dias		&	Rafa\l{} Filipow			& 	David Gauld\\
Ondrej Kalenda	&	Ljubi\v{s}a Ko\v{c}inac*		&	Adam Krawczyk\\
Adam Kwela		& 	Andrzej Nowik		&	Selma\"O{}zca\v{g}*\\
Yinhe Peng		& 	Szymon Plewik		&	Robert Rałowski	\\
Masami Sakai		&	Marion Scheepers		&  	Paul Szeptycki*	\\
Piotr Szewczak	&	Franklin Tall			& 	Secil Tokg\"oz	\\
Boaz Tsaban		& 	Tomasz Weiss		& 	Lyubomyr Zdomskyy\\
Shuguo Zhnag	&	Ondrej Zindulka		&	Szymon Zeberski
\end{tabular}
\end{center}

\noindent (* to be confirmed)

\medskip

Subject to time constraints, students and other participants may have an opportunity to contribute a short lecture on a topic of selection principles.

\noindent\textbf{Special request:} Since a part of this workshop is especially accessible to students, we would appreciate your forwarding this message to your students and to other students who may be interested in attending this conference.

\noindent The conference is organized by Cardinal Stefan Wyszyński University in Warsaw.

\noindent\textbf{Important:} This is the only message distributed widely. Please email Piotr Szewczak (p.szewczak@wp.pl) to be included in the mailing list for details and updates.

\medskip

On behalf of the organizing committee,

\medskip

\noindent Piotr Szewczak (Cardinal Stefan Wyszynski University and Bar-Ilan University)\\
\noindent Boaz Tsaban (Bar-Ilan University)\\
\noindent Lyubomyr Zdomskyy (Kurt Gödel Research Center)

\section{Long announcements}

\arXivl{1607.04688}
{On the definability of Menger spaces which are not $\sigma$-compact}
{Franklin D. Tall and Secil Tokgoz}
{Hurewicz proved completely metrizable Menger spaces are $\sigma$-compact. We
extend this to \v{C}ech-complete Menger spaces and consistently to projective
Menger metrizable spaces. On the other hand, it is consistent that there is a
co-analytic Menger space that is not $\sigma$-compact.}

\arXivl{1607.04781}
{Definable versions of Menger's conjecture}
{Franklin D. Tall}
{Menger's conjecture that Menger spaces are $\sigma$-compact is false; it is
true for analytic subspaces of Polish spaces and undecidable for more complex
definable subspaces of Polish spaces. For non-metrizable spaces, analytic
Menger spaces are $\sigma$-compact, but Menger continuous images of co-analytic
spaces need not be. The general co-analytic case is still open, but many
special cases are undecidable, in particular, Menger topological groups. We
also prove that if there is a Michael space, then productively Lindelof
Cech-complete spaces are $\sigma$-compact. We also give numerous
characterizations of proper K-Lusin spaces. Our methods include the Axiom of
Co-analytic Determinacy, non-metrizable descriptive set theory, and
Arhangel'skii's work on generalized metric spaces.}

\arXivl{1608.03546}
{Discrete subsets in topological groups and countable extremally
  disconnected groups}
{Evgenii Reznichenko and Ol'ga Sipacheva}
{It is proved that any countable topological group in which the filter of
neighborhoods of the identity element is not rapid contains a discrete set with
precisely one onisolated point and that the existence of a nondiscrete
countable extremally disconnected group implies the existence of a rapid
ultrafilter.}

\arXivl{1608.06210}
{$\sigma$-Ideals and outer measures on the real line}
{S. Garc\'ia-Ferreira, A. H. Tomita, and Y. F. Ortiz-Castillo}
{A {\it weak selection} on $\mathbb{R}$ is a function $f: [\mathbb{R}]^2 \to
\mathbb{R}$ such that $f(\{x,y\}) \in \{x,y\}$ for each $\{x,y\} \in
[\mathbb{R}]^2$. In this article, we continue with the study (which was
initiated in [ag]) of the outer measures $\lambda_f$ on the real line
$\mathbb{R}$ defined by weak selections $f$. One of the main results is to show
that $CH$ is equivalent to the existence of a weak selection $f$ for which: \[
\mathcal \lambda_f(A)= \begin{cases} 0 & \text{if $|A| \leq \omega$,}\\ \infty
& \text{otherwise.} \end{cases} \] Some conditions are given for a
$\sigma$-ideal of $\mathbb{R}$ in order to be exactly the family
$\mathcal{N}_f$ of $\lambda_f$-null subsets for some weak selection $f$. It is
shown that there are $2^\mathfrak{c}$ pairwise distinct ideals on $\mathbb{R}$
of the form $\mathcal{N}_f$, where $f$ is a weak selection. Also we prove that
Martin Axiom implies the existence of a weak selection $f$ such that
$\mathcal{N}_f$ is exactly the $\sigma$-ideal of meager subsets of
$\mathbb{R}$. Finally, we shall study pairs of weak selections which are
"almost equal" but they have different families of $\lambda_f$-measurable sets.}

\arXivl{1609.05822}
{Selectively (a)-spaces from almost disjoint families are necessarily countable under a certain parametrized weak diamond principle}
{Charles J.G. Morgan and Samuel G. Da Silva}
{The second author has recently shown [20] that any selectively (a) almost disjoint
family must have cardinality strictly less than $2^{\aleph_0}$, so under the Continuum
Hypothesis such a family is necessarily countable. However, it is also shown in the
same paper that $2^{\aleph_0} < 2^{\aleph_1}$ alone does not avoid the existence of
uncountable selectively (a) almost disjoint families. We show in this paper that a
certain effective parametrized weak diamond principle is enough to ensure countability
of the almost disjoint family in this context. We also discuss the deductive strength
of this specific weak diamond principle (which is consistent with the negation of the
Continuum Hypothesis, apart from other features).
\par
Houston Journal of Mathematics 42 (2016), 1031--1046.}

\arXivl{1206.0722}
{Notes on the od-Lindel\"of property}
{Mathieu Baillif}
{A space is od-compact (resp. od-Lindel\"of) provided any cover by open dense
sets has a finite (resp. countable) subcover. We first show with simple
examples that these properties behave quite poorly under finite or countable
unions. We then investigate the relations between Lindel\"ofness,
od-Lindel\"ofness and linear Lindel\"ofness (and similar relations with
`compact'). We prove in particular that if a $T_1$ space is od-compact, then
the subset of its non-isolated points is compact. If a $T_1$ space is
od-Lindel\"of, we only get that the subset of its non-isolated points is
linearly Lindel\"of. Though, Lindel\"ofness follows if the space is moreover
locally openly Lindel\"of (i.e. each point has an open Lindel\"of
neighborhood).}

\arXiv{1610.04800}
{Relating games of Menger, countable fan tightness, and selective
  separability}
{Steven Clontz}
{By adapting techniques of Arhangel'skii, Barman, and Dow, we may equate the
existence of perfect-information, Markov, and tactical strategies between two
interesting selection games. These results shed some light on Gruenhage's
question asking whether all strategically selectively separable spaces are
Markov selectively separable.}

\arXivl{1611.04998}
{Convergence in topological groups and the Cohen reals}
{Alexander Shibakov}
{We show that it is consistent to have an uncountable sequential group of
intermediate sequential order while no countable such groups exist. This is
proved by adding $\omega_2$ Cohen reals to a model of $\diamondsuit$.}

\section{Short announcements}\label{RA}

\arXiv{1607.04756}
{Hereditarily normal manifolds of dimension $> 1$ may all be metrizable}
{Alan Dow and Franklin D. Tall}

\arXiv{1607.07188}
{A long chain of P-points}
{Borisa Kuzeljevic and Dilip Raghavan}

\arXiv{1607.07669}
{The cofinal structure of precompact and compact sets in general metric
  spaces}
{Aviv Eshed, M. Vincenta Ferrer, Salvador Hern\'andez, Piotr Szewczak,
  Boaz Tsaban}

\arXiv{1607.07978}
{$\omega^\omega$-bases in topological and uniform spaces}
{Taras Banakh}

\arXiv{1608.03381}
{$I$-convergence classes of sequences and nets in topological spaces}
{Amar Kumar Banerjee and Apurba Banerjee}

\arXiv{1608.07144}
{The existence of continuous weak selections and orderability-type
  properties in products and filter spaces}
{Koichi Motooka, Dmitri Shakhmatov, Takamitsu Yamauchi}

\arXiv{1610.04506}
{Weakly linearly Lindel\"of monotonically normal spaces are Lindel\"of}
{I. Juh\'asz, V. V. Tkachuk, R. G. Wilson}

\arXiv{1611.07267}
{On the cardinality of almost discretely Lindel\"of spaces}
{Santi Spadaro}

\arXiv{1611.08289}
{Categorical properties on the hyperspace of nontrivial convergent
sequences}
{S. Garcia-Ferreira, R. Rojas-Hernandez, Y. F. Ortiz-Castillo}

\arXiv{1612.06651}
{First countable and almost discretely Lindel\"of $T_3$ spaces have
  cardinality at most continuum}
{Istv\'an Juh\'asz, Lajos Soukup, Zolt\'an Szentmikl\'ossy}

\ed