数理解析研究所講究録
第 1628 巻 2009 年 28-41

28

Asymptotic behavior of least energy solutions
for a biharmonic problem withnearly critical growth
高橋 太 (Futoshi Takahashi)
大阪市立大学理学研究科 (Osaka City Univ.)

E-mail: futoshi@sci.osaka-cu.ac.jp

1

Introduction

In this note, we concern the asymptotic behavior of blowing-up solutions to
the fourth order semilinear problem$(P_{\epsilon,K})\{\begin{array}{ll}\Delta^{2}u=c_{0}K(x)u^{p_{e}} in \Omega,u>0 in\Omega,u=\Delta u=0 on \partial\Omega\end{array}$as. Here,is a smooth bounded domain in $\mathbb{R}^{N}(N\geq 5),$ $c_{0}=$,is a small positive parameter,$(N-4)(N-2)N(N+2),$$p=(N+4)/(N-4)$ is the critical Sobolev exponent from the view point of
the Sobolev embedding $H^{2}\cap H_{0}^{1}(\Omega)arrow U^{+1}(\Omega)$ , andis a given
positive function.
When $K\equiv 1$ , Chou and Geng [1] obtained a result corresponding to theone of Han [5] on a strictly convex domain for solutions minimizing the
Sobolev quotient:$\epsilonarrow+0$$\Omega$$\epsilon>0$$p_{\epsilon}=p-\epsilon$$K\in C^{2}(\overline{\Omega})$$\Omega$$u_{\epsilon}$$\frac{\Vert u_{\epsilon}\Vert_{H^{2}\cap H_{0}^{1}(\Omega)}^{2}}{||u_{\epsilon}||_{L^{p\epsilon+1}(\Omega)}^{2}}arrow S$

as$\epsilonarrow 0$

.

29
Here
$\Vert u_{\epsilon}\Vert_{H^{2}\cap H_{0}^{1}(\Omega)}=(\int_{\Omega}|\triangle u|^{2}dx)^{1/2}$

is the norm of the Hilbert space, and $S= \inf\{\int_{\Omega}|\Delta u|^{2}dx|u\in$is the best Sobolev constant of the embedding. In particular, they proved that the blow up pointof solutions minimizing the Sobolev quotient is a critical point of the Robinfunction associated with the Green function under the Navier boundary condition.
, there always exists a function
Also whensatisfying$H^{2}\cap H_{0}^{1}(\Omega)$$H^{2}\cap H_{0}^{1}(\Omega),$$\Vert u\Vert_{Lp}+1(\Omega)=1\}$$H^{2}\cap H_{0}^{1}(\Omega)arrow L^{p+1}(\Omega)$$K\not\equiv 1$$\overline{u}_{\epsilon}$$\frac{\int_{\Omega}|\Delta\overline{u}_{\epsilon}|^{2}dx}{(\int_{\Omega}K(x)|\overline{u}_{\epsilon}|^{p_{e}+1}dx)^{2/(p_{g}+1)}}=\inf_{u\in H^{2}\cap H_{0}^{1}(\Omega)}\frac{\int_{\Omega}|\Delta u|^{2}dx}{(\int_{\Omega}K(x)|,u|^{p_{l}+1}dx)^{2/(pff+1)}}$.

$v\in H^{2}\cap$
We may assume
by solving the equation; see [9]. Thus, an appropriate constant multiple of Of\’e is a solution of, which we call a least energy solution to. In the following, we
will treat only least energy solutions to.
For non constant $K$ , least energy solutions
are known to blow up at
one point , which is a maximum point of $K$ in :$\overline{u}_{\epsilon}>0$$-\triangle v=|\Delta\overline{u}_{\epsilon}|,$$H_{0}^{1}(\Omega)$$(P_{\epsilon,K})$$(P_{\epsilon,K})$$(P_{\epsilon,K})$$\{u_{\epsilon}\}$$\overline{\Omega}$$x_{0}$$||u_{\epsilon}\Vert_{L^{\infty}(\Omega)}=u_{\epsilon}(x_{\epsilon})arrow\infty$and$x_{\epsilon}arrow x_{0}\in K^{-1}(m_{\frac{a}{\Omega}}xK)$

.

(1.1)

In what follows, we assume the function $K$ satisfies
Assumption (K)unique interior point
critical point of $K$ .

$K\in C^{2}(\overline{\Omega}),$$x_{0}\in\Omega$at theis a nondegenerate$0<K(x)\leq 1,$ $K$ attains

with $K(x_{0})=1$ , and$x_{0}$$\max_{\overline{\Omega}}K$

In the sequel, let $G=G(x, y)$ denote the Green function ofNavier boundary condition:$\{\begin{array}{l}\Delta^{2}G(\cdot, y)=\delta_{y} in \Omega,G(\cdot, y)=\Delta G(\cdot, y)=0 on \partial\Omega,\end{array}$and let$\Gamma(x, y)$be the fundamental solution of$\triangle^{2}$$\Gamma(x, y)=\{$ $\frac{1}{\frac{(N1}{\sigma_{4}}\log|x-y|-4)(N-2)\sigma_{N}}|x-y|^{4-N}-1,$’:,$N=4$ ,
$N\geq 5$$\Delta^{2}$

under the

30

here
is the volume of the $(N-1)$ dimensional unit sphere in. Finally,
let $R(x)= \lim_{yarrow x}[\Gamma(x, y)-G(x, y)]$ denote the Robin function ofwith
the Navier boundary condition. By the maximum principle, we see $R>0$
on and $R(x)arrow+\infty$ as tends to the boundary of .
$\mathbb{R}^{N}$$\sigma_{N}$$\Delta^{2}$$\Omega$$\Omega$$x$

Main result of this note reads as follows.
$N\geq 5$ be a smooth bounded domain. Let
Theorem 1 Let
be aleast energy solution toand letbe a point such thatfor
. Assume (K). Then after passing to a subsequence, $we$
have$\Omega\subset \mathbb{R}^{N},$$u_{\epsilon}$$\epsilon>0$$(P_{\epsilon,K})$$x_{\epsilon}\in\Omega$$u_{\epsilon}(x_{\epsilon})=\Vert u_{\epsilon}\Vert_{L\infty(\Omega)}$

(1)
(2)

$\{\begin{array}{ll}|x_{\epsilon}-x_{0}|=O(\Vert u_{\epsilon}\Vert_{L\infty(\Omega)}^{-2}) N=5,|x_{\epsilon}-x_{0}|=o(\Vert u_{\epsilon}\Vert_{L\infty)}^{-\frac{2}{N-4(\Omega}}) N\geq 6,\end{array}$$\Vert u_{\epsilon}\Vert_{L^{\infty}(\Omega)}^{\epsilon}arrow 1$$as\epsilonarrow 0$,(3) $\Vert u_{\epsilon}\Vert_{L\infty(\Omega)}u_{\epsilon}(x)arrow 2(N-4)(N-2)\sigma_{N}G(x, x_{0})$(4)as$\epsilonarrow 0,$$(x\neq x_{0})$$\{\begin{array}{ll}\lim_{\epsilonarrow 0}\epsilon\Vert u_{\epsilon}\Vert_{L(\Omega)}^{2_{\infty}}=\frac{2^{16}}{21}\pi R(x_{0}) N=5,\lim_{\epsilonarrow 0}\epsilon\Vert u_{\epsilon}\Vert_{L(\Omega)}^{2}\infty=-\frac{1}{4}\triangle K(x_{0})+480\pi^{3}R(x_{0}) N=6,\lim_{\epsilonarrow 0}\epsilon\Vert u_{\epsilon}\Vert_{(\Omega)}^{\frac{4}{L\infty N-4}}=-\frac{2}{(N-2)(N-4)}\Delta K(x_{0}) N\geq 7.\end{array}$Thus, the above theorem corresponds to the one proved by Hebey [6]for the second order Laplacian case problem. Our starting point of proof
is to establish a key pointwise estimate for ; see Lemma 2 below. To dothis, we rely on the blow up analysis with the Navier boundary condition
performed by Geng [4]. Although Geng assumed the strict convexity of thedomain and $K\equiv 1$ in [4], his blow up analysis works well if the solution
sequence considered is known a priori to blow up at the unique interior pointof . Note that in our case, the boundary blow up cannot occur since we, the unique maximum point of $K$ , for least energyknow
solutions. Therefore, we confirm that the blow up point
is indeed anisolated simple blow up point in the sense of [4], without any restriction of
the domain dimension. The needed pointwise estimate can be derived bom
this fact. For local blow up analysis (without any boundary condition) for
any solution sequence of subcritical biharmonic equations with nearly critical
growth, see the works of Djadli, Malchiodi and Ahmedou [2] and Felli [3].
See also the original work of YanYan Li [10] for the Laplacian case.
$u_{\epsilon}$$\Omega$$x_{\epsilon}arrow x_{0}\in\Omega$$x_{0}$

31

In Theorem 1, we observe that the asymptotics depend sensitively on thedimension of the domain: The geometric effect (the Robin function $R(x_{0})$ ) is
dominant in the lowest dimension $N=5$ , the effect of the coefficient function
is dominant when $N\geq 7$ , and they are mixed when $N=6$ . Thisphenomenon was also observed in the second order Laplacian case by Hebey[6].
$(\triangle K(x_{0})\cdot)$

2

Proof of Theorem 1

In this section, we will show the sketch of proof of Theorem 1. We will treat
the case $N\geq 6$ only for the sake of simplicity. Detailed arguments including
the case $N=5$ can be found in the forthcoming paper [8].
First, we recall the Pohozaev type identity for a biharmonic equation with
the Navier boundary condition. Let
be a solution of thefollowing equation$u\in C^{4}(\Omega)\cap C^{3}(\overline{\Omega})$$s$$\{\begin{array}{ll}\Delta^{2}u=f(x, u) in \Omega,u=\Delta u=0 on \partial\Omega,\end{array}$where is in
we have an identity:$f$$C^{1}(\overline{\Omega}\cross \mathbb{R})$. Denote $F(x, u)= \int_{0}^{u}f(x, s)ds$ for any$x\in\overline{\Omega}$. Then$\int_{\Omega}NF(x, u)-(\frac{N-4}{2})uf(x, u)+(x-y)\cdot\nabla_{x}F(x, u)dx$(2.1)$= \int_{\partial\Omega}((x-y)\cdot\nabla u)\frac{\partial v}{\partial\nu}ds_{x}$for any$x\in\partial\Omega$$y\in \mathbb{R}^{N}$.

, where $v=$ -Au and $\nu=\nu(x)$ is an outer unit normal at

For a least energy solution$u_{\epsilon}$of$(P_{\epsilon,K})$, the identity (2.1) becomes$\frac{c_{0}(N-4)^{2}}{2(2N-\epsilon(N-4))}\epsilon\int_{\Omega}K(x)u_{\epsilon}^{p_{\epsilon}+1}dx$$+ \frac{c_{0}(N-4)}{2N-\epsilon(N-4)}\int_{\Omega}(x-y)\cdot\nabla K(x)u_{\epsilon}^{p_{\epsilon}+1}dx=\int_{\partial\Omega}((x-y)\cdot\nabla u_{\epsilon})\frac{\partial v_{\epsilon}}{\partial\nu}ds_{x}$

(2.2)

where

$v_{\epsilon}=-\Delta u_{\epsilon}$. Also by differentiating (2.2) with respect to$y_{i}$$\frac{c_{0}(N-4)}{2N-\epsilon(N-4)}\int_{\Omega}\frac{\partial K}{\partial x_{i}}(x)u_{\epsilon}^{p_{e}+1}dx=\int_{\partial\Omega}\frac{\partial u_{\epsilon}}{\partial\nu}\frac{\partial v_{\epsilon}}{\partial\nu}\nu_{i}ds_{x}$, we have

(2.3)

32

for all $i=1,$ $\cdots N$ . Note that.
on
Next, define the scaled function$u_{\epsilon},$)$v_{\epsilon}>0$

in

and

$\Omega$$\partial\Omega$$-|\nabla v_{\epsilon}|\nu$$\tilde{u}_{\epsilon}(y):=\frac{1}{\Vert u_{\epsilon}\Vert}u_{\epsilon}(\frac{y}{\Vert u_{\epsilon}||^{R\epsilon_{\frac{-1}{4}}}}+x_{\epsilon})$where$\nabla u_{\epsilon}=-|\nabla u_{\epsilon}|\nu,$ $\nabla v_{\epsilon}=$$\Omega_{\epsilon}=\Vert u_{\epsilon}\Vert^{g_{L_{4}}\underline{-1}}(\Omega-x_{\epsilon})$$\Vert\cdot\Vert_{L(\Omega)}\infty$. It holds that

,

(2.4)

$y\in\Omega_{\epsilon}$, and in the following, we abbreviate, andsatisfies$0<\tilde{u}_{\epsilon}\leq 1,\tilde{u}_{\epsilon}(0)=1$$\Vert\cdot\Vert=$$\tilde{u}_{\epsilon}$$\Delta^{2}\tilde{u}_{\epsilon}=c_{0}\tilde{K}_{\epsilon}(y)\tilde{u}_{\epsilon}^{p_{e}}$$\{$$in\Omega_{\epsilon}on\partial\Omega_{\epsilon}$$\tilde{u}_{\epsilon}=\Delta\tilde{u}_{\epsilon}=0$

where$\tilde{K}_{\epsilon}(y)=K(\frac{y}{||u_{\epsilon}||^{R}\succ-1}+x_{\epsilon})$$\Vert u_{\epsilon}\Vertarrow\infty$,,. By (1.1) and (K), we know$x_{\epsilon}arrow x_{0}$as$\in\Omega$.

$\epsilonarrow 0$,

compact uniformly on
thus
as
and
By standard elliptic estimates and the uniqueness of the limit, we have$\tilde{K}_{\epsilon}arrow K(x_{0})=1$$\Omega_{\epsilon}arrow \mathbb{R}^{N}$$\tilde{u}_{\epsilon}arrow U$

as$\epsilonarrow 0$$\mathbb{R}^{N}$

compact uniformly in$\epsilonarrow 0$

.

(2.5)

$\mathbb{R}^{N}$, where$U(y)=( \frac{1}{1+|y|^{2}})^{\frac{N-4}{2}}$is the unique solution of$\{\begin{array}{l}\Delta^{2}U=c_{0}U^{p} in\mathbb{R}^{N},0<U\leq 1, U(0)=1,\lim_{|y|arrow\infty}U(y)=0.\end{array}$By (2.5), we easily see that there exists a constant $M\geq 1$ independent ofsuch that for any sufficiently small, there holds$\epsilon$$\epsilon$$1\leq\Vert u_{\epsilon}\Vert^{\epsilon}\leq M$

.

(2.6)

See [5]: Corollary 1, or [1]: Lemma 4.1.
through the
Also we have the following crucial pointwise estimate fortheory of isolated simple blow up points in [4]; see also [2] and [3].
$u_{\epsilon}$

33

Lemma 2 There exists a constant $C>0$ independent of such that for any
with, the following estimates hold true;$\epsilon$$r_{\epsilon}=R_{\epsilon}\Vert u_{\epsilon}\Vert^{-R\equiv}4\underline{1}arrow 0$$R_{\epsilon}arrow\infty$$u_{\epsilon}(x) \leq C\frac{\Vert u_{\epsilon}\Vert}{(1+\Vert u_{\epsilon}\Vert^{\varpi_{-}^{4}z}|x-x_{\epsilon}|^{2})^{\frac{N-4}{2}}}$$u_{\epsilon}(x) \leq\frac{C}{\Vert u_{\epsilon}\Vert}\frac{1}{|x-x_{\epsilon}|^{N-4}}$

for

,

,

$for|x-x_{\epsilon}|\leq r_{\epsilon}$$\{|x-x_{\epsilon}|>r_{\epsilon}\}\cap\Omega$

.

,

(2.7)

(2.8)

Proof. As stated in Introduction, we appeal to the blow up analysis in [4]to prove Lemma. We will see that the interior blow up point
is indeed anisolated simple blow up one. We refer [4] for the definition of isolated, andisolated simple blow up points. See also the original work by YanYan Li [10]for the Laplacian case problem.
First, by a standard argument originally due to R. Schoen (for example,[7] Lemma 3.1), we know that any interior blow up point is an isolated one;see [4] Proposition 2.1. Note that though the convexity of the domain isassumed in [4], the assumption is used only to assure that any blow up point
is in the interior of the domain . Also since
makes one point blow up in
our case, we do not need an argument using the Pohozaev identity to deal
$x_{0}$$\Omega$$u_{\epsilon}$with multiple blow up points and their interactions. Therefore, the coefficient
function $K$ does not have any effect on the validity of the proofs in [4]. Thus,
by Proposition 2.2 in [4], we have the estimate$\Vert\tilde{u}_{\epsilon}(\cdot)-(1+|y|^{2})^{-\frac{N-4}{2}}\Vert_{C^{4}(B_{R_{g}(0)})}\leq\delta_{\epsilon}$

for any$arrow 0$ .
By takingwithand$(2.7)$ holds when.
Next, Proposition 4.1 in [4] is valid for least energy solutions offor
any $N\geq 5$ , when $K$ is a positive function satisfying (K). Thus we have that
any interior isolated blow up point is an isolated simple one by Proposition
4.1 in [4], and by Proposition 3.2 in [4], we have the estimate$R_{\epsilon}arrow\infty$$(1+R_{\epsilon}^{2})^{-\frac{N-4}{2}},$$R_{\epsilon}\Vert u_{\epsilon}\Vert^{-R\epsilon_{\frac{-1}{4}}}arrow 0$$\delta,$$\delta_{\epsilon}\leq$$|x-x_{\epsilon}|\leq r_{\epsilon}=R_{\epsilon}\Vert u_{\epsilon}\Vert^{-R\in_{4^{\underline{-1}}}}$$(P_{\epsilon,K})$$u_{\epsilon}(x) \leq\frac{C}{\Vert u_{\epsilon}\Vert}\frac{1}{|x-x_{\epsilon}|^{N-4}}$(2.9), where $C$ and are positive constants independent
for anyof . From this, we check that the estimate$r_{\epsilon}\leq|x-x_{\epsilon}|\leq\rho$$\rho$$\epsilon$$u_{\epsilon}(x) \leq\frac{C}{\Vert u_{\epsilon}\Vert}\frac{1}{\rho^{N-4}}$for$\{|x-x_{\epsilon}|>\rho\}\cap\Omega$(2.10)

34
holds true. Indeed, from (2.9) we have$u_{\epsilon}(x) \leq\frac{C}{\Vert u_{\epsilon}\Vert}\frac{1}{\rho^{N-4}}$

for $|x-x_{\epsilon}|=\rho$ .

(2.11)

If there exists a point $x’\in\{|x-x_{\epsilon}|>\rho\}\cap\Omega$ such that,
we would have a maximum point in the region $\{|x-x_{\epsilon}|>\rho\}\cap\Omega$ . But thisand (2.11) would contradict the fact thatis an isolated simple blow up
point. Finally, (2.8) follows easily from (2.9), (2.10) and the boundedness ofロ
the domain.
$u_{\epsilon}(x’)> \frac{C}{||u_{\epsilon}||}\frac{1}{\rho^{N-4}}$$x_{0}$In terms of$\tilde{u}_{\epsilon}$in (2.4), the above lemma reads(2.12)$\tilde{u}_{\epsilon}(y)\leq\{\begin{array}{ll}CU(y) for |y|\leq R_{\epsilon},C\frac{1}{|y|^{N-4}} for \{|y|>R_{\epsilon}\}\cap\Omega_{\epsilon},\end{array}$where
is any sequence as in Lemma 2.
Fkom Lemma 2, we also obtain the following:$R_{\epsilon}arrow\infty$

Lemma 3 There exists a constant $C>0$ independent of$\epsilon$

such that$\int_{\partial\Omega}|\nabla u_{\epsilon}||\nabla v_{\epsilon}|ds\leq C\Vert u_{\epsilon}\Vert^{-2}$holds trzee.
Proof. This is done by using Lemma 2 and the fact: Let

$u$

solve

$-\triangle u=f$$\{$

and let$\omega’\subset\subset\omega$$in\Omega on\partial\Omega$

,

$u=0$

be a neighborhood of$\partial\Omega$. Then$\Vert u\Vert_{W^{1,q}(\Omega)}+||u\Vert_{C^{1,\alpha}(\omega^{l})}\leq C(\Vert f\Vert_{L^{1}(\Omega)}+\Vert f\Vert_{L\infty(\omega)})$$\alpha\in(0,1)$ .
holds forthe claimed estimate isis indeed proved in the proof.
We apply 2.13) to$q< \frac{N}{N-1},$(2.13)

See [5] Lemma 2; there the left hand side of
however the estimate (2.13)$\Vert u\Vert_{W^{1,q}(\Omega)}+\Vert\nabla u\Vert_{C^{0,\alpha}(\omega’))}$$($$\{\begin{array}{ll}-\triangle u_{\epsilon}=v_{\epsilon} in \Omega,-\Delta v_{\epsilon}=c_{0}K(x)u_{\epsilon}^{p_{\epsilon}} in \Omega,u_{\epsilon}=v_{\epsilon}=0 on\partial\Omega.\end{array}$

35
As a consequence, it appears that we need to estimate
to control bothandBy (2.6) and the fact $0<K(x)\leq 1$ , we have$\Vert c_{0}K(x)u_{\epsilon}^{p_{\epsilon}}\Vert_{L^{\infty}(\omega)}$$\Vert\nabla u_{\epsilon}\Vert_{L(\partial\Omega)}\infty$$\Vert c_{0}K(x)u_{\epsilon}^{p_{\epsilon}}\Vert_{L^{1}(\Omega)}$.

$\Vert\nabla v_{\epsilon}\Vert_{L(\partial\Omega)}\infty$

and$\int_{\Omega}c_{0}K(x)u_{\epsilon}^{P\epsilon}dx\leq C\int_{\Omega}u_{\epsilon}^{p_{\Xi}}dx=C\Vert u_{\epsilon}\Vert^{p_{\epsilon}-(\frac{-1}{4})N}Ri\int_{\Omega_{\epsilon}}\tilde{u}_{\epsilon}^{p_{\epsilon}}(y)dy$$=C \Vert u_{\epsilon}\Vert^{p_{\epsilon}-(\frac{-1}{4})N}\epsilon a(\int_{\mathbb{R}^{N}}U^{p}(y)dy+o(1))$$\leq C\Vert u_{\epsilon}\Vert^{-1+\epsilon(\frac{N-4}{4})}\leq C\Vert u_{\epsilon}\Vert^{-1}$if

is sufficiently small. Here we have used (2.5), (2.12) and the Lebesgue
convergence theorem.
On the other hand, since we may take a neighborhood ofsmall such
, we see by Lemma 2
that$\epsilon>0$$\partial\Omega$$x_{0}\not\in\omega$$c_{0}K(x)u_{\epsilon}^{p_{\epsilon}}(x)\leq C\frac{\Vert u_{\epsilon}||^{-p_{e}}}{|x-x_{0}|^{(N-4)p_{\epsilon}}}$$\leq C|$for any$|^{-p_{g}}\leq C|$

剛-1

small such that
. Thus we have
. These estimates with (2.13) leads to$x\in\omega$$C\Vert u_{\epsilon}\Vert^{-1}$

圏

, if

$\epsilon>0$$1<p_{\epsilon}$$\Vert\nabla u_{\epsilon}\Vert_{L^{\infty}(\partial\Omega)}\leq C\Vert u_{\epsilon}\Vert^{-1}$and$\Vert c_{0}K(x)u_{\epsilon}^{p_{\epsilon}}\Vert_{L}\infty(\omega)\leq$$\Vert\nabla v_{\epsilon}\Vert_{L(\partial\Omega)}\infty\leq C\Vert u_{\epsilon}\Vert^{-1}$from which we obtain Lemma 3.

,ロ

Now, we will prove the estimates$|x_{\epsilon}-x_{0}|=o(\Vert u_{\epsilon}\Vert^{-\frac{2}{N-4}})$,(2.14)$N\geq 6$

under the assumption (K).
Indeed, by Taylor expansion, we have$K(x)=1+ \frac{1}{2}\sum_{i,j=1}^{N}b_{ij}(x_{i}-x_{i}^{0})(x_{j}-x_{j}^{0})+O(|x-x_{0}|^{3})$,(2.15)

and$\frac{\partial K}{\partial x_{i}}(x)=\sum_{j=1}^{N}b_{ij}(x_{j}-x_{j}^{0})+O(|x-x_{0}|^{2})$(2.16)

36

for all $i=1,$
we have$\cdots,$$N$, where we set$b_{ij}= \frac{\partial^{2}K}{\partial x_{i}\partial x_{j}}(x_{0})$. Inserting (2.16) into (2.3),$\frac{c_{0}(N-4)}{2N-\epsilon(N-4)}\int_{\Omega}\sum_{j=1}^{N}b_{ij}(x_{j}-x_{j}^{0})u_{\epsilon}^{p_{\epsilon}+1}dx+\int_{\Omega}o(|x-x_{0}|^{2})u_{\epsilon}^{p_{\zeta}+1}dx$(2.17)$= \int_{\partial\Omega}\frac{\partial u_{\epsilon}}{\partial\nu}\frac{\partial v_{\epsilon}}{\partial\nu}\nu_{i}ds_{x}$$N$ . The right hand side of (2.17) isfor $i=1,$
Now, by the change of variables (2.4), we have$\cdots,$$O(\Vert u_{\epsilon}\Vert^{-2})$$\int_{\Omega}o(|x-x_{0}|^{2})u_{\epsilon}^{p_{\epsilon}+1}dx=\Vert u_{\epsilon}\Vert^{p_{\epsilon}+1-(\frac{-1}{4})N}\epsilon\epsilon\int_{\Omega_{\epsilon}}o(|\frac{y}{\Vert u_{\epsilon}||^{\epsilon a_{4}^{\underline{-1}}}}$by Lemma 3.

十 $x_{\epsilon}-x_{0}|^{2})\tilde{u}_{\epsilon}^{p_{e}+1}dy$ .

Splitting the integral as$\int_{\Omega_{\epsilon}}o(|\frac{y}{\Vert u_{\epsilon}||^{\epsilon\epsilon_{\frac{-1}{4}}}}+x_{\epsilon}-x_{0}|^{2})\tilde{u}_{\epsilon}^{p_{\epsilon}+1}dy$$= \int_{\{y\in\Omega_{\epsilon}:|y|\leq||u_{\epsilon}||^{Rg}Z^{\underline{-1}}|x_{\epsilon}-xo|\}}(\cdots)dy+\int_{\{y\in\Omega_{e}:|y|>||u_{e}||^{R}|x_{e}-x_{0}|\}}\succ-1(\cdots)dy$$=:I_{1}+I_{2}$,and estimating$I_{1}=O(|x_{\epsilon}-x_{0}|^{2})$,$I_{2} \leq C\Vert u_{\epsilon}\Vert^{-(\frac{-1}{2})}\epsilon\epsilon\int_{\Omega_{\epsilon}}|y|^{2}\tilde{u}_{\epsilon}^{p_{\zeta}+1}dy$,$=C \Vert u_{\epsilon}\Vert^{-\frac{4}{N-4}}(\int_{\mathbb{R}^{N}}|y|^{2}U^{p+1}dy+o(1))=O(\Vert u_{\epsilon}\Vert^{-\frac{4}{N-4})}$$\Vert u_{\epsilon}\Vert^{p_{e}+1-(\frac{-1}{4})N}Rfi=\Vert u_{\epsilon}\Vert^{(\frac{N-4}{4})\epsilon}=O(1)$by (2.6), (2.5), (2.12) and the Lebesgue convergence theorem, we have$\int_{\Omega}o(|x-x_{0}|^{2})u_{\epsilon}^{p_{\epsilon}+1}dx=O(|x_{\epsilon}-x_{0}|^{2})+O(\Vert u_{\epsilon}\Vert^{-\frac{4}{N-4}})$.

(2.18)

37

The same argument leads to$\int_{\Omega}o(|x-x_{0}|^{3})u_{\epsilon}^{p_{\epsilon}+1}dx=O(|x_{\epsilon}-x_{0}|^{3})+O(\Vert u_{\epsilon}\Vert^{-\frac{6}{N-4}})$.

(2.19)

Now,$\sum_{j=1}^{N}b_{ij}\int_{\Omega}(x_{j}-x_{j}^{0})u_{\epsilon}^{p_{\epsilon}+1}dx$$= \sum_{j=1}^{N}b_{ij}\int_{\Omega_{\epsilon}}(\frac{y_{j}}{\Vert u_{\epsilon}\Vert^{R\epsilon_{4}^{\underline{-1}}}}+(x_{\epsilon})_{j}-x_{j}^{0})(\Vert u_{\epsilon}\Vert\tilde{u}_{\epsilon})^{p_{\epsilon}+1}(y)\Vert u_{\epsilon}\Vert^{-(\frac{-1}{4})N}g_{\xi}dy$$= \sum_{j=1}^{N}L^{-\underline{1}}4$$+ \sum_{j=1}^{N}b_{ij}\Vert u_{\epsilon}\Vert^{p_{g}+1-(\frac{-1}{4})N}\epsilon\epsilon((x_{\epsilon})_{j}-x_{j}^{0})\int_{\Omega_{\epsilon}}\tilde{u}_{\Xi}^{p_{\epsilon}+1}dy$$=:J_{1}+J_{2}$.

(2.20)By (2.5), (2.12) and the Lebesgue convergence theorem, we see$\int_{\Omega_{\epsilon}}y_{j}\overline{u}_{\epsilon}^{p_{\epsilon}+1}dy=\int_{\mathbb{R}^{N}}y_{j}U^{p+1}(y)dy+o(1)=o(1)$for any $j=1,$$\cdots,$$N$. Therefore,$J_{1}$in (2.20) is$J_{1}=C\Vert u_{\epsilon}\Vert^{-(\frac{2}{N-4})+(\frac{N-3}{4})\epsilon}\cross 0(1)$ $=o(\Vert u_{\epsilon}\Vert^{-\frac{2}{N-4}})$by (2.6). Similarly, we have
$J_{2}= \Vert u_{\epsilon}\Vert^{\epsilon(\frac{N-4}{4})}\sum_{j=1}^{N}b_{ij}((x_{\epsilon})_{j}-x_{j}^{0})\int_{\Omega_{\epsilon}}\tilde{u}_{\epsilon}^{p_{\epsilon}+1}dy=O(1)\cross\sum_{j=1}^{N}b_{ij}((x_{\epsilon})_{j}-x_{j}^{0})$.

Returning to (2.17) with these, we get that$\sum_{j=1}^{N}b_{ij}((x_{\epsilon})_{j}-x_{j}^{0})+O(|x_{\epsilon}-x_{0}|^{2})=O(\Vert u_{\epsilon}\Vert^{-2})+o(\Vert u_{\epsilon}\Vert^{-\frac{2}{N-4}})$.

(2.21)

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is a nondegenerate critical point of $K$ , the matrix
is invertible. Hence $bom(2.21)$ , we have (2.14).
Next we prove Theorem 1 (2):By our assumption that$x_{0}$$(b_{ij})_{1\leq i,j\leq N}=( \frac{\partial^{2}K}{\partial x_{i}\partial x_{j}}(x_{0}))$$\Vert u_{\epsilon}\Vert^{\epsilon}arrow 1$

as,(2.22)$\epsilonarrow 0$

by using (2.14).
In fact, inserting (2.15) and (2.16) into (2.2), we have$\frac{c_{0}(N-4)^{2}}{2(2N-\epsilon(N-4))}\epsilon\int_{\Omega}u_{\epsilon}^{p_{\epsilon}+1}dx$$+ \{\frac{c_{0}(N-4)}{2N-\epsilon(N-4)}+\frac{c_{0}(N-4)^{2}}{2(2N-\epsilon(N-4))}\frac{\epsilon}{2}\}\int_{\Omega}\sum_{i,j=1}^{N}b_{ij}(x_{i}-x_{i}^{0})(x_{j}-x_{j}^{0})u_{\epsilon}^{p_{\epsilon}+1}dx$(2.23)$+ \int_{\Omega}o(|x-x_{0}|^{3})u_{\epsilon}^{p_{e}+1}dx=\int_{\partial\Omega}|\nabla u_{\epsilon}||\nabla v_{\epsilon}|((x-x_{0})\cdot\nu)ds_{x}$when $N\geq 6$ . Hence by (2.23), (2.18), (2.19) and Lemma 3, we have$O(1)\cross\epsilon+O(|x_{\epsilon}-x_{0}|^{2})+O(\Vert u_{\epsilon}\Vert^{-\frac{4}{N-4})}=O(\Vert u_{\epsilon}\Vert^{-2})$.

This in turn implies$\epsilon\leq C\Vert u_{\epsilon}\Vert^{-\frac{4}{N-4}}$,(2.24)

when $N\geq 6$

for some constant $C>0$ , here we have used (2.14). By the mean value
theorem, it holds$|\Vert u_{\epsilon}\Vert^{\epsilon}-1|=|\Vert u_{\epsilon}\Vert^{t\epsilon}\epsilon\log\Vert u_{\epsilon}\Vert|$

for some $t\in(0,1)$ . Therefore by (2.6) and (2.24), it holds$|\Vert u_{\epsilon}\Vert^{\epsilon}-1|=O(\Vert u_{\epsilon}\Vert^{-\frac{4}{N-4}}\log\Vert u_{\epsilon}\Vert))$$N\geq 6$.

Thus we obtain (2.22).
Once (2.22) is established, we can check that the following lemma along
the line of [1]: Proposition 5.1, or [5]: Proposition 1.
Lemma 4 We have$\triangle^{2}(\Vert u_{\epsilon}\Vert u_{\epsilon})arrow c_{0}\frac{2\sigma_{N}}{N(N+2)}\delta_{x_{0}}$39in the sense of Radon measures of St, and$\Vert u_{\epsilon}\Vert u_{\epsilon}arrow c_{0}\frac{2\sigma_{N}}{N(N+2)}G(\cdot, x_{0})$$in$ $C^{3,\alpha}(\omega)$$in$ $C^{1,\alpha}(\omega)$$\Vert u_{\epsilon}\Vert v_{\epsilon}arrow c_{0}\frac{2\sigma_{N}}{N(N+2)}(-\Delta G)(\cdot, x_{0})$for some.

$\alpha\in(0,1)$, where$\omega$,

is any open neighborhood of$\partial\Omega$, not containing$x_{0}$

Finally, we will prove Theorem 1 (4) when $N\geq 6$ .
By (2.5), (2.12), (2.14), (2.22) and the Lebesgue convergence theorem,
we have the followings:$\int_{\Omega}u_{\epsilon}^{p_{\zeta}+1}dxarrow\int_{\mathbb{R}^{N}}U^{p+1}dy=\frac{\sigma_{N}}{2}\frac{\Gamma(\frac{N}{2})^{2}}{\Gamma(N)}$,(2.25)$\sum_{i,j=1}^{N}b_{ij}\int_{\Omega}(x_{i}-x_{i}^{0})(x_{j}-x_{j}^{0})u_{\epsilon}^{p_{\epsilon}+1}dx$$= \sum_{i,j=1}^{N}b_{ij}\Vert u_{\epsilon}\Vert^{\epsilon(\frac{N-4}{4})_{\cross}}$$\int_{\Omega_{e}}(\frac{y_{i}y_{j}}{\Vert u_{\epsilon}\Vert^{z\epsilon_{2}^{\underline{-1}}}}+\frac{y_{i}((x_{\epsilon})_{j}-x_{j}^{0})+(x_{\epsilon})_{i}-x_{i}^{0})}{\Vert u_{\epsilon}\Vert^{R\epsilon_{\frac{y_{j-1}(}{4}}}}+((x_{\epsilon})_{i}-x_{i}^{0})((x_{\epsilon})_{j}-x_{j}^{0}))\tilde{u}_{\epsilon}^{p_{\epsilon}+1}dy$(2.26)$= \sum_{i,j=1}^{N}b_{ij}\Vert u_{\epsilon}\Vert^{-\frac{4}{N-4}+\epsilon(\frac{N-2}{4})}\int_{\Omega_{\epsilon}}y_{i}y_{j}\overline{u}_{\epsilon}^{p_{\epsilon}+1}dy+o(\Vert u_{\epsilon}\Vert^{-\frac{4}{N-4}})$

if $N\geq 6$ , and$\int_{\Omega_{\epsilon}}y_{i}y_{j}\tilde{u}_{\epsilon}^{p_{E}+1}dyarrow\int_{\mathbb{R}^{N}}y_{i}y_{j}U^{p+1}dy=\frac{1}{N}\int_{\mathbb{R}^{N}}|y|^{2}U^{p+1}dy\delta_{ij}$,$= \frac{\sigma_{N}}{N}\int_{0}^{\infty}\frac{r^{N+1}}{(1+r^{2})^{N}}dr\delta_{ij}=\frac{\sigma_{N}}{2(N-2)}\frac{\Gamma(\frac{N}{2})^{2}}{\Gamma(N)}\delta_{ij}$where(2.27)

is Kronecker’s delta.
Furthermore, Lemma 4leads to$\delta_{ij}$,$\Vert u_{\epsilon}\Vert^{\frac{4}{N-4}}\int_{\partial\Omega}|\nabla u_{\epsilon}||\nabla v_{\epsilon}|((x-x_{0})\cdot\nu)ds_{x}arrow 0$

if $N\geq 7$ ,(2.28)

40

and$\Vert u_{\epsilon}\Vert^{\frac{4}{N-4}}\int_{\partial\Omega}|\nabla u_{\epsilon}||\nabla v_{\epsilon}|((x-x_{0})\cdot\nu)ds_{x}$$arrow(\frac{2c_{0}\sigma_{N}}{N(N+2)}I^{2}\int_{\partial\Omega}|\nabla G||\nabla\Delta G|((x-x_{0})\cdot\nu)ds_{x}$$=( \frac{2c_{0}\sigma_{N}}{N(N+2)}I^{2}(N-4)R(x_{0})=2^{9}\sigma_{6}^{2}R(x_{0})$if $N=6$ ,,(2.29)

where we have used a formula in [1] Lemma 3.1:$\int_{\partial\Omega}|\nabla G||\nabla\triangle G|((x-x_{0})\cdot\nu)ds_{x}=(N-4)R(x_{0})$for any.
Thus, multiplying (2.23) by 1
(2.27), and (2.28) or (2.29), we obtain$x_{0}\in\Omega$$u_{\epsilon}\Vert^{\frac{4}{N-4}}$

when $N\geq 6$ , using (2.25), (2.26),$\lim_{\epsilonarrow 0}\epsilon\Vert u_{\epsilon}\Vert^{\frac{4}{N-4}}=\frac{-\frac{(N-4)c_{0}}{2N}\sum^{N}b_{ii}\Gamma()}{\frac{co(N^{2}\frac{\sigma_{N}}{2(N-2)4N-4)}}{}\underline{\sigma}_{2(N)}\alpha_{\frac{\Gamma(}{\Gamma}}^{\underline{N}}=)^{2}}$$=- \frac{2}{(N-2)(N-4)}\triangle K(x_{0})$,if $N\geq 7$ ,$\lim_{\epsilonarrow 0}\epsilon\Vert u_{\epsilon}\Vert^{\frac{4}{N-4}}=\frac{2^{9}\pi^{6}R(x_{0})-\frac{4}{\pi^{3}15}\pi^{3}\Delta K(x_{0})}{\frac{16}{15}}$$=480 \pi^{3}R(x_{0})-\frac{1}{4}\triangle K(x_{0})$, in particular,
RecallThis proves Theorem 1 when $N\geq 6$ .
$\sigma_{N}=\frac{2\pi^{N/2}}{\Gamma(N/2)}$$\sigma_{6}=\pi^{3}$,

if $N=6$ .

.
口

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