Uniqueness results through a priori estimates part I: a three dimensional Neumann problem

Meijun Zhu Department of Mathematics The University of British Columbia Vancouver, B.C. Canada V6T 1Z2 mzhu@math.ubc.ca

Abstract

A uniqueness result about the Neumann problem: −Δu + λu = u^{5} in Ω, ∂u/∂ν = 0 on ∂Ω is obtained, where Ω ⊂ ℝ^{3} is a bounded smooth convex domain and λ > 0 is close to 0.

0 Introduction

This is the first part of our study on the uniqueness properties of some Nonlinear elliptic problems when some parameters in front of linear terms are close to the first eigenvalue of the corresponding linear problems. In this part, we are interested in the following Neumann boundary problem:

\{ \begin{matrix}−Δu + λu = u^{p}, u > 0, & in Ω \\ \frac{∂u}{∂ν} = 0 & on ∂Ω \end{matrix}  (0.1)

where p > 1, λ > 0 and Ω ⊂ ℝ^{n} is a smooth bounded domain (that is: a bounded domain with smooth boundary). Let t = λ^{−1/(p−1)}, v = tu and d^{2} = 1/λ. Equation (0.1) takes the following form

\{ \begin{matrix}−d^{2}Δv + v = v^{p}, v > 0, & in Ω \\ \frac{∂v}{∂ν} = 0 & on ∂Ω. \end{matrix}  (0.2)

Equation (0.2) arises naturally in various studies of pattern formations in mathematical biology, and were systematically studied by Lin, Ni, and Takagi, see [17] and their most recent work [19] and the references therein.

It is more convenient for us to state results on (0.1), even though to state a result on (0.2) may have more direct meaning in mathematical biology. When λ is large, various results concerning the existence of one, or multiple peaks solutions and asymptotic behaviors of those solutions have appeared recently. Here, we consider (0.1) when λ is positive and close to 0. It is well-known that (0.1) has no solution if λ ≤ 0.

When p is a subcritical exponent, that is, p < \frac{n+2}{n−2}, Lin, Ni and Takagi [17] showed that (0.1) has a unique solution if λ is sufficiently small. Such kind of uniqueness results about radially symmetric solution of (0.1) were also obtained by Lin and Ni in [16] when Ω is an annulus and p > 1 or when Ω is a ball and p > \frac{n+2}{n−2}. It was then conjectured by Lin and Ni in [16] that for any p > 1, there exists a λ_{0} > 0 such that (0.1) has only a constant solution if λ < λ_{0}. Notice: 0 is the first eigenvalue to the following Neumann problem

\{ \begin{matrix}−Δu + λu = 0 & in Ω \\ \frac{∂u}{∂ν} = 0 & on ∂Ω. \end{matrix}

Intuitively, if a solution of (0.1) is obtained from minimizing the quotient energy, the L^{∞} norm of this solution will tend to zero as λ tends to zero. The uniqueness will follow from this fact easily (see Section 1 for more details). This kind of idea was implicitly addressed by Adimurthi and Yadava in [1].

Unfortunately, the above conjecture of Lin-Ni is not true in general, especially when p is the critical Sobolev exponent, that is, p = \frac{n+2}{n−2}. In this case, when Ω is a unit ball and n = 4, 5, 6, it was shown by Adimurthi and Yadava [2] that (0.1) has at least two radial solutions if λ > 0 and is close to 0 (see also [5]). A very interesting question concerning n = 3 arises:

Question : \{ \begin{matrix}Is Lin-Ni’s conjecture still true in the \\ case of critical power when n = 3? \end{matrix}

In the same paper [2], through shooting method of ODE, Adimurthi and Yadava was able to show that the radial solution of (0.1) is unique when Ω is a unit ball, n = 3, p = 5 and λ is close to 0. More recently, similar results were obtained for n ≥ 7 by Adimurthi and Yadava in [3]. However due to the Neumann boundary condition, even though Ω is a unit ball, no theory can guarantee all solutions of (0.1) are radially symmetric. Also, obviously those ODE analyses can not be applied to other domains.

The main purpose of this paper is, by using PDE’s analyses, to show that the answer to the above question is Yes provided the domain is convex. Therefore, from now on, we restrict ourselves on n = 3 and consider the following equation:

\{ \begin{matrix}−Δu + λu = u^{5}, u > 0, & in Ω \\ \frac{∂u}{∂ν} = 0 & on ∂Ω \end{matrix}  (0.3)

where Ω ∈ ℝ^{3} is a bounded domain with smooth boundary and λ > 0.

Our main result can be stated as the following theorem.

Theorem 0.1 Assume that Ω is a smooth bounded convex domain in ℝ^{3}, then there exists a constant λ_{0} > 0 such that, if 0 < λ < λ_{0}, the solution to (0.3) is unique. More specifically, the solution must be a constant solution.

The main step in the proof of Theorem 0.1 is to show that L^{∞} norms of solutions of (0.3) are uniformly bounded as λ → 0. In view of those counterexamples given in [2] and [5], we would like to point out that such L^{∞} estimates do not hold when dimension n = 4, 5, 6. Since we do not assume that solutions of (0.3) are obtained from minimizing the corresponding energy functional, we can not use those energy dependent estimates (such as the application of sharp Sobolev constants, Moser iteration method and others). Instead, in order to get such a priori L^{∞} estimates, we use so-called energy independent a priori estimates, which were initially used by Schoen to give an alternative proof of Yamabe problem in [20] and later on were widely extended by other analysts, see [21], [12]-[14], [10], [15] and references therein.

The basic idea to obtain the L^{∞} a priori estimates in this paper is the following: Let u_{λ} be a solution of (0.3). Suppose ||u_{λ}||_{L^{∞}} → ∞ as λ → 0, we can show ( this is the main part of this paper) that there exist some discrete points \{q_{j}^{∗}\}_{j=1}^{k} ⊂ \overline{Ω}, such that, for any compact set K ⊂ \overline{Ω} \ \{q_{1}^{∗}, ..., q_{k}^{∗}\}, after passing to some subsequence of {u_{λ}}, ||u_{λ}||_{L^{∞}} · u_{λ}(x) → G(x, {q_{j}^{∗}}_{j=1}^{k}) in C^{2}(K), where G(x, {q_{j}^{∗}}_{j=1}^{k}) is some Green’s function in Ω with Neumann boundary condition and some discrete poles at {q_{j}^{∗}}_{j=1}^{k}. However, from the geometric property of a convex domain, we can show that no such Green’s function exists. We thus derive a contradiction. The convexity assumption of Ω will also be used when we discuss the local asymptotic behaviors of u_{λ}, see the proof of Proposition 2.2 below. We do not know whether one can remove the assumption of the convexity of the domain in our theorem.

In the second part of this study (in a forthcoming paper [22]), we will consider the following Dirichlet problem:

\{ \begin{matrix}−Δu − λu = u^{\frac{n+2}{n−2}} , u > 0, & in Ω \\ u = 0 & on ∂Ω \end{matrix}

where Ω is a bounded domain in ℝ^{n} (n ≥ 3). We will establish the uniqueness result when λ (< λ_{1}) is close to λ_{1}, where λ_{1} denotes the first eigenvalue of −Δ with Dirichlet boundary condition.

We organize this paper as follows. In Section 1, under the assumptions on the a priori L^{∞} bound, we give the proof of Theorem 0.1. In Section 2, we focus on obtaining those a priori estimates. Some frequently used Pohozaev type identities are put in the appendix with proofs. Later on, throughout this paper, we always assume Ω ⊂ ℝ^{3} and 0 < λ < 1 and use C, C_{0}, C_{1}, ⋯, to represent various positive constants.

Acknowledgments: I would like to thank Y.Y.Li for teaching me those powerful energy independent a priori estimates and for pointing out an error in the proof of Lemma 2.2 in an early version. The observation of the difference between n = 3 and n ≥ 4 stems from the joint work [15]. This work was completed when I was a postdoctoral fellow at the University of British Columbia. I would like to thank R. Froese, N. Ghoussoub, C. Gui and Pacific Institution for mathematical sciences (PIms) for their generous supports.

1 Small L^{∞} norm implies uniqueness

In this section we give the proof of Theorem 0.1. First of all, we assume that the following a priori estimates hold for solutions to equation (0.3).

Theorem 1.1 Assume that Ω ⊂ ℝ^{3} is a smooth bounded convex domain and u_{λ} solves (0.3), then there exist constants C > 0 and λ_{∗} > 0, such that

||u_{λ}||_{L^{∞}(Ω)} ≤ C for all 0 < λ < λ_{∗}.

The proof of Theorem 1.1 is the main part of this paper, we relegate it to next section.

Observe that the only solution to

\{ \begin{matrix}−Δu = u^{5}, u ≥ 0, & in Ω \\ \frac{∂u}{∂ν} = 0 & on ∂Ω \end{matrix}  (1.1)

is the trivial one. A quick consequence of Theorem 1.1 is the following corollary.

Corollary 1.1 Assume that Ω ⊂ ℝ^{3} is a smooth bounded convex domain and u_{λ} solves (0.3), then ||u_{λ}||_{L^{∞}} → 0 as λ → 0.

Proof. We prove this corollary by contradiction. If not, there exists a constant C_{1} > 0, such that, after passing to some subsequence of λ (we still denote it as λ), ||u_{λ}||_{L^{∞}} ≥ C_{1}. From Theorem 1.1 we know that ||u_{λ}||_{L^{∞}} ≤ C. Using standard elliptic estimates, we know u_{λ} → u_{0} in C^{2}(\overline{Ω}) where u_{0} satisfies (1.1). Since ||u_{λ}||_{L^{∞}} ≥ C_{1}, we know ||u_{0}||_{L^{∞}} ≥ C_{1}. This contradicts to the above observation!

Now we are ready to give the proof of Theorem 0.1 based on the above corollary. The method we use here is different to that in [18] and standard bifurcation theory ( see e.g. [7]). It seems to me that this method is more straightforward.

Proof of Theorem 0.1. We prove it through a contradiction argument. Suppose not, there are sequences λ_{i} → 0, {u_{λ_{i}}} and {v_{λ_{i}}} such that u_{λ_{i}} > 0 and v_{λ_{i}} > 0 solve (0.3) for λ = λ_{i} and ||u_{λ_{i}} − v_{λ_{i}}||_{L^{∞}} ≠ 0.

Without loss of generality, we assume ||u_{λ_{i}} − v_{λ_{i}}||_{L^{∞}} = max\overline{Ω}(u_{λ_{i}} − v_{λ_{i}}). Denote A_{λ_{i}} = max\overline{Ω}(u_{λ_{i}} − v_{λ_{i}}) and z_{λ_{i}} = (u_{λ_{i}} − v_{λ_{i}})/A_{λ_{i}}. Then z_{λ_{i}} satisfies

\{ \begin{matrix}−Δz_{λ_{i}} + λ_{i}z_{λ_{i}} = c_{λ_{i}}z_{λ_{i}} & in Ω \\ \frac{∂z_{λ_{i}}}{∂ν} = 0 & on ∂Ω, \end{matrix}

where c_{λ_{i}}(x) = 5ξ_{λ_{i}}^{4} and ξ_{λ_{i}} is a positive function between u_{λ_{i}} and v_{λ_{i}}. Thus, c_{λ_{i}}(x) → 0 as i → ∞ due to Corollary 1.1. Noticing ||z_{λ_{i}}||_{L^{∞}} = 1, by standard elliptic estimates, we know that z_{λ_{i}} → z_{0} in C^{2}(\overline{Ω}), and z_{0} satisfies

\{ \begin{matrix}−Δz_{0} = 0 & in Ω \\ \frac{∂z_{0}}{∂ν} = 0 & on ∂Ω. \end{matrix}

It follows that z_{0} = C. Since max\overline{Ω} z_{λ_{i}} = 1, we know z_{0} = 1.

On the other hand, since u_{λ_{i}} and v_{λ_{i}} satisfy (0.3), we have

∫_{Ω} −Δu_{λ_{i}}v_{λ_{i}} + λ_{i} ∫_{Ω} u_{λ_{i}}v_{λ_{i}} = ∫_{Ω} v_{λ_{i}}u_{λ_{i}}^{5}

and

∫_{Ω} −Δv_{λ_{i}}u_{λ_{i}} + λ_{i} ∫_{Ω} u_{λ_{i}}v_{λ_{i}} = ∫_{Ω} u_{λ_{i}}v_{λ_{i}}^{5}.

Combining with the boundary condition, we have

∫_{Ω} u_{λ_{i}}v_{λ_{i}}(u_{λ_{i}}^{4} − v_{λ_{i}}^{4}) = 0.

Therefore, for all λ_{i}, there exists x_{λ_{i}} ∈ \overline{Ω} such that z_{λ_{i}}(x_{λ_{i}}) = 0. It follows that z_{0} = 0. Contradiction!

2 A priori estimates

We devote to prove Theorem 1.1 in this section by using energy independent a priori estimates. Some of the definitions and properties in this section have already appeared in [20], [21], [12]-[15]. We follow the approach in [12] closely.

2.1 Definition and basic properties

We introduce some terminologies which will be used to classify those “bad” points (blowup points) of solution to (0.3). We denote B_{r}(x) as the ball of radius r centered at x.

Definition 2.1 Assume that u_{λ} solves (0.3). A point ȳ ∈ \overline{Ω} is called an isolated blowup point of \{u_{λ}\} if there exist r̄ > 0, C̃ > 0, and a sequence y_{λ} ∈ \overline{Ω} tending to ȳ, such that, y_{λ} is a local maximum of u_{λ}, u_{λ}(y_{λ}) → +∞ and

u_{λ}(y) ≤ C̃|y − y_{λ}|^{− \frac{1}{2}} for all y ∈ Br̄(y_{λ}) ∩ \overline{Ω}.  (2.1)

Let y_{λ} → ȳ be an isolated blowup point of {u_{λ}} and set

ū_{λ}(r) = \frac{1}{|∂B_{r} ∩ Ω|} ∫_{∂B_{r}(y_{λ})∩Ω} u_{λ}, 0 < r < r̄.  (2.2)

Definition 2.2 ȳ ∈ \overline{Ω} is called an isolated simple blowup point, if ȳ is an isolated blowup point of \{u_{λ}\}, such that, there exists a 0 < ρ < \frac{r̄}{3} (independent of λ), r^{\frac{1}{2}} ū_{λ}(r) has precisely one critical point in (0, ρ) for small λ.

We present some basic properties on isolated, or isolated simple blowup points.

Lemma 2.1 Assume that u_{λ} satisfies (0.3) and y_{λ} → ȳ ∈ \overline{Ω} is an isolated blowup point, then for any 0 < r < \frac{1}{3} r̄, we have the following Harnack inequality:

max_{y∈B_{2r}\B_{\frac{r}{2}} ∩\overline{Ω}} u_{λ}(y) ≤ C_{1} min_{y∈B_{2r}\B_{\frac{r}{2}} ∩\overline{Ω}} u_{λ}(y),  (2.3)

where C_{1} is a positive constant independent of r and λ.

Proof. Set v_{λ}(y) = r^{\frac{1}{2}} u_{λ}(ry + y_{λ}) for y ∈ B_{3}(0). Then v_{λ} satisfies

\{ \begin{matrix}−Δv_{λ} + r^{2}λv_{λ} = v_{λ}^{5} & in B_{3} \\ \frac{∂v_{λ}}{∂ν} = 0 & on B_{3} ∩ ∂Ω_{r} \end{matrix}  (2.4)

where Ω_{r} = (Ω − y_{λ})/r.

Due to (2.1), we know that

0 < v_{λ} ≤ C̃|y|^{− \frac{1}{2}} ∀y ∈ B_{3}(0).

One might straighten part of the boundary B_{3} ∩ ∂Ω_{r} if B_{3} ∩ ∂Ω_{r} is not empty ( see e.g. [17] or the proof of Lemma 2.3 below for more details). Then applying the standard Harnack inequality to (2.4) in B_{9/4} \ B_{1/4} ∩ \overline{Ω}_{r} ( see e.g. [9]), we have

max_{B_{9/4}\B_{1/4}∩\overline{Ω}_{r}} v_{λ} ≤ C_{1} min_{B_{9/4}\B_{1/4}∩\overline{Ω}_{r}} v_{λ},

which yields (2.3).

Lemma 2.2 Suppose that u_{λ} solves (0.3) and y_{λ} → ȳ ∈ \overline{Ω} is an isolated blowup point. Then for any R >> 1, after passing to some subsequence, we have that

||u_{λ}(y_{λ})^{−1}u_{λ}(u_{λ}^{−2}(y_{λ})x + y_{λ}) − (1 + \frac{1}{3}|x|^{2})^{− \frac{1}{2}} ||_{C^{2}(B_{2R}(0))} → 0  (2.5)

or

||u_{λ}(y_{λ})^{−1}u_{λ}(u_{λ}^{−2}(y_{λ})x + y_{λ}) − (1 + \frac{1}{3}|x|^{2})^{− \frac{1}{2}} ||_{C^{2}(B_{2R}^{+}(0))} → 0  (2.6)

as λ → 0.

Proof. Let Ω_{λ} = (Ω − y_{λ}) · u_{λ}^{2}(y_{λ}) and

ξ_{λ}(x) = u_{λ}^{−1}u_{λ}(u_{λ}^{−2}(y_{λ})x + y_{λ}), ∀|x| < r̄u_{λ}^{2}(y_{λ}) and x ∈ Ω_{λ}.

Then ξ_{λ}(x) satisfies the following equation:

\{ \begin{matrix}−Δξ_{λ} + λu_{λ}^{−4}(y_{λ})ξ_{λ} = ξ_{λ}^{5}, & in \{x ∈ Ω_{λ} : |x| ≤ r̄u_{λ}^{2}(y_{λ})\} \\ \frac{∂ξ_{λ}}{∂ν} = 0, & on ∂Ω_{λ} ∩ B_{r̄u_{λ}^{2}(y_{λ})} \\ 0 < ξ_{λ}(x) ≤ C̃|x|^{− \frac{1}{2}} & in \{x ∈ Ω_{λ} : |x| ≤ r̄u_{λ}^{2}(y_{λ})\}. \end{matrix}  (2.7)

From Lemma 2.1, we know that for any 0 < r ≤ 1,

max_{|x|=r,x∈\overline{Ω}_{λ}} ξ_{λ} ≤ C_{1} min_{|x|=r,x∈\overline{Ω}_{λ}} ξ_{λ}.

8 Three dimensional Neumann problem Noticing u_{λ}^{−4}(y_{λ}) → 0 as λ → 0, we have

−Δξ_{λ} + o(1)ξ_{λ} ≥ 0 ∀|x| ≤ 1.

We set η_{λ} = (1 + |x|^{2})^{−1}ξ_{λ}. Then as λ is sufficiently small η_{λ} satisfies

Δη_{λ} + ∑_{i=1}^{3} \frac{4x_{i}}{1 + |x|^{2}} · ∂_{i}η_{λ} ≤ 0 ∀|x| ≤ 1 and x ∈ Ω_{λ}.

It then follows from the maximum principle that

ξ_{λ}(0) = η_{λ}(0) ≥ min_{|x|=r,x∈\overline{Ω}_{λ}} η_{λ} ≥ \frac{1}{2} min_{|x|=r,x∈\overline{Ω}_{λ}} ξ_{λ},

which yields that

max_{|x|=r,x∈\overline{Ω}_{λ}} ξ_{λ} ≤ Cξ_{λ}(0) = C.

Combining with the third line of (2.7), we have

ξ_{λ}(x) ≤ C in {x ∈ Ω_{λ} : |x| ≤ r̄u_{λ}^{2}(y_{λ})}.

Therefore, by some standard elliptic estimates we know that after passing to some subsequence ξ_{λ} → ξ in C_{loc}^{2}(ℝ_{T}^{3} ), where

T = \overline{lim_{λ→0}}dist(y_{λ}, ∂Ω) · u_{λ}^{2}(y_{λ}), ℝ_{T}^{3} = \{x = (x_{1}, x_{2}, x_{3}) : x_{3} > −T \}

and ξ satisfies:

\{ \begin{matrix}−Δξ = ξ^{5} in ℝ^{3} \\ ξ(0) = 1, 0 ≤ ξ ≤ 1, \end{matrix}

or

\{ \begin{matrix}−Δξ = ξ^{5} & in ℝ_{T}^{3} \\ \frac{∂ξ}{∂ν} = 0 & on ∂ℝ_{T}^{3} \\ ξ(0) = 1, 0 ≤ ξ ≤ 1. \end{matrix}

It follows from [6] that T = 0 or T = ∞ and thus

ξ(x) = (1 + \frac{1}{3}|x|^{2})^{− \frac{1}{2}} .

Lemma 2.2 is established.

For an isolated simple blowup point, we have the following lemma.

Lemma 2.3 Assume that {u_{λ}} and {y_{λ}} are those subsequences given in Lemma 2.2, y_{λ} → ȳ is an isolated simple blowup point. Then there exist 0 < δ << 1 and 0 < ρ_{0} = ρ_{0}(δ) < ρ, such that, as λ sufficiently small, for Ru_{λ}^{−2}(y_{λ}) ≤ |y − y_{λ}| ≤ ρ_{0} and y ∈ \overline{Ω},

u_{λ}(y) ≤ C_{2}u_{λ}(y_{λ})^{2δ−1}|y − y_{λ}|^{−1+δ},  (2.8)

where C_{2} is some positive constant independent of λ.

Proof. Denote r_{λ} = Ru_{λ}^{−2}(y_{λ}). From Lemma 2.2, we know

u_{λ}(y) ≤ Cu_{λ}(y_{λ})R^{−1} on ∂B_{r_{λ}}(y_{λ}) ∩ \overline{Ω}.  (2.9)

Also from Lemma 2.2, we know that r^{\frac{1}{2}} ū_{λ}(r) takes some maximal value in B_{r_{λ}}(y_{λ}). Therefore, in view of the definition of isolated simple blowup, we know that

r^{\frac{1}{2}} ū_{λ}(r) is strictly decreasing for r_{λ} < r < ρ.  (2.10)

For all r_{λ} ≤ |y − y_{λ}| < ρ,

|y − y_{λ}|^{\frac{1}{2}} u_{λ}(y) ≤ C|y − y_{λ}|^{\frac{1}{2}} ū_{λ}(|y − y_{λ}|) (by Harnack)
≤ Crλ^{\frac{1}{2}} ū_{λ}(r_{λ})  ( by ( 2.10))
≤ Crλ^{\frac{1}{2}} u_{λ}(y_{λ})R^{−1}  ( by (2.9))
= CR^{− \frac{1}{2}} .

(2.11)

It follows that

u_{λ}(y)^{2} ≤ CR^{−1}|y − y_{λ}|^{−1} for r_{λ} ≤ |y − y_{λ}| < ρ and y ∈ \overline{Ω}.  (2.12)

We then discuss the proof in the following two cases.

Case 1: ȳ ∉ ∂Ω, that is: dist(ȳ, ∂Ω) ≥ 2ρ_{1} for some positive constant ρ_{1} > 0. In this case, the proof can be carried out in the same way as in [12]. We include

it here for completeness. We consider the operator:

L_{λ}φ = Δφ + u_{λ}(y)^{4}φ − λφ.

Since L_{λ}u_{λ} = 0 and u_{λ} > 0, the maximum principle holds for L_{λ} in B_{ρ_{1}}(y_{λ}) (see e.g. [4]).

For 0 < μ < 1, a direct computation shows

Δ(|y − y_{λ}|^{−μ}) = −μ(1 − μ)|y − y_{λ}|^{−2−μ}, ∀|y − y_{λ}| > 0.

10 Three dimensional Neumann problem

Combining with (2.12), we have

L_{λ}(|y − y_{λ}|^{−μ}) ≤ {−μ(1 − μ) + O(R^{−2})}|y − y_{λ}|^{−2−μ}, ∀y ∈ B_{ρ_{1}}(y_{λ}) \ {y_{λ}}. For any fixed δ < 1/100, as λ small enough, we have ∀y ∈ B_{ρ_{1}}(y_{λ}) \ {y_{λ}},

\{ \begin{matrix}L_{λ}(|y − y_{λ}|^{−δ}) ≤ 0 \\ L_{λ}(|y − y_{λ}|^{δ−1}) ≤ 0. \end{matrix}  (2.13)

Set M_{λ} = max_{∂B_{ρ_{1}}(y_{λ})} u_{λ}, and

ϕ_{λ}(y) = M_{λ}ρ_{1}^{δ}|y − y_{λ}|^{−δ} + Au_{λ}(y_{λ})^{2δ−1}|y − y_{λ}|^{δ−1}, r_{λ} ≤ |y − y_{λ}| ≤ ρ_{1}, where A will be given later.

Notice, on ∂B_{r_{λ}}(y_{λ}), ϕ_{λ}(y) ≥ Au_{λ}(y_{λ})R^{−1+δ} ≥ Au_{λ}(y_{λ})R^{−1}. Using (2.9), we can choose A sufficient large (independent of λ) such that

ϕ_{λ}(y) ≥ u_{λ}(y) for all |y − y_{λ}| = r_{λ}.

Also, it is easy to check

ϕ_{λ}(y) ≥ u_{λ}(y) for all |y − y_{λ}| = ρ_{1}.

Using (2.13), we know ϕ_{λ} is a supersolution for L_{λ} in the annulus r_{λ} ≤ |y −y_{λ}| ≤ ρ_{1}. From the maximum principle, we conclude

u_{λ}(y) ≤ ϕ_{λ}(y), for all r_{λ} ≤ |y − y_{λ}| ≤ ρ_{1}.  (2.14)

Combining with (2.10) and Lemma 2.1, we have for any r_{λ} < θ < ρ_{1},

ρ_{1}^{2}M_{λ} ≤ Cρ_{1}^{2}ū_{λ}(ρ_{1})
≤ Cθ^{2}ū_{λ}(θ)
≤ Cθ^{2}{M_{λ}ρ_{1}^{δ}θ^{−δ} + Au_{λ}(y_{λ})^{2δ−1}θ^{δ−1}}.

Choosing θ > 0 small enough such that

Cρ_{1}^{δ}θ^{2−δ} < \frac{1}{2}ρ_{1}^{2},

we know

M_{λ} ≤ Cu_{λ}(y_{λ})^{2δ−1}.

It then follows from (2.14), the definition of ϕ_{λ} and δ < 1/100 that

u(y) ≤ C_{2}u_{λ}(y_{λ})^{2δ−1}|y − y_{λ}|^{−1+δ} for Ru_{λ}^{−2}(y_{λ}) ≤ |y − y_{λ}| ≤ ρ_{1}.

Choosing ρ_{0} = ρ_{1}, we establish Lemma 2.3 in case 1.

Case 2: ȳ ∈ ∂Ω.

We first straighten the boundary ∂Ω near ȳ as in [17]. Without loss of generality, we assume that ȳ is the coordinate origin and y_{3}-axis is the inner normal to ∂Ω at ȳ. Therefore, there exists a smooth function φ(y′) ≥ 0, y′ = (y_{1}, y_{2}), defined for |y′| < γ_{1} for some small γ_{1} > 0, satisfying: φ(0) = 0 and ∂_{j}φ(0) = 0 for j = 1, 2, such that ∂Ω is given by (y_{1}, y_{2}, φ(y′)) in a neighborhood of the origin.

For x ∈ ℝ^{3} and |x| << 1, we define: y = Φ(x) := (Φ_{1}(x), Φ_{2}(x), Φ_{3}(x)) with

\{ \begin{matrix}Φ_{j}(x) = x_{j} − x_{3} · ∂_{j}φ(x′), j = 1, 2, \\ Φ_{3}(x) = x_{3} + φ(x′). \end{matrix}  (2.15)

Since, in virtue of (2.15), the differential map of Φ satisfies DΦ(0) = Identity, we know that Φ has the inverse in a neighborhood of {0}. If we denote x = Ψ(y) := Φ^{−1}(y) and f_{λ}(x) = u_{λ}(y) in a neighborhood of ȳ, f_{λ}(x) satisfies

\{ \begin{matrix}∑_{i,j=1}^{3} a_{ij}(x)\frac{∂^{2}f_{λ}}{∂x_{i}∂x_{j}} + ∑_{j=1}^{3} b_{j}(x)\frac{∂f_{λ}}{∂x_{j}} − λf_{λ} + f_{λ}^{5} = 0 & in B_{2ϵ}^{+}(0) \\ \frac{∂f_{λ}}{∂x_{3}} = 0 & on B_{2ϵ}(0) ∩ \{x_{3} = 0\}, \end{matrix}

(2.16)

where

\{ \begin{matrix}a_{ij}(x) = ∑_{l=1}^{3} \frac{∂Ψ_{i}}{∂y_{l}} (Φ(x)) · \frac{∂Ψ_{j}}{∂y_{l}} (Φ(x)) \\ b_{j}(x) = (ΔΨ_{j})(Φ(x)) \end{matrix}

and B_{2ϵ}^{+}(0) = B_{2ϵ}(0) ∩ {(x_{1}, x_{2}, x_{3}) ∈ ℝ^{3} | x_{3} > 0}. We refer Lin-Ni-Takagi [17] to readers for more details. Easily to see that

\{ \begin{matrix}a_{ij}(x) = δ_{ij} + o_{ϵ}(1) & |x| < ϵ \\ b_{j} = o_{ϵ}(1) & |x| < ϵ \end{matrix}  (2.17)

where o_{ϵ}(1) → 0 as ϵ → 0. Choosing ϵ small enough such that Φ(B_{2ϵ}^{+}) ⊂ {y : |y − y_{λ}| ≤ ρ}, and denoting Ξ_{λ}^{+} = B_{ϵ}^{+} \ Φ^{−1}(B_{r_{λ}}(y_{λ})), x_{λ} = Φ^{−1}(y_{λ}), we have, from (2.12), that

f_{λ}(x)^{2} ≤ CR^{−1}|x − x_{λ}|^{−1} x ∈ Ξ_{λ}^{+}.  (2.18)

Now, we consider the operator

F_{λ}ϕ = ∑_{i,j=1}^{3} a_{ij}(x)\frac{∂^{2}ϕ}{∂x_{i}∂x_{j}} + ∑_{j=1}^{3} b_{j}(x)\frac{∂ϕ}{∂x_{j}} − λϕ + f_{λ}^{4}ϕ.

Since F_{λ}f_{λ} = 0 and f_{λ} > 0, we know that the maximum principle holds for F_{λ}. Denote x̄_{λ} as the reflection of x_{λ} with respect to x_{3} = 0 hyperplane. Using (2.17) and (2.18), we can see for any 0 < μ < 1, x ∈ Ξ_{λ}^{+},

F_{λ}(|x − x_{λ}|^{−μ}) ≤ [−μ(1 − μ) + o_{ϵ}(1) + O(R^{−2})] · |x − x_{λ}|^{−μ−2}.

and

F_{λ}(|x − x̄_{λ}|^{−μ}) ≤ [−μ(1 − μ) + o_{ϵ}(1)] · |x − x̄_{λ}|^{−μ−2} + O(R^{−2})|x − x_{λ}|^{−2} · |x − x̄_{λ}|^{−μ}.

Notice: |x − x_{λ}| ≤ |x − x̄_{λ}| for x ∈ Ξ_{λ}^{+}. Fix δ < 1/100. Combining the above two inequalities together, we know for ϵ sufficiently small, as λ small, that

\{ \begin{matrix}F_{λ}(|x − x_{λ}|^{−δ} + |x − x̄_{λ}|^{−δ}) ≤ 0 \\ F_{λ}(|x − x_{λ}|^{δ−1} + |x − x̄_{λ}|^{δ−1}) ≤ 0. \end{matrix}  (2.19)

Set N_{λ} = max_{∂B_{ϵ}} f_{λ}, and

φ_{λ}(x) = N_{λ}ϵ^{δ}(|x−x_{λ}|^{−δ}+|x−x̄_{λ}|^{−δ})+Bu_{λ}(y_{λ})^{2δ−1}(|x−x_{λ}|^{δ−1}+|x−x̄_{λ}|^{δ−1}), x ∈ Ξ_{λ}^{+}

where B will be given later.

On Φ(∂B_{r_{λ}}(y_{λ}) ∩ Ω), using (2.18), we know that there exists a constant C > 0 independent of λ such that φ_{λ}(x) ≥ CBu(y_{λ})R^{−1}. Using (2.9) and choosing B sufficient large, we know

φ_{λ}(x) ≥ f_{λ}(x) on Φ(∂B_{r_{λ}}(y_{λ}) ∩ Ω).

Easy to check

φ_{λ}(x) ≥ f_{λ}(x) on ∂B_{ϵ}^{+}(0).

Since φ_{λ}(x) > 0 satisfies

\{ \begin{matrix}F_{λ}φ_{λ} ≤ 0 in Ξ_{λ}^{+} \\ \frac{∂φ_{λ}}{∂x_{3}} = 0 on x_{3} = 0, \end{matrix}

we know from the maximum principle that

φ_{λ}(x) ≥ f_{λ}(x) in Ξ_{λ}^{+}.

As in case 1, noticing |x − x̄_{λ}|^{−1+δ} ≤ |x − x_{λ}|^{−1+δ} for x ∈ Ξ_{λ}^{+}, we have

f_{λ}(x) ≤ Cu_{λ}(y_{λ})^{2δ−1}|x − x_{λ}|^{−1+δ}.

It follows that for y ∈ Ψ(Ξ_{λ}^{+}),

u_{λ}(y) ≤ Cu_{λ}(y_{λ})^{2δ−1}|y − y_{λ}|^{−1+δ}.

Choosing ρ_{0} small such that B_{2ρ_{0}}(ȳ) ⊂ Ψ(B_{ϵ}), we then establish Lemma 2.3 in case Proposition 2.1 Assume that {u_{λ}} is the same subsequence given in Lemma 2.3 and y_{λ} → ȳ is an isolated simple blowup point, then, there exists a constant C > 0, such that, as λ small enough,

u_{λ}(y)u_{λ}(y_{λ}) ≤ C|y − y_{λ}|^{−1}, ∀|y − y_{λ}| ≤ ρ and y ∈ \overline{Ω}.  (2.20)

Proof. We first prove that (2.20) holds for |y − y_{λ}| = μ ≤ ρ_{0} and y ∈ Ω, where ρ_{0} is given in Lemma 2.3..

For any 0 < μ ≤ ρ_{0} and e ∈ ∂B_{1}(0) such that y_{λ} + μe ∈ Ω, we set w_{λ}(y) := w_{λ}(y, μ, e) = u_{λ}(y_{λ} + μe)^{−1}u_{λ}(y). Then w_{λ}(y) satisfies

\{ \begin{matrix}−Δw_{λ} + λw_{λ} = u_{λ}(y_{λ} + μe)^{4}w_{λ}^{5} & in Ω \\ \frac{∂w_{λ}}{∂ν} = 0 & on ∂Ω. \end{matrix}  (2.21)

It follows from Lemma 2.1 that for all K ⊂⊂ B_{ρ}(ȳ) \ {ȳ} there exists a constant C = C(K, μ) such that

w_{λ}(y) ≤ C in K.

Due to Lemma 2.3 we know u_{λ}(y_{λ} + μe)^{4} → 0 as λ → 0. It follows that w_{λ} → w in C_{loc}^{2}(B_{ρ}(ȳ) ∩ Ω \ {ȳ}) and w satisfies

\{ \begin{matrix}−Δw = 0 & in Ω \ \{ȳ\} \\ \frac{∂w}{∂ν} = 0 & on ∂Ω \ \{ȳ\}. \end{matrix}  (2.22)

Notice that for any fixed 0 < μ < ρ,

lim_{λ→0}u_{λ}(y_{λ} + μe)^{−1}r^{\frac{1}{2}} ū_{λ}(r) = r^{\frac{1}{2}} w̄(r)

where

w̄(r) = \frac{1}{|∂B_{r}(0) ∩ Ω|} ∫_{∂B_{r}(0)∩Ω} w.

We know from the assumption of isolated simple blowup that r^{\frac{1}{2}} w̄(r) is nonincreasing for 0 < r < ρ. If w(y) is regular at point {ȳ}, then w̄ = C > 0 (since w_{λ}(y_{λ}+μe) = 1, C ≠ 0). This shows that r^{\frac{1}{2}} w̄(r) is strictly increasing for 0 < r < ρ, contradiction! Therefore, we know that w has a singular point at ȳ.

We consider the following two cases:

Case 1: ȳ ∉ ∂Ω, that is dist(ȳ, ∂Ω) ≥ σ_{0} for some positive constant σ_{0}. Without loss of generality, we assume σ_{0} ≥ ρ_{0}.

In this case, w(y) satisfies:

\{ \begin{matrix}−Δw(y) = 0 & in B_{σ_{0}}(ȳ) \ \{ȳ\} \\ w(y) = g(y) & on ∂B_{σ_{0}}(ȳ) \\ w(y) → ∞ as y → ȳ \end{matrix}

where g(y) ∈ C^{2}(∂B_{σ_{0}}(ȳ)). It is well known (see e.g. Proposition 9.1 in [15]) that

w(y) = \frac{b}{|y − ȳ|} + h(y)

where b > 0 and h(y) ∈ C^{1}(B_{σ_{0}}(ȳ)) is a harmonic function. b may depend on e. From Harnack inequality, we know that there exist b_{∗}(μ), b^{∗}(μ) > 0 such that 0 < b_{∗}(μ) ≤ b ≤ b^{∗}(μ) < 0. Therefore,

lim_{λ→0} ∫_{B_{σ_{0}}(y_{λ})} Δw_{λ} = − lim_{λ→0} ∫_{∂B_{σ_{0}}(y_{λ})} \frac{∂w_{λ}}{∂ν} = ∫_{∂B_{σ_{0}}(ȳ)} \frac{∂w}{∂ν} = −4bπ ≤ −4b_{∗}(μ)π. (2.23)

On the other hand, from (2.21) we know that

\begin{aligned}− ∫_{B_{σ_{0}}(y_{λ})} Δw_{λ} & = ∫_{B_{σ_{0}}(y_{λ})}[u_{λ}(y_{λ} + μe)^{−1}u_{λ}^{5} − λw_{λ}] \\ & ≤ u_{λ}(y_{λ} + μe)^{−1} · ∫_{B_{σ_{0}}(y_{λ})} u_{λ}^{5}. \end{aligned}  (2.24)

Set x = u_{λ}(y_{λ})^{2} · (y − y_{λ}) and r_{λ} = Ru_{λ}^{−2}(y_{λ}). Using Lemma 2.2, we have

∫_{|y−y_{λ}|≤r_{λ}} u_{λ}^{5} = Cu_{λ}(y_{λ})^{5} ∫_{|y−y_{λ}|≤r_{λ}}(1 + \frac{1}{3}|x|^{2})^{− \frac{5}{2}} dy
= Cu_{λ}(y_{λ})^{5} ∫_{|x|≤R}(1 + \frac{1}{3}|x|^{2})^{− \frac{5}{2}} · u_{λ}(y_{λ})^{−6}dx
≤ Cu_{λ}(y_{λ})^{−1}.

Using Lemma 2.3, we have

∫_{r_{λ}≤|y−y_{λ}|≤σ_{0}} u_{λ}^{5} ≤ C ∫_{r_{λ}≤|y−y_{λ}|≤σ_{0}}(u_{λ}(y_{λ})^{2δ−1}|y − y_{λ}|^{−1+δ})^{5}dy
≤ Cr_{λ}^{−2+5δ}u_{λ}(y_{λ})^{10δ−5}
= CR^{−2+5δ}u_{λ}(y_{λ})^{−1}

Therefore

∫_{|y−y_{λ}|<σ_{0}} u_{λ}^{5} ≤ Cu_{λ}(y_{λ})^{−1}.  (2.25)

It follows from (2.23), (2.24) and (2.25) that

u_{λ}(y_{λ} + μe) · u_{λ}(y_{λ}) ≤ C(μ), |y − y_{λ}| = μ, y ∈ Ω.

Case 2: ȳ ∈ ∂Ω.

In this case, for any 0 < σ ≤ ρ_{0}, w satisfies

−Δw(y) = 0  in B_{σ}(ȳ) ∩ Ω
w(y) = g_{1}(y)  on ∂B_{σ}(ȳ) ∩ Ω
\frac{∂w(y)}{∂ν} = 0  on B_{σ}(ȳ) ∩ ∂Ω \ \{ȳ\}
w(y) → ∞ as y → ȳ

where g_{1}(y) ∈ C^{2}(∂B_{σ}(ȳ)). We need a lemma to classify the singular point.

Lemma 2.4 Let Ω_{1} ⊂ ℝ^{n}(n ≥ 3) be a smooth bounded domain, ȳ on ∂Ω_{1}. Suppose w ∈ C^{2}(\overline{Ω}_{1} \ \{ȳ\}) satisfies

−Δw = 0, w > 0,  in B_{σ_{∗}}(ȳ) ∩ Ω_{1}
\frac{∂w}{∂ν} = 0  on B_{σ_{∗}}(ȳ) ∩ ∂Ω_{1} \ \{ȳ\}
w(y) → ∞ as y → ȳ

Then

w(y) = \frac{a_{∗}}{|y − ȳ|^{n−2}} + h_{∗}(y)

where a_{∗} > 0 and h_{∗} is a harmonic function in B_{σ_{∗}}(ȳ) ∩ Ω_{1}.

The proof of this lemma can be carried out in the same way as the proof of Proposition 9.1 in [15]. We omit details here.

Due to the above lemma, we know that

w(y) = \frac{b_{1}}{|y − ȳ|} + h_{1}(y)

where b_{1} > 0 and h_{1}(y) ∈ C^{1}(B_{σ}(ȳ) ∩ Ω) is a harmonic function. Furthermore, |∂h_{1}/∂ν| ≤ C = C(∂Ω). Observing that there exists a constant C_{2} > 0, such that for all σ > 0, |∂B_{σ}(y_{λ}) ∩ Ω|/|∂B_{σ}(y_{λ})| > C_{2} uniformly for σ, we know that as σ small enough, there exists m_{1} > 0 such that

lim_{λ→0} ∫_{B_{σ_{0}}(y_{λ})∩Ω} Δw_{λ} = − lim_{λ→0} ∫_{∂B_{σ_{0}}(y_{λ})∩Ω} \frac{∂w_{λ}}{∂ν} = ∫_{∂B_{σ_{0}}(ȳ)∩Ω} \frac{∂w}{∂ν} < −m_{1}.  (2.26)

We then follow case 1 closely and can show that (2.20) holds for |y − y_{λ}| = μ ≤ ρ_{0} and y ∈ Ω.

Secondly, we prove Proposition 2.1 for r_{λ} < |y − y_{λ}| ≤ ρ_{0}. Suppose the contrary, then there exist a sequence ỹ_{λ} satisfying r_{λ} ≤ |ỹ_{λ} − y_{λ}| ≤ ρ_{0} and |ỹ_{λ} − y_{λ}| → 0 as λ → 0, such that

lim_{λ→0} u_{λ}(ỹ_{λ})u_{λ}(y_{λ})|ỹ_{λ} − y_{λ}| = ∞.

Set r̃_{λ} = |ỹ_{λ} − y_{λ}|/ρ_{0}, and ũ_{λ}(z) = r̃_{λ}^{1/2}u_{λ}(y_{λ} + r̃_{λ}z). Then {ũ_{λ}} satisfies

\{ \begin{matrix}−Δũ_{λ} + λr̃_{λ}^{2}ũ_{λ} = ũ_{λ}^{5} & in Ωr̃_{λ} \\ \frac{∂ũ_{λ}}{∂ν} = 0 & on ∂Ωr̃_{λ} \end{matrix}

where Ωr̃_{λ} = (Ω − y_{λ})/r̃_{λ}. It is easy to see that {0} is an isolated simple blowup point. If dist(0, ∂Ωr̃_{λ}) ≥ δ_{0,0} > 0, we repeat the above process of case 1, and have

ũ_{λ}(0)ũ_{λ}(z) ≤ C ∀|z| = ρ_{0}.

Choosing z = (ỹ_{λ} − y_{λ})/r̃_{λ}, we have from the above that

ũ_{λ}(0)ũ_{λ}(\frac{ỹ_{λ} − y_{λ}}{r̃_{λ}}) ≤ C.  (2.27)

If dist(0, ∂Ωr̃_{λ}) → 0, noticing that Ωr̃_{λ} tends to ℝ_{+}^{3}, we can repeat the process of case 2 and have (2.27) again. Therefore,

lim_{λ→0} u_{λ}(ỹ_{λ})u_{λ}(y_{λ})|ỹ_{λ} − y_{λ}| < ∞.

Contradiction!

When |y − y_{λ}| ≤ r_{λ}, (2.20) follows from Lemma 2.2. When ρ_{0} ≤ |y − y_{λ}| ≤ ρ, (2.20) follows from Harnack inequality. We complete the proof of Proposition 2.1.

2.2 Isolated blowup points must be isolated simple blowup points

In this subsection, we focus on showing the following proposition.

Proposition 2.2 Let u_{λ} solve (0.3) and ȳ be an isolated blowup point. If Ω is convex, then ȳ is an isolated simple blowup point.

Here the difference between n = 3 and n ≥ 4 will come in. Technically, we can see this proposition might not hold for high dimensional domain since estimate (2.33) below does not hold for n ≥ 4. However, In view of those counterexamples in [2] or [5], we know such kind of technical difference is essential and this proposition is not true any more when n = 4, 5, 6. See more details from the example in [15].

Also, we would like to emphasis here that the assumption of convexity about Ω will be used in the proof of the above proposition.

Proof of Proposition 2.2. Without loss of generality, we assume that u_{λ} is the subsequence satisfying (2.5) or (2.6) for R >> 1. It follows from Lemma 2.2 that r^{\frac{1}{2}} ū_{λ}(r) has precisely one critical point in the interval 0 < r < r_{λ} = Ru_{λ}(y_{λ})^{−2}. If ȳ is not an isolated simple blowup point, we know that r^{\frac{1}{2}} ū_{λ}(r) has the second critical point r = s_{λ} with s_{λ} ≥ r_{λ} and lim_{λ→0} s_{λ} = 0.

Set

\{ \begin{matrix}Ω_{λ} = (Ω − y_{λ})/s_{λ} \\ ξ_{λ}(y) = sλ^{\frac{1}{2}} u_{λ}(s_{λ}y + y_{λ}) ∀ y ∈ Ω_{λ}. \end{matrix}

Then ξ_{λ} satisfies

−Δξ_{λ}(y) + λs_{λ}^{2}ξ_{λ}(y) = ξ_{λ}(y)^{5}  in Ω_{λ}
|y|^{\frac{1}{2}} ξ_{λ}(y) ≤ C  in Ω_{λ} ∩ B_{2}(0)
\frac{∂ξ_{λ}(y)}{∂ν} = 0  on ∂Ω_{λ}

lim_{λ→∞}ξ_{λ}(0) = ∞

r^{\frac{1}{2}} ξ̄_{λ}(r) has precisely one critical point in 0 < r < 1

\frac{d}{dr}\{r^{\frac{1}{2}} ξ̄_{λ}(r)\}|_{r=1} = 0.

(2.28)

It follows that {0} is an isolated simple blowup point of {ξ_{λ}}. Denote

T = lim_{λ→0} inf \frac{dist(y_{λ}, ∂Ω)}{s_{λ}}.

Without loss of generality, we can assume that ξ_{λ} satisfies (2.5) or (2.6) for some R_{1} >> 1. We know from Proposition 2.1 and Harnack inequality that for all compact set K ⊂ ℝ_{T}^{3} \ {0},

C(K)^{−1} ≤ ξ_{λ}(0)ξ_{λ}(y) ≤ C(K) on K,

where ℝ_{T}^{3} = {x = (x_{1}, x_{2}, x_{3}) | x_{3} > −T }. It follows, for some constant a > 0, that

ξ_{λ}(0)ξ_{λ}(y) → H(y) ≡ a|y|^{−1} + b(y) in C_{loc}^{2}(ℝ_{T}^{3} \ {0}),  (2.29)

where b(y) satisfies Δb = 0 on ℝ_{T}^{3} .

If T = ∞, then from the maximum principle and the fact : lim inf_{|y|→∞} b(y) ≥ 0, we know that b(y) is nonnegative, hence a constant. We denote it as b.

If T = 0, then ∂b/∂ν = 0 on ∂ℝ_{+}^{3}. We reflect b(x) into lower half space and conclude that b(y) = b.

If 0 < T < ∞, we have, via the maximum principle and the boundary condition: ∂H/∂ν = 0 on ∂ℝ_{T}^{3} , that

b(y) = a|y − y_{0}|^{−1} + C

where C ≥ 0, y_{0} = (0, 0, −2T). Thus

H(y) = a|y|^{−1} + a_{1} + h(y) ∀y ∈ ℝ_{T}^{3} \ {0}  (2.30)

where a_{1} > 0, h(y) ∈ C^{2}(B_{T/2}(0)) and h(0) = 0.

It is easy to see from (2.28) and (2.29) that

\frac{d}{dr}\{r^{1/2}H(r)\}|_{r=1} = 0.

Therefore, we have

b = a > 0.

In the following, we discuss in two cases depending on the locations of {0} in Ω_{λ}. Case 1. d(0, ∂Ω_{λ}) ≥ σ_{1} > 0 for some fixed σ_{1} > 0. In this case, we apply the corresponding Pohozaev identity (3.2) to (2.28) on

B_{σ}(0) ⊂ B_{σ_{1}}(0) for some small σ < σ_{1} and have (see Appendix for more details on obtaining such Pohozaev identity)

−λs_{λ}^{2} ∫_{B_{σ}} ξ_{λ}^{2}dy = ∫_{∂B_{σ}} B(y, σ, ξ_{λ}, ∇ξ_{λ}) − σ ∫_{∂B_{σ}}(\frac{λs_{λ}^{2}ξ_{λ}^{2}}{2} − \frac{ξ_{λ}^{6}}{6} )  (2.31)

where

B(y, σ, ξ_{λ}, ∇ξ_{λ}) = \frac{1}{2}ξ_{λ} \frac{∂ξ_{λ}}{∂ν} − \frac{σ}{2} |∇ξ_{λ}|^{2} + σ(\frac{∂ξ_{λ}}{∂ν})^{2}.

It follows from (2.29) and (2.30) that

lim_{σ→0^{+}} lim_{λ→0}ξ_{λ}^{2}(0) ∫_{∂B_{σ}} B(y, σ, ξ_{λ}, ∇ξ_{λ}) = lim_{σ→0^{+}} ∫_{∂B_{σ}} B(y, σ, H, ∇H)

≤ −\frac{(min(a,a_{1}))^{2}}{2} |∂B_{1}| < 0,

(2.32)

and

lim_{λ→0}ξ_{λ}^{2}(0) · ∫_{∂B_{σ}} σ(\frac{λs_{λ}^{2}ξ_{λ}^{2}}{2} − \frac{ξ_{λ}^{6}}{6} ) = 0.

Using Lemma 2.2 and Proposition 2.1, we can calculate ∫_{B_{σ}} ξ_{λ}^{2} as follows (we set x = ξ_{λ}(0)^{2}y):

\begin{aligned}∫_{|y|≤r_{λ}} ξ_{λ}^{2} & ≤ Cξ_{λ}^{2}(0) ∫_{|y|≤r_{λ}}(1 + \frac{1}{3}|x|^{2})^{−1}dy \\ & ≤ Cξ_{λ}^{2}(0) ∫_{|x|≤R_{1}}(1 + \frac{1}{3}|x|^{2})^{−1}ξ_{λ}(0)^{−6}dx \\ & ≤ o(1)ξ_{λ}^{−2}(0) \end{aligned}  (2.33)

and

\begin{aligned}∫_{r_{λ}≤|y|≤σ} ξ_{λ}^{2}(y) & ≤ C ∫_{r_{λ}≤|y|≤σ}(ξ_{λ}(0)^{−1}|y|^{−1})^{2} \\ & ≤ Cξ_{λ}(0)^{−2}. \end{aligned}

Multiplying (2.31) by ξ_{λ}(0)^{2} and sending σ, λ to 0, then the absolute value of left hand side of (2.31) tends to 0 but the absolute value of right hand side does not. Contradiction! We thus establish Proposition 2.2 in case 1.

Case 2. dist(0, ∂Ω_{λ}) → 0 as λ → 0.

Noticing {0} is an isolated simple blowup point, by a slight modification of the proof of Proposition 2.1, we know that there exists ρ_{1} > 0, such that

ξ_{λ}(y)ξ_{λ}(0) ≤ C|y|^{−1} ∀y ∈ Ω_{λ} and |y| < ρ_{1}.  (2.34)

We discuss in two different subcases:

Subcase 1. {0} ∈ ∂Ω_{λ}.

We apply the corresponding Pohozaev identity (3.8) to (2.28) on Ω_{λ}^{+} := B_{σ}(0) ∩ Ω_{λ} for some σ < ρ_{1} and have (see Appendix for more details to get such identity)

−λs_{λ}^{2} ∫_{Ω_{λ}^{+}} ξ_{λ}^{2} + \frac{λs_{λ}^{2}}{2} ∫_{∂Ω_{λ}^{+}} ξ_{λ}^{2}y · ν − \frac{1}{6} ∫_{∂Ω_{λ}^{+}} ξ_{λ}^{6}y · ν + \frac{1}{2} ∫_{Γ_{1}} |∇ξ_{λ}|^{2}y · ν = B_{1}(y, σ, ξ_{λ}, ∇ξ_{λ})

(2.35)

where

B_{1}(y, σ, ξ_{λ}, ∇ξ_{λ}) = ∫_{Γ_{2}}[σ(\frac{∂ξ_{λ}}{∂ν})^{2} − \frac{σ}{2} |∇ξ_{λ}|^{2} + \frac{1}{2}ξ_{λ} \frac{∂ξ_{λ}}{∂ν}],

Γ_{1} = ∂Ω_{λ}∩B_{σ}, Γ_{2} = Ω_{λ}∩∂B_{σ} and ν is the exterior unit normal to the corresponding boundary surface.

Since Ω is convex, we know that Ω_{λ} is convex. Therefore

y · ν ≥ 0, ∀y ∈ ∂Ω_{λ}^{+}.

Similar to the calculations in case 1, we know:

lim_{σ→0^{+}} lim_{λ→0}ξ_{λ}^{2}(0) · B_{1}(y, σ, ξ_{λ}, ∇ξ_{λ}) = −\frac{(min(a, a_{1}))^{2}}{4} |∂B_{1}| < 0,

lim_{σ→0^{+}} lim_{λ→0}ξ_{λ}^{2}(0) · ∫_{Γ_{2}} ξ_{λ}^{6}y · ν = 0

and

λs_{λ}^{2} ∫_{Ω_{λ}^{+}} ξ_{λ}^{2} ≤ Cλs_{λ}^{2}ξ_{λ}(0)^{−2}.

We claim:

∫_{Γ_{1}} ξ_{λ}^{6}y · ν ≤ o_{λ}(1)ξ_{λ}^{−2}(0)

where o_{λ}(1) → 0 as λ → 0.

Multiplying both sides of (2.35) by ξ^{2}(0) and sending λ, σ to zero, we reach a contradiction since the left hand side will be nonnegative and the right hand side will be a negative number, thus establish Proposition 2.2 in this subcase.

We are left to show the above claim. By choosing a suitable coordinate, we can assume that y_{3} axis is the inner norm of ∂Ω_{λ} at point {0}.

We write

∫_{Γ_{1}} ξ_{λ}^{6}y · ν = ∫_{Γ_{1}∩{y: |y|≤r̄_{λ}}} ξ_{λ}^{6}y · ν + ∫_{Γ_{1}∩{y: r̄_{λ}<|y|<σ}} ξ_{λ}^{6}y · ν := Π_{1} + Π_{2}

where r̄_{λ} = R_{1}ξ_{λ}(0)^{−2}.

Since Ω_{λ} tends to ℝ_{+}^{3}, it is not difficult to see that for any y = (y_{1}, y_{2}, y_{3}) ∈ ∂Ω_{λ} ∩ B_{σ}(0), as λ tends to 0,

y · ν = o_{λ}(1)(|y_{1}|^{2} + y_{2}|^{2}), y_{3} = o_{λ}(1)\sqrt{y_{1}^{2} + y_{2}^{2}}.  (2.36)

Therefore, by using Proposition 2.1 we have

\begin{aligned}Π_{2} & ≤ o_{λ}(1)ξ^{−6}(0) ∫_{r̄_{λ}^{2}/2≤|y_{1}|^{2}+|y_{2}|^{2}≤σ^{2}} (|y_{1}|^{2} + |y_{2}|^{2})^{−2}dy_{1}dy_{2} \\ & ≤ o_{λ}(1)ξ_{λ}(0)^{−6} · r̄_{λ}^{−2} = o_{λ}(1)ξ_{λ}^{−2}(0). \end{aligned}

Also, by using Lemma 2.2 and (2.36), we have

Π_{1} ≤ o_{λ}(1) ∫_{Γ_{1}∩\{y: |y|≤r̄_{λ}\}} ξ_{λ}^{6}(0)(1 + \frac{1}{3}ξ_{λ}^{4}(0)|y|^{2})^{−3}|y|^{2}ds
≤ o_{λ}(1)ξ_{λ}^{−2}(0) ∫_{r≤2R_{1}}(1 + \frac{1}{3}r^{2})^{−3} · r^{3}dr
≤ o_{λ}(1)ξ_{λ}^{−2}(0).

The claim follows from the above two inequalities.

Subcase 2: {0} ∉ ∂Ω_{λ}.

Denote O_{λ} as the closest point to the blowup point y_{λ} on ∂Ω_{λ}. We choose a coordinate such that it centers at O_{λ} and y_{3} axis is the inner norm of ∂Ω_{λ} at O_{λ}.

Applying the corresponding Pohozaev identity (3.8) to (2.28) on Ω_{λ}^{+} := B_{σ}(0) ∩ Ω_{λ}, we have

−λs_{λ}^{2} ∫_{Ω_{λ}^{+}} ξ_{λ}^{2} + \frac{λs_{λ}^{2}}{2} ∫_{∂Ω_{λ}^{+}} ξ_{λ}^{2}y · ν − \frac{1}{6} ∫_{∂Ω_{λ}^{+}} ξ_{λ}^{6}y · ν + \frac{1}{2} ∫_{Γ_{1}} |∇ξ_{λ}|^{2}y · ν = B_{1}(y, σ, ξ_{λ}, ∇ξ_{λ})

(2.37)

where Γ_{1}, B_{1}(y, σ, ξ_{λ}, ∇ξ_{λ}) and ν are defined in the same way as in subcase 1.

Similarly we know

y · ν ≥ 0, ∀y ∈ ∂Ω_{λ}^{+},

lim_{σ→0^{+}} lim_{λ→0}ξ_{λ}^{2}(0) · B_{1}(y, σ, ξ_{λ}, ∇ξ_{λ}) = −\frac{(min(a, a_{1}))^{2}}{4} |∂B_{1}| < 0,

lim_{σ→0^{+}} lim_{λ→0}ξ_{λ}^{2}(0) · ∫_{Γ_{2}} ξ_{λ}^{6}y · ν = 0

and

λs_{λ}^{2} ∫_{B_{σ}^{+}} ξ_{λ}^{2} ≤ Cλs_{λ}^{2}ξ_{λ}(0)^{−2}.

In order to estimate ∫_{Γ_{1}} ξ_{λ}^{6}y · ν, we write

∫_{Γ_{1}} ξ_{λ}^{6}y · ν = ∫_{Γ_{1}∩{y: |y−y_{λ}|≤r̄_{λ}}} ξ_{λ}^{6}y · ν + ∫_{Γ_{1}∩{y: r̄_{λ}<|y−y_{λ}|<σ}} ξ_{λ}^{6}y · ν := Υ_{1} + Υ_{2}

where r̄_{λ} = R_{1}ξ_{λ}(0)^{−2}.

It is easy to see that ∀P = (y_{1}, y_{2}, y_{3}) ∈ ∂Ω_{λ},

\sqrt{y_{1}^{2} + y_{2}^{2}} ≤ |P y_{λ}|  (2.38)

where |P y_{λ}| means the distance from P to y_{λ}.

By using Proposition 2.1, (2.36) and (2.38) we have

\begin{aligned}Υ_{2} & ≤ o_{λ}(1)ξ^{−6}(0) ∫_{r̄_{λ}^{2}/2≤|y_{1}|^{2}+|y_{2}|^{2}≤σ^{2}} (|y_{1}|^{2} + |y_{2}|^{2})^{−2}dy_{1}dy_{2} \\ & ≤ o_{λ}(1)ξ_{λ}(0)^{−6} · r̄_{λ}^{−2} = o_{λ}(1)ξ_{λ}^{−2}(0). \end{aligned}

By using Lemma 2.2, (2.36) and (2.38), we have

Υ_{1} ≤ o_{λ}(1) ∫_{Γ_{1}∩\{y: |y−y_{λ}|≤r̄_{λ}\}} ξ_{λ}^{6}(0)(1 + \frac{1}{3}ξ_{λ}^{4}(0)|y − y_{λ}|^{2})^{−3}|y|^{2}ds
≤ o_{λ}(1) ∫_{Γ_{1}∩\{y: |y|≤r̄_{λ}\}} ξ_{λ}^{6}(0)(1 + \frac{1}{3}ξ_{λ}^{4}(0)|y|^{2})^{−3}|y|^{2}ds
≤ o_{λ}(1)ξ_{λ}^{−2}(0) ∫_{r≤2R_{1}}(1 + \frac{1}{3}r^{2})^{−3} · r^{3}dr
≤ o_{λ}(1)ξ_{λ}^{−2}(0).

We then can derive a contradiction as in subcase 1. We hereby establish Proposition 2.2.

2.3 Finite isolated simple blowup points

In this subsection, we will show that: if u_{λ} blows up, it has only isolated blowup points and these points must stay away from each other uniformly. Combing this fact with previous subsection we know that u_{λ} has only finite isolated and simple blowup points.

By a slight modification of the proof of Lemma 3.1 in [21] (see also [15]), we have the following proposition.

Proposition 2.3 Assume 0 ≤ λ ≤ 1 and u is a solution to (0.3). For any ϵ > 0 and R > 0, there exist C^{∗}(ϵ, R), C_{∗}(ϵ, R) > 0, such that: if

max_{\overline{Ω}} u(y) > C^{∗}(ϵ, R),

then there exist finite points in \overline{Ω} denoted as

BL(u) = {q_{1}, q_{2}, ..., q_{k}}

such that

i) Each q_{α} is a local maximum point of u. Let r_{α} = Ru^{−2}(q_{α}), then B_{r_{α}}(q_{α}) ∩ B_{r_{β}}(q_{β}) = ∅ for α ≠ β and

||u^{−1}(q_{α})u(u^{−2}(q_{α})x + q_{α}) − v(x)||_{C^{2}(B_{2R}(0))} < ϵ,

or

||u^{−1}(q_{α})u(u^{−2}(q_{α})x + q_{α}) − v(x)||_{C^{2}(B_{2R}^{+}(0))} < ϵ,

where v(x) = (1 + \frac{1}{3}|x|^{2})^{− \frac{1}{2}} .

ii)

u(y) ≤ C_{∗}\{dist(y, BL(u))\}^{− \frac{1}{2}} , ∀y ∈ \overline{Ω}.

We now state our main proposition in this subsection.

Proposition 2.4 Assume that 0 ≤ λ ≤ 1, u_{λ} is a solution to (0.3) and Ω is convex. Then for ϵ > 0 sufficiently small and R > 1 sufficiently large, there exists a constant δ^{∗} = δ^{∗}(ϵ, R, C^{∗}), such that : if max_{\overline{Ω}}u_{λ}(x) > C^{∗}, we have

|q_{α} − q_{β}| ≥ δ^{∗}, for all 1 ≤ α ≠ β ≤ k,

where C^{∗}, q_{α} = q_{α}(u_{λ}), q_{β} = q_{β}(u_{λ}), k = k(u_{λ}) are the ones defined in Proposition 2.3.

Proof. We prove it by contradiction. Suppose the contrary, then for some small ϵ > 0, large R, λ_{i} → λ_{0} ∈ [0, 1] and {u_{i}(y)} satisfying

\{ \begin{matrix}−Δu_{i} + λ_{i}u_{i} = u_{i}^{5}, u_{i} > 0, & in Ω \\ \frac{∂u_{i}}{∂ν} = 0 & on ∂Ω, \end{matrix}  (2.39)

with max\overline{Ω} u_{i} ≥ C^{∗}, we have

lim_{i→∞} min_{α≠β}|q_{α}(u_{i}) − q_{β}(u_{i})| = 0.

Without loss of generality, we can assume that

δ_{i} := |q_{1}(u_{i}) − q_{2}(u_{i})| = min_{j≠l}|q_{j}(u_{i}) − q_{l}(u_{i})| → 0.

Also we have, in view of Proposition 2.3 (i), that B_{Ru_{i}(q_{1})^{−2}}(q_{1}) ∩ B_{Ru_{i}(q_{2})^{−2}}(q_{2}) = ∅. It follows that u_{i}(q_{1}), u_{i}(q_{2}) → ∞.

Let w_{i}(x) = δi^{\frac{1}{2}} u_{i}(δ_{i}x + q_{1}), Ω_{i} = (Ω − q_{1})/δ_{i}. It follows that w_{i}(x) satisfies

\{ \begin{matrix}−Δw_{i}(x) + δ_{i}^{2}λ_{i}w_{i}(x) = w_{i}(x)^{5} & in B_{\frac{1}{δ_{i}}} (0) ∩ Ω_{i} \\ w_{i}(x) > 0 & in B_{\frac{1}{δ_{i}}} (0) ∩ Ω_{i} \\ \frac{∂w_{i}}{∂ν} = 0 on ∂Ω_{i}. \end{matrix}  (2.40)

From Proposition 2.3, we know

δ_{i} > max{Ru_{i}(q_{1})^{−2}, Ru_{i}(q_{2})^{−2}}.  (2.41)

and

u_{i}(y) ≤ C_{∗}dist(y, \{q_{1}, q_{2}, ..., q_{k}\})^{− \frac{1}{2}} for all y ∈ Ω.  (2.42)

Denoting x_{α} = (q_{α} − q_{1})/δ_{i} for 2 ≤ α ≤ k, we have (using (2.41) and (2.42)) that

w_{i}(0), w_{i}(x_{2}) ≥ R;

Each x_{α} is a local maximum point of w_{i}; w_{i}(x)dist(x, \{x_{1}, x_{2}, ..., x_{k}\})^{\frac{1}{2}} ≤ C_{∗} ∀ x ∈ Ω_{i}.

As in [15], we claim that

w_{i}(0) → ∞,  and w_{i}(x_{2}) → ∞.  (2.43)

First, we assume {0}, x_{2} ∈ Ω_{i}. If one of them tends to infinity along a subsequence, say w_{i}(0) → ∞, we know that {0} is an isolated blowup point, and therefore, is an isolated simple blowup point. Then w_{i}(x_{2}) has to tend to infinity along the same sequence since otherwise, by Harnack inequality, {w_{i}} would be uniformly bounded near {x_{2}} along a further subsequence. In turn, using Proposition 2.1 and Harnack inequality, we know that {w_{i}} uniformly goes to zero near {x_{2}}, which violates w_{i}(x_{2}) ≥ R. On the other hand, if both of {w_{i}(0)} and {w_{i}(x_{2})} stay bounded, we know from a similar argument as the above that {w_{i}} is locally bounded. It then follows from standard elliptic estimates that after passing to a subsequence, w_{i} → w in C^{3}(ℝ^{3}) and x_{2} → x̄ with |x̄| = 1, where w satisfies −Δw = w^{5}, w > 0 in ℝ^{3} and ∇w(0) = ∇w(x̄) = 0. This is a contradiction since there is no positive solution to the equation with two distinct critical points according to the Liouville type theorem of Caffarelli, Gidas and Spruck [6].

If one of {0}, x_{2} ∈ ∂Ω_{i}, modifying the above proof slightly by using the boundary condition ∂w_{i}/∂ν = 0, we can show (2.43) still holds. We omit detail here.

According to our definition, we know {0} and x_{2} → x̄ are two isolated blowup points for w_{i}(x). Therefore, from Proposition 2.2 we know that these two points are exactly two isolated simple blowup points for w_{i}(x).

After passing to a subsequence, we denote S̃ as the set of blowup points of w_{i}(x), and have

min{|x_{m} − x_{l}| : x_{m}, x_{l} ∈ S̃} ≥ 1.

Without loss of generality, we assume that w_{i} is the subsequence satisfying (2.5) or (2.6) for some R >> 1 and assume (passing to some further subsequence) that

w_{i}(0) = min_{x∈S̃} w_{i}(x).  (2.44)

Define

T = lim_{i→∞} inf \frac{dist(q_{1}, ∂Ω)}{δ_{i}}.

We discuss in two cases depending on the locations of {0} in Ω_{i}

Case 1. T > 0.

In this case, we know that {0} ∈ Ω_{i} and Ω_{i} tends to ℝ_{T}^{3} . From Proposition 2.1, we have

w_{i}(0)w_{i}(x) ≤ C(K) K ⊂⊂ ℝ_{T}^{3} \ S̃.  (2.45)

Thus

lim_{i→∞}w_{i}(0)w_{i}(x) = h^{∗}(x) in C_{loc}^{2}(ℝ_{T}^{3} \ S̃)

where h^{∗}(x) is a regular harmonic function in ℝ_{T}^{3} \ S̃ and ∂h^{∗}/∂ν = 0 on ∂ℝ_{T}^{3} if T < ∞. It follows that

h^{∗}(x) = a_{2}|x|^{−1} + a_{3}|x − x̄|^{−1} + b^{∗}(x) x ∈ ℝ_{T}^{3} \ S̃  (2.46)

where a_{2}, a_{3} ≥ 0, and b^{∗} is some regular harmonic function in ℝ_{T}^{3} \ {S̃ \ {0, x̄}} and ∂b^{∗}/∂ν ≥ 0 if T < ∞. Since {0} is an isolated simple blowup point, we know that r^{1/2}\overline{h^{∗}}(r) is nonincreasing near r = 0. It follows that h^{∗} has a singular point at \{0\}, that is a_{2} > 0. Similar, we know a_{3} > 0. Also, noticing h^{∗} ≥ 0 and h^{∗} → ∞ as x → S̃, we know that b^{∗} ≥ 0 as |x| → ∞ or x → S̃. It follows from the maximum principle that b^{∗}(x) ≥ 0 for all x ∈ ℝ_{T}^{3} \ {S̃ \ {0, x̄}}. Therefore, for some A > 0,

h^{∗}(x) = a_{2}|x|^{−1} + A + O(|x|) as |x| → 0.  (2.47)

Choosing σ < 1 such that B_{σ}(0) ⊂ Ω_{i}, we apply (3.2) to (2.40) on B_{σ}(0) and get

−δ_{i}^{2}λ_{i} ∫_{B_{σ}(0)} w_{i}^{2}dx = ∫_{∂B_{σ}(0)} B(x, σ, w_{i}, ∇w_{i}) − σ ∫_{∂B_{σ}(0)}(\frac{δ_{i}^{2}λ_{i}w_{i}^{2}}{2} − \frac{w_{i}^{6}}{6} ). (2.48)

From (2.45), (2.46) and (2.47), it is easy to see : as σ small enough

lim_{i→∞}w_{i}(0)^{2} ∫_{∂B_{σ}} B(σ, y, w_{i}, ∇w_{i}) = ∫_{∂B_{σ}} B(σ, y, h^{∗}, ∇h^{∗}) < 0.  (2.49)

On the other hand, as in the proof of Proposition 2.2, we have

lim_{i→∞} w_{i}(0)^{2} · δ_{i}^{2}λ_{i} ∫_{B_{σ}} w_{i}^{2} = 0.

From (2.45), we have

lim_{i→∞} w_{i}(0)^{2} · σ ∫_{∂B_{σ}(0)}(\frac{δ_{i}^{2}λ_{i}w_{i}^{2}}{2} − \frac{w_{i}^{6}}{6} ) = 0.

Combining these two identities with (2.48), we have

lim_{i→∞}w_{i}(0)^{2} ∫_{∂B_{σ}} B(σ, x, w_{i}, ∇w_{i}) ≥ 0.  (2.50)

Obviously we get a contradiction from (2.49) and (2.50), thus establish Proposition 2.4 in case 1.

Case 2. T=0.

In this case, Ω_{i} tends to ℝ_{+}^{3}. From Proposition 2.1, we have

w_{i}(0)w_{i}(x) ≤ C(K) K ⊂⊂ ℝ_{+}^{3} \ S̃.  (2.51)

Thus

lim_{i→∞}w_{i}(0)w_{i}(x) = h̃^{∗}(x) in C_{loc}^{2}(ℝ_{+}^{3} \ S̃)

where h̃^{∗}(x) is a regular harmonic function in ℝ_{+}^{3} \ S̃. It follows that

h̃^{∗}(x) = b_{1}|x|^{−1} + b_{2}|x − x̄|^{−1} + b̃^{∗}(x) x ∈ ℝ_{+}^{3} \ S̃  (2.52)

where b_{1}, b_{2} ≥ 0 and b̃^{∗} is some regular harmonic function in ℝ_{+}^{3} \ {S̃ \ {0, x̄}} and ∂b̃^{∗}/∂x_{3} = 0 on {x_{3} = 0}. Similar argument to case 1, we know that b_{1}, b_{2} > 0 and b̃^{∗}(x) ≥ 0 for all x ∈ ℝ_{+}^{3} \ {S̃ \ {0, x̄}}.

We then discuss in two different subcases as in the proof of Proposition 2.2. Subcase 1. {0} ∈ ∂Ω_{i}. For all σ < 1, we apply (3.8) to (2.40) on B_{σ}(0) ∩ Ω_{i}, and have

−δ_{i}^{2}λ_{i}·∫_{Ω^{+}} w_{i}^{2}+ \frac{δ_{i}^{2}λ_{i}}{2} ∫_{∂Ω^{+}} w_{i}^{2}x·ν− \frac{1}{6} ∫_{∂Ω^{+}} w_{i}^{6}x·ν+ \frac{1}{2} ∫_{Γ_{1}} |∇w_{i}|^{2}x·ν = B_{1}(x, σ, w_{i}, ∇w_{i})

(2.53)

where

B_{1}(x, σ, w_{i}, ∇w_{i}) = ∫_{Γ_{2}}[σ(\frac{∂w_{i}}{∂ν})^{2} − \frac{σ}{2} |∇w_{i}|^{2} + \frac{1}{2}w_{i} \frac{∂w_{i}}{∂ν}],

Ω^{+} = Ω_{i} ∩ B_{σ}(0), Γ_{1} = ∂Ω_{i} ∩ B_{σ}(0) and Γ_{2} = Ω_{i} ∩ ∂B_{σ}(0).

As in the proof of Proposition 2.2, we have

x · ν ≥ 0 ∀x ∈ ∂Ω^{+},

lim_{σ→0^{+}} lim_{λ_{i}→0}w_{i}(0)^{2} ∫_{∂B_{σ}} B_{1}(x, σ, w_{i}, ∇w_{i}) < 0,

lim_{σ→0^{+}} lim_{λ_{i}→0}w_{i}(0)^{2} · ∫_{Γ_{2}} \frac{w_{i}^{6}}{6} x · ν = 0,

λ_{i} ∫_{Ω_{+}} w_{i}^{2} ≤ Cλ_{i}w_{i}(0)^{−2}

and

∫_{Γ_{1}} w_{i}^{6}x · ν ≤ o(1)w_{i}^{−2}(0)

where o(1) → 0 as i → ∞.

Bringing all these estimates into (2.53) we reach a contradiction as i → ∞, thus complete the proof of this proposition in subcase 1.

Subcase 2. {0} ∉ ∂Ω_{i}.

In this case, we can derive a contradiction in the same way as the proof of subcase 2 of Proposition 2.2. We leave these details to interested readers.

We hereby establish the Proposition 2.4.

2.4 Conclusion via Green’s function

Now we can prove Theorem 1.1 through a contradiction argument. Suppose the contrary, there exists a sequence {u_{λ}} such that ||u_{λ}||_{L^{∞}} → ∞. According to Proposition 2.2 and 2.4, we know: after passing to some subsequence of λ, we can find k points in \overline{Ω}: BL(u_{λ}) = \{q_{1}, ⋯, q_{k}\}, q_{α} → q_{α}^{∗}, which are isolated and simple blowup points for α = 1, 2, ⋯, k. Without loss of generality, we assume u_{λ}(q_{1}) = min_{q∈BL(u_{λ})} u_{λ}(q). Let w_{λ}(y) = u_{λ}(q_{1}) · u_{λ}(y). From Proposition 2.1, we know for any K ⊂⊂ Ω \ BL(u_{λ}), after passing to some subsequence (still denote it as w_{i})

w_{λ}(y) ≤ C in K.

Therefore, after passing to some further subsequence (still denote it as w_{i})

w_{λ}(y) → w_{0} in C^{2}(K)

and w_{0} > 0 satisfies

\{ \begin{matrix}−Δw_{0} = 0 in Ω \ \{q_{1}^{∗}, q_{2}^{∗}, ... q_{k}^{∗}\} \\ \frac{∂w_{0}}{∂ν} = 0 on ∂Ω \ \{q_{1}^{∗}, q_{2}^{∗}, ... q_{k}^{∗}\}. \end{matrix}

It is well known (see for example the proof of Proposition 9.1 in [15]) that

w_{0} = ∑_{α=1}^{k} \frac{c_{α}}{|y − q_{α}^{∗}|} + H^{∗}(y) in Ω

where c_{α} ≥ 0. It follows that H^{∗}(y) satisfies:

\{ \begin{matrix}ΔH^{∗} = 0 in Ω \\ \frac{∂H^{∗}}{∂ν} = − ∑_{α=1}^{k} c_{α} · \frac{∂}{∂ν} (|y − q_{α}^{∗}|^{−1}) on ∂Ω. \end{matrix}

Since there is at least one blowup point of u_{λ}, we know, in view of subsection 2.2, that there is at least one isolated simple blowup point. It follows that at least one of c_{α} is nonzero. Noticing Ω is convex, we know that

c_{α} · \frac{∂}{∂ν} (|y − q_{α}^{∗}|^{−1}) ≤ 0 on ∂Ω

and − ∑_{α=1}^{k} c_{α} · \frac{∂}{∂ν} (|y − q_{α}^{∗}|^{−1}) ≢ 0. From the maximum principle and Hopf Lemma, we then know that such H^{∗}(y) does not exist. Contradiction!

3 Appendix: Pohozaev type identities

In this appendix, we present some Pohozaev type identities which are used in the current context. Here we assume that Ω is a smooth bounded domain in ℝ^{3}.

Proposition 3.1 Assume that v satisfies

\{ \begin{matrix}−Δv + αv = v^{5}, v ≥ 0, & in Ω \\ \frac{∂v}{∂ν} = 0 & on ∂Ω. \end{matrix}  (3.1)

If B_{σ}(x_{0}) ⊂ Ω, then

−α ∫_{B_{σ}(x_{0})} v^{2} = ∫_{∂B_{σ}(x_{0})} B(y, σ, v, ∇v) − σ ∫_{∂B_{σ}(x_{0})}(\frac{αv^{2}}{2} − \frac{v^{6}}{6} )  (3.2)

where

B(y, σ, v, ∇v) = \frac{1}{2}v \frac{∂v}{∂ν} − \frac{σ}{2} |∇v|^{2} + σ(\frac{∂v}{∂ν})^{2},  (3.3)

ν denotes the exterior unit normal on ∂B_{σ}(x_{0}).

Proof. It is elementary to check (see for example [11]) that for any f, g ∈ C^{2}(B_{σ}(x_{0})), we have

2Δf(∇f · ∇g) = div[2(∇f · ∇g)∇f − |∇f|^{2}∇g] + |∇f|^{2}Δg − 2∇f · g′′ · ∇f, (3.4)

where g′′ = (\frac{∂^{2}g}{∂y_{i}∂y_{j}}), thus the last term denotes a quadratic form.

Let g = \frac{1}{2}|y − x_{0}|^{2}, then ∇g = y − x_{0}, Δg = 3, ∇f · g′′ · ∇f = |∇f|^{2}. Integrating both sides of (3.4) on B_{σ}(x_{0}) , we have

∫_{B_{σ}(x_{0})} 2Δf(∇f·(y−x_{0})) = ∫_{B_{σ}(x_{0})} div[2(∇f·(y−x_{0}))∇f−|∇f|^{2}(y−x_{0})]+∫_{B_{σ}(x_{0})} |∇f|^{2}.

(3.5)

Noticing that

(∇f · (y − x_{0}))\frac{∂f}{∂ν} = |\frac{∂f}{∂ν} |^{2}(y − x_{0}) · ν = σ|\frac{∂f}{∂ν} |^{2} on ∂B_{σ}(x_{0}),

we have

∫_{B_{σ}(x_{0})} div[2(∇f · (y − x_{0}))∇f − |∇f|^{2}(y − x_{0})] = ∫_{∂B_{σ}(x_{0})} 2σ|\frac{∂f}{∂ν} |^{2} − σ|∇f|^{2}.

Also

∫_{B_{σ}(x_{0})} |∇f|^{2} = − ∫_{B_{σ}(x_{0})} Δff + ∫_{∂B_{σ}(x_{0})} f \frac{∂f}{∂ν} .

Bringing these two identities into (3.5), we have

\begin{matrix}∫_{B_{σ}(x_{0})} 2Δf∇f · (y − x_{0}) + ∫_{B_{σ}(x_{0})} Δff \\ = ∫_{∂B_{σ}(x_{0})}\{2σ|\frac{∂f}{∂ν} |^{2} − σ|∇f|^{2} + f \frac{∂f}{∂ν} \} ∀ f ∈ C^{2}(B_{σ}(x_{0})). \end{matrix}

In particular

\begin{matrix}∫_{B_{σ}(x_{0})} 2Δv∇v · (y − x_{0}) + ∫_{B_{σ}(x_{0})} Δvv \\ = ∫_{∂B_{σ}(x_{0})}\{2σ|\frac{∂v}{∂ν} |^{2} − σ|∇v|^{2} + v \frac{∂v}{∂ν} \}. \end{matrix}  (3.6)

Now, using the fact that v(y) satisfies (3.1), we have

2 ∫_{B_{σ}(x_{0})}((y − x_{0})_{i}∂_{i}v)Δv
= −2 ∫_{B_{σ}(x_{0})}(y − x_{0})_{i}∂_{i}v)(v^{5} − αv)
= ∫_{B_{σ}(x_{0})} v^{6} − \frac{σ}{3} ∫_{∂B_{σ}(x_{0})} v^{6}
−3α ∫_{B_{σ}(x_{0})} v^{2} + ασ ∫_{∂B_{σ}(x_{0})} v^{2},

and

∫_{B_{σ}(x_{0})} vΔv = − ∫_{B_{σ}(x_{0})} v(v^{5} − αv)

where we denote (y − x_{0})_{i} as the i-th component of y − x_{0}. Bringing these two identities to (3.6), we have (3.2).

Proposition 3.2 Assume that v satisfies

\{ \begin{matrix}−Δv + αv = v^{5}, v ≥ 0, & in Ω \\ \frac{∂v}{∂ν} = 0 & on ∂Ω \end{matrix}  (3.7)

and x_{0} ∈ ∂Ω. Then

\begin{matrix}−α ∫_{Ω_{+}} v^{2} + \frac{α}{2} ∫_{∂Ω_{+}} v^{2}(y − x_{0}) · ν − \frac{1}{6} ∫_{∂Ω_{+}} v^{6}(y − x_{0}) · ν + \frac{1}{2} ∫_{Γ_{1}} |∇v|^{2}(y − x_{0}) · ν \\ = B_{1}(y, σ, v, ∇v) \end{matrix}

(3.8)

where

B_{1}(y, σ, v, ∇v) = ∫_{Γ_{2}}[σ(\frac{∂v}{∂ν})^{2} − \frac{σ}{2} |∇v|^{2} + \frac{1}{2}v \frac{∂v}{∂ν}],

Ω_{+} = Ω ∩ B_{σ}(x_{0}), Γ_{1} = ∂Ω ∩ B_{σ}(x_{0}), Γ_{2} = Ω ∩ ∂B_{σ}(x_{0}) and ν denotes the exterior unit normal to the corresponding boundary surface.

Proof. From (3.5) and integrating by part, we have

\begin{matrix}2 ∫_{Ω_{+}} Δv(∇v · (y − x_{0})) − ∫_{Ω_{+}} |∇v|^{2} \\ = ∫_{∂Ω_{+}}[2(∇v · (y − x_{0}))(∇v · ν) − |∇v|^{2}((y − x_{0}) · ν)]. \end{matrix}  (3.9)

Using (3.7), we have

L. H. S. of (3.9) = 2 ∫_{Ω_{+}}(αv − v^{5})∇v · (y − x_{0}) − ∫_{∂Ω_{+}} v · \frac{∂v}{∂ν} + ∫_{Ω_{+}}(αv − v^{5}) · v
= −2α ∫_{Ω_{+}} v^{2} + ∫_{∂Ω_{+}}[(αv^{2} − \frac{1}{3}v^{6})(y − x_{0}) · ν − v \frac{∂v}{∂ν}]

and

R. H. S. of (3.9) = ∫_{Γ_{2}}[2σ(\frac{∂v}{∂ν})^{2} − σ|∇v|^{2}] − ∫_{Γ_{1}} |∇v|^{2}(y − x_{0}) · ν.

Bringing these two identities to (3.9), we have the proposition.

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