ON A SINGULARLY PERTURBED NEUMANN

PROBLEM WITH THE CRITICAL EXPONENT

Nassif Ghoussoub Department of Mathematics

The University of British Columbia Vancouver, B.C. Canada V6T 1Z2

Changfeng Gui Department of Mathematics University of Connecticut

Storrs, CT 06269

Meijun Zhu Department of Mathematics The University of Oklahoma Norman, Oklahoma 73019

Abstract

We consider the following singularly perturbed Neumann problem with the critical exponent:

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

and show that as ϵ tends to zero, the peak of the solutions tends to the boundary of the domain provided that the energy of the solutions remains below a certain natural level. This phenomenon is very different from the subcritical case.

0 Introduction

Recently the following singularly perturbed Neumann problem has attracted the attention of many analysts:

\{ \begin{matrix}−ϵ^{2}Δu + u = u^{p}, u > 0, & in Ω \\ \frac{∂u}{∂ν} = 0 & on ∂Ω \end{matrix}  (0.1)

where 1 < p ≤ \frac{n+2}{n−2}, Ω ⊂ ℝ^{n} is a smooth bounded domain and n ≥ 3. Equation (0.1) arises from the Gierer-Meinhardt model in biological pattern formation. We refer to [13] for the background of this model and some history.

It is a common phenomena that the solutions to the above equation develop peaks when ϵ tends to zero. In the subcritical case (p < \frac{n+2}{n−2}), solutions to (0.1) with either boundary peaks or interior peaks have been constructed. See [15], [16], [5], [10], [17], [6] and the references there in. Solutions with mixed boundary and interior peaks have also been constructed in [7]. In the critical case (p = \frac{n+2}{n−2}), various existence results on solutions with boundary peaks have been proven in [14],

[4] and others. However, never has a solution with interior peaks been found in this case. One may wonder if (0.1) can ever have such a solution in the critical case.

In this paper, we will provide a negative answer to the above question. Now we assume that u_{ϵ} is a solution to (0.1) with p = \frac{n+2}{n−2}, i.e.,

\{ \begin{matrix}−ϵ^{2}Δu + u = u^{\frac{n+2}{n−2}} , u > 0, & in Ω \\ \frac{∂u}{∂ν} = 0 & on ∂Ω. \end{matrix}  (0.2)

We shall prove that the maximum point (the peak) will converge to the boundary of the domain Ω.

It is well-known that the solutions correspond to the critical points of the normalized energy functional

I_{ϵ}(u) = \frac{∫_{Ω} |∇u|^{2} + ϵ^{−2} ∫_{Ω} |u|^{2}}{(∫_{Ω} |u|^{\frac{2n}{n−2}} )^{\frac{n−2}{n}}} .

Let S be the best constant in the Sobolev inequality in ℝ^{n}, i.e.

S = inf_{u∈D^{1,2}(ℝ^{n})} \frac{∫_{ℝ^{n}} |∇u|^{2}}{(∫_{ℝ^{n}} |u|^{\frac{2n}{n−2}} )^{\frac{n−2}{n}}} .

Our main result can be stated as the following.

Theorem 0.1 Assume that Ω ⊂ ℝ^{n} is a smooth bounded domain and u_{ϵ} solves (0.2) with

I_{ϵ}(u_{ϵ}) < (\frac{3}{2})^{\frac{2}{n}} · S − δ_{∗}  (0.3)

for some fixed δ_{∗} > 0. If x_{ϵ} is a local maximum point of u_{ϵ} satisfying u_{ϵ}(x_{ϵ}) → ∞, then necessarily

dist(x_{ϵ}, ∂Ω) → 0 as ϵ → 0.

Let λ = ϵ^{−2} and normalize u_{ϵ} to u_{λ} = k_{λ}u_{ϵ} so that

∫_{Ω} u_{λ}^{2n/(n−2)} = 1.

Then u_{λ} satisfies

\{ \begin{matrix}−Δu_{λ} + λu_{λ} = ξ_{λ}uλ^{\frac{n+2}{n−2}} , u_{λ} > 0, & in Ω \\ \frac{∂u_{λ}}{∂ν} = 0 & on ∂Ω. \end{matrix}  (0.4)

Remark 0.1 Under condition (0.3) the solution can have only one peak as ϵ tends to zero. It would be interesting to check whether the theorem still holds under the weaker condition that the normalized energy ξ_{λ} is simply bounded above by a constant C.

Below we briefly describe our approach.

We suppose that there is an interior peak for the solution, i.e., x_{λ} → x_{0} ∈ Ω, where x_{λ} is a local maximum point and u_{λ}(x_{λ}) → ∞ as λ → ∞. The standard approach consists of showing that u_{λ}(x) is close to some canonical bubble: i.e., a function concentrated near x_{0}, thus u_{λ}(x) is sufficiently small when x is away from x_{0}. Then one may try to use a Pohozaev type identity to derive a contradiction. Unfortunately, due to the Neumann boundary condition, even under the quite strong energy constraint (0.3), it is hard to obtain the needed estimates by directly using energy-dependent estimates as in [9] or [11] via conformal map.

Our new strategy to prove Theorem 0.1 is to use properties of the following Green’s function:

\{ \begin{matrix}−ΔG + G = 0 & in ℝ^{n} \ \{0\} \\ G < C & ∀ x ∈ B_{1}(0)^{c}. \end{matrix}

To see why such a Green’s function comes into our analysis, we make the following scaling

w_{λ}(x) = λ^{− \frac{n−2}{4}} u_{λ}(λ^{− \frac{1}{2}} x + x_{λ}), ∀x ∈ Ω_{λ} := (Ω − x_{λ}) · λ^{\frac{1}{2}} .

Using energy estimates and (0.3) we can show that ξ_{λ} → S and λμ_{λ}^{2} → 0 as λ → ∞. Thus w_{λ} satisfies

−Δw_{λ} + w_{λ} = ξ_{λ}wλ^{\frac{n+2}{n−2}} , w_{λ} > 0, in Ω_{λ}
\frac{∂w_{λ}}{∂ν} = 0  on ∂Ω_{λ}

∫_{Ω_{λ}} wλ^{\frac{2n}{n−2}} = 1,

and w_{λ}(0) → ∞ as λ → ∞ (since λμ_{λ}^{2} → 0). Away from the origin, we know that w_{λ} carries small energy, thus w_{λ} is very small. Combining this with the maximum principle and Harnack inequality in any compact set K ⊂⊂ ℝ^{n} \ {0} we are able to show that w_{λ}(x)/ max_{∂B_{1}(0)} w_{λ}(x) → G in C_{loc}^{2}(ℝ^{n}\{0}). We will use the asymptotic behavior of G near the origin and a Pohozaev type identity to derive a contradiction.

1 Proof of Theorem 0.1

Theorem 0.1 will be proved by contradiction. We suppose that there is a sequence of solutions u_{λ} to (0.4) and a local maximum point x_{λ} of u_{λ} such that u_{λ}(x_{λ}) → ∞ and x_{λ} → x_{0} as λ → ∞ for some interior point x_{0} ∈ Ω.

1.1 A rescaling of the problem

First of all, we establish a relation between λ and maximum value of u_{λ}. We then use a suitable rescaling to reduce equation (0.4) into another one with bounded coefficients so that local elliptic estimates can be applied.

Since u_{λ} has an interior blow up point, using classification results about −Δu = u^{(n+2)/(n−2)}, one easily sees that there exists a δ_{1} > 0 such that

lim_{λ→∞} ξ_{λ} ≥ \frac{1}{2^{2/n}} S + δ_{1}.

The standard facts and 0.3 about the levels where Palais-Smale condition for functional

I_{λ}(u) = \frac{∫_{Ω} |∇u|^{2} + λ ∫_{Ω} u^{2}}{(∫_{Ω} u^{\frac{2n}{n−2}} )^{\frac{n−2}{n}}}

fails yield the following (see for example [8] Lemma 1.1).

Lemma 1.1

lim_{λ→∞} ξ_{λ} = S.

From Lemma 1.1 one can easily check that x_{λ} is the only blowup point of u_{λ}, thus x_{λ} is the maximum point of u_{λ}. Recall μ_{λ} = u_{λ}(x_{λ})^{− \frac{2}{n−2}} .

Lemma 1.2 For any λ ≥ 1,

λμ_{λ}^{2} ≤ (\frac{3}{2})^{2/n}S − δ_{∗}.

Proof. Let

v_{λ}(x) = μλ^{\frac{n−2}{2}} u_{λ}(μ_{λ}x + x_{λ}), ∀ x ∈ Ω_{λ} := \frac{Ω − x_{λ}}{μ_{λ}} .  (1.1)

Then v_{λ}(x) ≤ 1 and

λμ_{λ}^{2} ∫_{Ω_{λ}} v_{λ}^{2} = λ ∫_{Ω} u_{λ}^{2}, ∫_{Ω_{λ}} vλ^{\frac{2n}{n−2}} = ∫_{Ω} uλ^{\frac{2n}{n−2}} = 1.

It follows that

λμ_{λ}^{2} = λμ_{λ}^{2} ∫_{Ω_{λ}} vλ^{\frac{2n}{n−2}} ≤ λμ_{λ}^{2} ∫_{Ω_{λ}} v_{λ}^{2} = λ ∫_{Ω} u_{λ}^{2} ≤ (\frac{3}{2})^{2/n}S − δ_{∗}.

Lemma 1.3 There exists a constant C > 0 such that for any δ > 0,

lim_{λ→∞} inf ∫_{B_{δ}(x_{λ})∩Ω} uλ^{\frac{2n}{n−2}} ≥ C > 0.

Proof. Let v_{λ}(x) be given by (1.1), then v_{λ} satisfies

\{ \begin{matrix}−Δv_{λ} + λμ_{λ}^{2}v_{λ} = ξ_{λ}vλ^{\frac{n+2}{n−2}} , 0 < v_{λ} ≤ 1, & in Ω_{λ} \\ \frac{∂v_{λ}}{∂ν} = 0 & on ∂Ω_{λ}. \end{matrix}

Therefore, via standard elliptic estimates, we know that up to some subsequence, v_{λ} → v in C^{2}(B_{1}(0)) with v satisfying

\{ \begin{matrix}−Δv + av = Sv^{\frac{n+2}{n−2}} , 0 < v ≤ 1, & in B_{1}(0) \\ v(0) = 1 \end{matrix}

where a = lim sup_{λ→∞} λμ_{λ}^{2}. It follows that ∫_{B_{1}(0)} v^{2n/(n−2)} ≥ 2C > 0 for some constant C, thus ∫_{B_{1}(0)} v_{λ}^{2n/(n−2)} ≥ C > 0 for λ sufficiently large. Thus as λ → ∞

∫_{B_{δ}(x_{λ})∩Ω} uλ^{\frac{2n}{n−2}} ≥ ∫_{B_{1}(0)} vλ^{\frac{2n}{n−2}} ≥ C.

Lemma 1.4 For any δ > 0,

lim_{λ→∞} inf ∫_{B_{δ}(x_{λ})∩Ω} uλ^{\frac{2n}{n−2}} = 1.

Proof. We only need to show that the above lemma holds for any 0 < δ < dist(x_{0}, ∂Ω)/2. We prove it by contradiction. Suppose that for some 0 < δ < dist(x_{0}, ∂Ω)/2 and some subsequence (still denote it as λ) we have

lim_{λ→∞} ∫_{B_{δ}(x_{λ})} uλ^{\frac{2n}{n−2}} = ϑ < 1.

Let η(x) be some cutoff function

η(x) = \{ \begin{matrix}1, & x ∈ B_{δ/2}(x_{λ}) \\ 0, & x ∈ Ω \ B_{δ}(x_{λ}), \end{matrix}

satisfying |∇η| + |∇^{2}η| ≤ C(δ).

Multiplying equation (0.4) by η^{2}u_{λ}^{r} for some r > 1 and integrating by parts, we have

∫_{Ω} ∇u_{λ} · ∇(η^{2}u_{λ}^{r}) = ξ_{λ} ∫_{Ω} η^{2}u_{λ}^{\frac{n+2}{n−2} +r} − λ ∫_{Ω} u_{λ}^{1+r}η^{2}.

Direct calculation shows that

\begin{aligned}∫_{Ω} ∇u_{λ} · ∇(η^{2}u_{λ}^{r}) = & \frac{4r}{(r+1)^{2}} ∫_{Ω} |∇(uλ^{\frac{r+1}{2}} η)|^{2} + \frac{r−1}{(r+1)^{2}} ∫_{Ω} u_{λ}^{r+1}Δη^{2} \\ & −\frac{4r}{(r+1)^{2}} ∫_{Ω} u_{λ}^{r+1}|∇η|^{2}. \end{aligned}

Combining the above two identities, we have that

∫_{Ω} |∇(uλ^{\frac{r+1}{2}} η)|^{2} ≤ \frac{(r+1)^{2}}{4r} ∫_{Ω} ∇u_{λ} · ∇(u_{λ}^{r}η^{2}) − \frac{r−1}{4r} ∫_{Ω} u_{λ}^{r+1}Δη^{2} + ∫_{Ω} u_{λ}^{r+1}|∇η|^{2}
≤ \frac{ξ_{λ}(r+1)^{2}}{4r} ∫_{Ω} u_{λ}^{\frac{n+2}{n−2} +r}η^{2} − \frac{r−1}{4r} ∫_{Ω} u_{λ}^{r+1}Δη^{2} + ∫_{Ω} u_{λ}^{r+1}|∇η|^{2}.

We now choose r small so that r + 1 < \frac{2n}{n−2}. Since ∫_{Ω} u_{λ}^{2n/(n−2)} = 1, we have

∫_{Ω} |∇(uλ^{\frac{r+1}{2}} η)|^{2} ≤ \frac{ξ_{λ}(r + 1)^{2}}{4r} ∫_{Ω} u_{λ}^{\frac{n+2}{n−2} +r}η^{2} + C_{1}

for some constant C_{1} = C_{1}(δ, r, Ω).

Note that u_{λ}^{(r+1)/2}η has compact support in Ω. Using Hölder inequality and the sharp Sobolev inequality, we have

∫_{Ω} u_{λ}^{\frac{n+2}{n−2} +r}η^{2} = ∫_{B_{δ}(x_{λ})}(u_{λ}^{r+1}η^{2}) · uλ^{\frac{4}{n−2}}

≤ (∫_{B_{δ}(x_{λ})} uλ^{\frac{2n}{n−2}} )^{\frac{2}{n}} · (∫_{Ω}(u_{λ}^{r+1}η^{2})^{\frac{n}{n−2}} )^{\frac{n−2}{n}}

≤ ϑ^{\frac{2}{n}} S^{−1} · ∫_{Ω} |∇(uλ^{\frac{r+1}{2}} η)|^{2}.

Since ϑ < 1 and ξ_{λ} → S as λ → ∞, we can further choose r > 1 sufficiently close to 1 such that when λ is large enough

\frac{ξ_{λ}(r + 1)^{2}}{4r} · ϑ^{\frac{2}{n}} S_{−1} ≤ 1 − ϵ.

It follows that

∫_{Ω} |∇(uλ^{\frac{r+1}{2}} η)|^{2} ≤ C_{2}(ϵ, δ, r, Ω),

thus

∫_{B_{δ/2}(x_{λ})} u_{λ}^{\frac{r+1}{2} · \frac{2n}{n−2}} ≤ C_{3}(ϵ, δ, r, Ω)

for large λ. Since ∫_{B_{δ/2}(x_{λ})} u_{λ}^{2} ≤ ∫_{Ω} u_{λ}^{2} → 0 as λ → ∞, we conclude, via an interpolation inequality, that

∫_{B_{δ/2}(x_{λ})} uλ^{\frac{2n}{n−2}} → 0 as λ → ∞.

This contradicts Lemma 1.3.

Here are some easy consequences of the above lemma.

Corollary 1.1 For any x ≠ x_{0} ∈ \overline{Ω} and any δ < dist(x, x_{0})/2, we have

lim_{λ→∞} sup ∫_{B_{δ}(x)∩Ω} uλ^{\frac{2n}{n−2}} = 0.

Let ϱ = dist(x_{0}, ∂Ω). For the rest of the paper, we shall denote by D^{r} the set D^{r} = \{x ∈ \overline{D} | dist(x, ∂D) ≤ r\} where D is a bounded smooth domain.

Corollary 1.2

lim_{λ→∞} sup ∫_{∂Ω} uλ^{\frac{2(n−1)}{n−2}} = 0.

Proof. It follows from equation (0.4) and Corollary 1.1 that

lim sup_{λ→∞} ∫_{Ω^{\frac{ϱ}{2}}} |∇u_{λ}|^{2} + lim sup_{λ→∞} ∫_{Ω^{\frac{ϱ}{2}}} |u_{λ}|^{2} = 0.

The rest follows from the trace inequality.

The following general lemma should be a well-known. We include a proof for completeness.

Lemma 1.5 Assume that u satisfies (0.4). There exists then a constant E such that if ||u||_{L^{\frac{2n}{n−2}} (Ω^{ϱ/2})} + ||u||_{L^{\frac{2(n−1)}{n−2}} (∂Ω)} ≤ E, then

||u||_{L^{∞}(Ω^{\frac{ϱ}{4}} )} ≤ C(||u||_{L^{\frac{2n}{n−2}} (Ω^{ϱ/2})} + ||u||_{L^{\frac{2(n−1)}{n−2}} (∂Ω)})

for some positive constant C = C(E, sup_{λ} ξ_{λ}).

Proof. Without loss of generality, we assume ϱ = 4. Let P be any point on ∂Ω, η be some cutoff function with compact support in B_{2}(P). Multiplying (0.4) by u^{k}η^{2} for some k > 1 and integrating by parts, we obtain

\begin{aligned}∫_{Ω} ∇u · ∇(u^{k}η^{2}) & ≤ ξ_{λ} ∫_{Ω} u^{\frac{n+2}{n−2} +k}η^{2} \\ & ≤ C ∫_{Ω} u^{\frac{n+2}{n−2} +k}η^{2}. \end{aligned}

Direct calculations show that

\begin{aligned}∫_{Ω} ∇u · ∇(η^{2}u^{k}) = & \frac{4k}{(k+1)^{2}} ∫_{Ω} |∇(u^{\frac{k+1}{2}} η)|^{2} + \frac{k−1}{(k+1)^{2}} ∫_{Ω} u^{k+1}Δη^{2} \\ & −\frac{4k}{(k+1)^{2}} ∫_{Ω} u^{k+1}|∇η|^{2} + \frac{k−1}{(k+1)^{2}} ∫_{∂Ω} u^{k+1}∇η^{2} · ν. \end{aligned}

Combining the two identities above, we get that ∫_{Ω} |∇(u^{\frac{k+1}{2}} η)|^{2} ≤ \frac{(k+1)^{2}}{4k} ∫_{Ω} ∇u · ∇(u^{k}η^{2}) − \frac{k−1}{4k} ∫_{Ω} u^{k+1}Δη^{2} + ∫_{Ω} u^{k+1}|∇η|^{2} − \frac{k−1}{4k} ∫_{∂Ω} u^{k+1}∇η^{2} · ν ≤ C · \frac{(k+1)^{2}}{4k} ∫_{Ω} u^{\frac{n+2}{n−2} +k}η^{2} + \frac{k−1}{4k} ∫_{Ω} u^{k+1}|Δη^{2}| + ∫_{Ω} u^{k+1}|∇η|^{2} + \frac{k−1}{4k} ∫_{∂Ω} u^{k+1}|∇η^{2} · ν|.

(1.2)

Set

R_{i} = 1 + \frac{1}{2^{i−1}} , i = 1, 2, 3, · · · .

We choose some smooth cutoff function η_{i} satisfying

\{ \begin{matrix}η_{i}(x) = 1, |x − P | < R_{i+1}; η_{i}(x) = 0, |x − P | > R_{i}; \\ |∇η_{i}| ≤ C · 2^{i}, |∇^{2}η_{i}| ≤ C · 4^{i}. \end{matrix}

Taking η = η_{i} in (1.2) and using the Sobolev embedding theorem, we have

\begin{matrix}\{∫_{Ω}(u^{(k+1)/2}η_{i})^{\frac{2n}{n−2}} \}^{\frac{n−2}{n}} + \{∫_{∂Ω}(u^{(k+1)/2}η_{i})^{\frac{2(n−1)}{n−2}} \}^{\frac{n−2}{n−1}} \\ ≤ C · \frac{(k+1)^{2}}{4k} ∫_{Ω} u^{\frac{n+2}{n−2} +k}η_{i}^{2} + 4^{i}C ∫_{Ω∩B_{R_{i}}(P)} u^{k+1} + 4^{i}C ∫_{∂Ω∩B_{R_{i}}(P)} u^{k+1}. \end{matrix}  (1.3)

It can be then seen from the proof of Lemma 1.4 that there exists some constant E such that if ||u||_{L^{\frac{2n}{n−2}} (Ω^{ϱ/2})} + ||u||_{L^{\frac{2(n−1)}{n−2}} (∂Ω)} ≤ E, then

∫_{Ω^{ϱ/3}} u^{τ} + ∫_{∂Ω} u^{τ} ≤ C(E)  (1.4)

for some τ > \frac{2n}{n−2}.

Choose τ_{0} ∈ (\frac{2n}{n−2}, τ) close to \frac{2n}{n−2} and set r_{0} = \frac{(n−2)τ_{0}}{4} and β = \frac{n(r_{0}−1)}{(n−2)r_{0}}. We can check that β > 1. Also as τ_{0} is close to \frac{2n}{n−2}, β is close to 1. Therefore, we can make β < \frac{(n−1)}{n−2} after we choose a suitable τ_{0}. Thus

(∫_{∂Ω∩B_{R_{i+1}}(P)} u^{(k+1)β})^{\frac{1}{β}} ≤ C( ∫_{∂Ω∩B_{R_{i}}(P)}(u^{(k+1)/2}η_{i})^{\frac{2(n−1)}{n−2}} )^{\frac{n−2}{n−1}} .  (1.5)

Also by using Hölder’s inequality, we know that

∫_{Ω∩B_{R_{i}}(P)} u^{\frac{n+2}{n−2} +k} ≤ ( ∫_{Ω∩B_{R_{i}}(P)} u^{(k+1)r_{0}/(r_{0}−1)}dvg̃)^{(r_{0}−1)/r_{0}}(∫_{Ω∩B_{R_{i}}(P)} u^{τ_{0}})^{\frac{1}{r_{0}}} .

Together with (1.4) we obtain

∫_{Ω∩B_{R_{i}}(P)} u^{\frac{n+2}{n−2} +k} ≤ C( ∫_{Ω∩B_{R_{i}}(P)} u^{(k+1)r_{0}/(r_{0}−1)})^{(r_{0}−1)/r_{0}}.  (1.6)

Set p_{0} = 2r_{0}/(r_{0} − 1) < \frac{2n}{n−2}, p_{i} = βp_{i−1} = 2nβ^{i−1}/(n − 2), q_{i} = p_{i}(r_{0} − 1)/r_{0} = 2β^{i}, for i ≥ 1. Taking k = q_{i} − 1 (for i ≥ 1) in (1.3), and using (1.5) and (1.6), we obtain

\begin{aligned}& ∥u∥_{p_{i+1},Ω∩B_{R_{i+1}}(P)}^{q_{i}} + ∥u∥_{q_{i+1},∂Ω∩B_{R_{i+1}}(P)}^{q_{i}} \\ ≤ & (4^{i}C + \frac{q_{i}^{2}C}{q_{i}−1})(∥u∥_{p_{i},Ω∩B_{R_{i}}(P)}^{q_{i}} + ∥u∥_{q_{i},∂Ω∩B_{R_{i}}(P)}^{q_{i}}), \end{aligned}

Since β > 1, we have (a^{β} + b^{β})^{1/β} ≤ a + b. It follows that

\begin{aligned}& (∥u∥_{p_{i+1},Ω∩B_{R_{i+1}}(P)}^{q_{i+1}} + ∥u∥_{q̄_{i+1},∂Ω∩B_{R_{i+1}}(P)}^{q_{i+1}})^{1/q_{i+1}} \\ ≤ & (4^{i}C + \frac{q_{i}^{2}C}{q_{i}−1})^{1/q_{i}}(∥u∥_{p_{i},Ω∩B_{R_{i}}(P)}^{q_{i}} + ∥u∥_{q̄_{i},∂Ω∩B_{R_{i}}(P)}^{q_{i}})^{1/q_{i}}. \end{aligned}  (1.7)

Since

(C4^{i} + \frac{Cq_{i}^{2}}{q_{i} − 1})^{\frac{1}{q_{i}}} ≤ [C(4^{i} + 2β^{i})]^{\frac{1}{2β_{i}}} ≤ C^{\frac{1}{2β_{i}}} (4 + β)^{\frac{i}{2β_{i}}} ,

we know that

∏_{i=1}^{∞}(4^{i}C + \frac{q_{i}^{2}C}{q_{i} − 1})^{1/q_{i}} ≤ C < ∞,

It follows (since q_{1} = 2β < \frac{2(n−1)}{n−2}) that

∥u∥_{p_{i+1},Ω∩B_{R_{i+1}}(P)} ≤ C(∥u∥_{p_{1},Ω∩B_{1}(P)}^{2β} + ∥u∥_{q_{1},∂Ω∩B_{1}(P)}^{2β})^{1/(2β)}
≤ C(∥u∥_{\frac{2n}{n−2} ,Ω∩B_{1}(P)}^{2β} + ∥u∥_{\frac{2(n−1)}{n−2} ,∂Ω∩B_{1}(P)}^{2β})^{1/(2β)}
≤ C_{1} · (∥u∥_{\frac{2n}{n−2} ,Ω∩B_{1}(P)} + ∥u∥_{\frac{2(n−1)}{n−2} ,∂Ω∩B_{1}(P)}).

Sending i to ∞, we have

∥u∥_{L^{∞}(Ω^{1}∩B_{1}(P))} ≤ C_{1} · (∥u∥_{p,Ω∩B_{1}(P)} + ∥u∥_{q,∂Ω∩B_{1}(P)}).

Since P is arbitrarily chosen, we have thereby completed the proof of Lemma 1.5.

With the aide of Lemma 1.5 and by using Corollary 1.1-1.2 one can actually show the following.

Corollary 1.3

||u_{λ}||_{L^{∞}(Ω^{\frac{ϱ}{2}} )} → 0 as λ → ∞.

Let now η be a cutoff function such that η(x) = 1 for |x − x_{0}| ≤ ϱ/2, η(x) = 0 for |x − x_{0}| ≥ ϱ/2 + r and |∇η(x)| ≤ C/r for some small r > 0. It is easy to see that as λ → ∞ and r → 0,

∫_{Ω} |∇(u_{λ} · η)|^{2} ≤ ∫_{Ω} |∇u_{λ}|^{2} + o(1) ≤ ξ_{λ} + o(1)

where o(1) → 0 as λ → ∞ and r → 0.

On the other hand, from the sharp Sobolev inequality and Lemma 1.4 we know that

∫_{Ω} |∇(u_{λ} · η)|^{2} ≥ S(∫_{Ω} |u_{λ} · η|^{2n/n−2})^{(n−2)/n} = S + o(1)

where o(1) → 0 as λ → ∞ and r → 0. Notice that ξ_{λ} → S. We therefore conclude that as λ → ∞,

λ ∫_{Ω} u_{λ}^{2} → 0.  (1.8)

Finally, as in the proof of Lemma 1.2, we have

Lemma 1.6

lim_{λ→∞} λμ_{λ}^{2} = 0.

Denote β = λ^{−1/2}, v_{β}(x) = β^{\frac{n−2}{2}} u_{λ}(βx + x_{λ}) for x ∈ Ω_{β} = \frac{Ω−x_{λ}}{β} . Due to Lemma 1.6 we know that μ_{λ}/β → 0 as λ → ∞. It follows that v_{β}(0) → ∞ as β → 0 and v_{β} satisfies

\{ \begin{matrix}−Δv_{β} + v_{β} = ξ_{λ}vβ^{\frac{n+2}{n−2}} , v_{β} > 0, & in Ω_{β} \\ \frac{∂v_{β}}{∂ν} = 0 & on ∂Ω_{β} \end{matrix}  (1.9)

Moreover, we have

∫_{Ω_{β}} vβ^{\frac{2n}{n−2}} = 1.

Also, it is easy to see that Ω_{β} tends to ℝ^{n} as β → 0.

1.2 Global Analysis

Denote B_{1}(0)^{c} = ℝ^{n} \ B_{1}(0). Since v_{β}(0) → ∞ as β → 0, we can show as in Lemma 1.3 that ∫_{B_{1}(0)} v_{β}^{2n/(n−2)} > C > 0 for all β. Using (1.8) we have ∫_{Ω_{β}} v_{β}^{2} → 0 as β → 0. It follows as in the proof of Lemma 1.4 that ∫_{B_{1}(0)} v_{β}^{2n/(n−2)} → 1 as β → 0. Therefore

∫_{B_{1}(0)^{c}∩Ω_{β}} vβ^{\frac{2n}{n−2}} → 0  as β → 0.  (1.10)

By Corollary 1.3 we know that

||v_{β}||_{L^{∞}(Ω_{β}^{1})} → 0 as β → 0,

where Ω_{β}^{1} = {x ∈ Ω_{β} | dist(x, ∂Ω_{β}) < 1}. Also, similarly to the proof of Lemma 1.5, we know from (1.10) that as β → 0,

||v_{β}||_{L^{∞}(B_{1}^{c}(0)\Ω_{β}^{1})} → 0.

Therefore,

||v_{β}||_{L^{∞}(B^{c}_{1}(0))} → 0 as β → 0.  (1.11)

We now rewrite equation (1.9) as

−Δv_{β} + c_{β}v_{β} = 0 ∀x ∈ B_{1}(0)^{c} ∩ Ω_{β},

where c_{β} = c_{β}(x) = 1 − ξ_{λ}v_{β}^{4/(n−2)}(x). From (1.11) we know that there exists a constant β_{0} such that for β ≤ β_{0}, c_{β}(x) > 0 for all x ∈ B_{1}(0)^{c}. Since ∂v_{β}/∂ν = 0 on ∂Ω_{β}, we get from the maximum principle and Hopf lemma:

Lemma 1.7

max_{\overline{Ω_{β}}\B_{1}(0)} v_{β} = max_{∂B_{1}(0)} v_{β}, ∀β ≤ β_{0}.

Now we let

c_{β} = max_{∂B_{1}(0)} v_{β}(x)

and define

m_{β}(x) = \frac{v_{β}(x)}{c_{β}}, x ∈ Ω_{β}.

Since in the above discussion B_{1}(0) can be replaced by B_{ρ}(0) for any 0 < ρ < 1, we know that v_{β}(x) → 0 for x ≠ x_{0}. Then the combination of Harnack inequality and Lemma 1.7 yields that m_{β}(x) ≤ C(K) for any K ⊂⊂ ℝ^{n} \ {0} and m_{β}(x) ≤ 1 for |x| ≥ 1. Therefore m_{β}(x) → G(x) in C_{loc}^{2}(ℝ^{n} \ {0}) and G(x) satisfies

\{ \begin{matrix}−ΔG + G = 0 & in ℝ^{n} \ \{0\} \\ G < 1 & ∀x ∈ B_{1}(0)^{c}. \end{matrix}  (1.12)

Since there is no bounded function satisfying −Δu + u = 0 in ℝ^{n}, we know that G(x) has a singular point at the origin. Thus by the Harnack inequality G(x) → ∞ as x → 0.

Lemma 1.8 Let G be given by (1.12), then for any constant 0 < γ ≤ 1/4, there exists some constant C > 0 such that

G(x) ≤ Ce^{−γ|x|}, ∀|x| ≥ 2.

Proof. For any point x with |x| ≥ 2, denote A = B_{3|x|−1}(3x) \ B_{|x|}(3x). For γ > 0, we choose a test function ϕ(z) as

ϕ(z) = e^{γ(r−3|x|+1)} + e^{γ(|x|−r)}, ∀z ∈ A,

where r = |z − 3x|. One can check that

−Δϕ(z) + ϕ(z) = −e^{γ(r−3|x|+1)}(γ^{2} + \frac{2γ}{r} − 1) − e^{γ(|x|−r)}(γ^{2} − \frac{2γ}{r} − 1).

Notice for all z ∈ A, r ≥ |z| ≥ 2. Therefore if γ ≤ 1/4,

−Δϕ(z) + ϕ(z) ≥ 0, ∀z ∈ A.

Since G(z) ≤ ϕ(z) for z ∈ ∂A, it follows from the maximum principle that G(x) ≤ ϕ(x) = Ce^{−γ|x|}.

It is well known that G(x) = G(|x|) and

lim_{x→0} G(x)|x|^{n−2} = a > 0.

Moreover, when n ≥ 3 we have

G(x) = \frac{a}{|x|^{n−2}} + H(x)

with a > 0 and H(x) satisfying

ΔH ≥ 0 in ℝ^{n}.

Together with Lemma 1.8 we know that H(x) < 0 as |x| → ∞. It follows from the maximum principle that H(0) < 0. Therefore

G(x) = \frac{a}{|x|^{n−2}} − b + E(x) as |x| → 0  (1.13)

where a, b > 0, E(0) = 0 and for any ϵ ∈ (0, 1) there exists a constants C(ϵ) > 0 such that

|x|^{n−4+ϵ}|E(x)| + |x|^{n−3+ϵ}|∇E(x)| ≤ C(ϵ).

See for example, Appendix B in [11] for more details about the estimates on E(x).

From (1.9) we know that v_{β} satisfies the following Pohozaev identity (see for example, [18]):

− ∫_{B_{σ}(0)} v_{β}^{2} = ∫_{∂B_{σ}(0)} B(x, σ, v_{β}, ∇v_{β}) + σ ∫_{∂B_{σ}(0)}(\frac{(n − 2)ξ_{λ}vβ^{\frac{2n}{n−2}}}{2n} − \frac{v_{β}^{2}}{2} ) (1.14)

where

B(x, σ, v_{β}, ∇v_{β}) = \frac{n − 2}{2}v_{β} · \frac{∂v_{β}}{∂ν} − \frac{σ}{2} · |∇v_{β}|^{2} + σ(\frac{∂v_{β}}{∂ν})^{2}.

Now we divide both sides of (1.14) by c_{β}^{2} and send first β then σ to zero. We can get

lim_{β,σ→0} ∫_{∂B_{σ}(0)} \frac{B(x, σ, v_{β}, ∇v_{β})}{c_{β}^{2}} = lim_{σ→0} ∫_{∂B_{σ}(0)} B(x, σ, G, ∇G) = \frac{(n − 2)^{2}ab}{2} · |∂B_{1}|.

When n = 3, we have

lim_{β,σ→0} σ ∫_{∂B_{σ}(0)}(\frac{(n − 2)ξ_{λ}vβ^{\frac{2n}{n−2}}}{2n} − \frac{v_{β}^{2}}{2} )/c_{β}^{2} = lim_{σ→0}(−\frac{σ}{2} ) · ∫_{∂B_{σ}(0)} G^{2} = 0

and

lim_{β,σ→0} ∫_{B_{σ}(0)} v_{β}^{2}/c_{β}^{2} ≥ 0.

Therefore we conclude that:

\frac{ab}{2} · |∂B_{1}| ≤ 0,

which is a contradiction.

When n > 3, noticing a/(2|x|^{n−2}) ≤ G(x) ≤ 2a/|x|^{n−2} in B_{σ}(0) for small σ, we have

lim_{β→0} ∫_{∂B_{σ}(0)} \frac{v_{β}^{2}}{2c_{β}^{2}} = \frac{σ}{2} · ∫_{∂B_{σ}(0)} G^{2} ≤ \frac{4a^{2}|∂B_{1}|}{σ^{n−4}}  (1.15)

and for any 0 < τ < 1,

lim_{β→0} ∫_{B_{σ}(0)} \frac{v_{β}^{2}}{c_{β}^{2}} ≥ ∫_{B_{σ}(0)\B_{τσ}(0)} G^{2} ≥ \{ \begin{matrix}\frac{4a^{2}|∂B_{1}|}{4} · ln \frac{1}{τ}, & n = 4 \\ \frac{4a^{2}|∂B_{1}|}{4(n−4)σ^{n−4}} · (τ^{4−n} − 1), & n > 4. \end{matrix} (1.16)

Now we choose τ < (4n − 15)^{1/(4−n)} when n > 4 and τ < e^{−4} when n = 4. Then by (1.15)-(1.16), one can see that (1.14) does not hold when β → 0 and σ is sufficiently small. This is a contradiction. This completes the proof of Theorem 0.1.

Remark 1.1 The main difference between our analysis and that in [12] is due to the unique property of G(x), which allows us to establish our theorem for all dimension n ≥ 3. Some related topics can be found in [19].

ACKNOWLEDGMENT

This work was carried out when M.Zhu was a postdoctoral fellow at the University of British Columbia. He would like to thank the Department of Mathematics at UBC and the Pacific Institute for the Mathematical Sciences (PIMS) for their generous support. We thank the anonymous referee for his comments which lead

to the removal of the dimensional restriction in the early version.

References

[1] Caffarelli, L.; Gidas, B.; Spruck,J. Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth. Comm. Pure Appl. Math. 1989, 42 (3), 271-297.

[2] Gidas, B.; Ni, W.-M.; Nirenberg, L. Symmetry and related properties via the maximum principle. Comm. Math. Phys. 1979, 68 (3), 209-243.

[3] Gidas, B.; Spruck, J. Global and local behavior of positive solutions of nonlinear elliptic equations. Comm. Pure Appl. Math. 1981, 34 (4), 525-598.

[4] Gui, C.; Ghoussoub, N. Multi-peak solutions for a semilinear Neumann problem with critical exponent. Math. Z. 1998, 229 (3) 443-474.

[5] Gui, C. Multi-peak solutions for a semilinear Neumann problem. Duke Math. J. 1996, 84 (3), 739-769.

[6] Gui, C.; Wei, J. Multiple interior peak solutions for some singularly perturbed Neumann problems. J. Diff. Equ. 1999, 158 (1) 1-27.

[7] Gui, C.; Wei, J. Multiple mixed boundary and interior peak solutions for some singularly perturbed Neumann problems. Canadian Journal of Mathematics 2000 52 (3), 522-538.

[8] Grossi, M. A class of solutions for the Neumann problem −Δu + λu = u^{\frac{N+2}{N−2}} . Duke Math. J. 1995, 79 (2), 309-334.

[9] Han, Z.C. Asymptotic approach to singular solutions for nonlinear elliptic equations involving critical Sobolev exponent. Annales de l’Institut Henri Poincare, Analyse nonlineaire 1991, 8 (2), 159-174.

[10] Li, Y.Y. On a singularly perturbed equation with Neumann boundary condition. Comm. in PDEs. 1998, 23 (3-4), 487-545.

[11] Li, Y.Y.; Zhu, M. Sharp Sobolev trace inequality on Riemannian manifolds with boundary. Comm. Pure Appl. Math. 1997, 50 (5), 449-487.

[12] Li, Y.Y.; Zhu, M. Yamabe type equations in dimension three. Communications in Contemporary Math. 1999, 1 (1), 1-50.

[13] Ni, W.-M. Diffusion, cross-diffusion, and their spike-layer steady states. AMS Notes 1998, No. 45 (1), 9-18.

[14] Ni, W.-M.; Pan, X.; Takagi, I. Singular behavior of least-energy solutions of a semilinear Neumann problem involving critical Sobolev exponents. Duke Math. J., 1992, 67 (1), 1-20.

[15] Ni, W.-M.; Takagi, I. On the shape of least energy solution to a semilinear Neumann problem. Comm. Pure Appl. Math. 1991, 41 (7), 819-851.

[16] Ni, W.-M.; Takagi, I. Locating the peaks of least energy solutions to a semilinear Neumann problem. Duke Math. J. 1993, 70 (2), 247-281.

[17] Wei, J.; Winter, M. Multi-peak Solutions for a Wide Class of Singular Perturbation Problems. J. London Math. Soc. 1999, 56 (2), 585-606.

[18] Zhu, M. Uniqueness results through a priori estimates, Part I: a 3-dimensional Neumann problem, J. Differential Equations. 1999, 154 (2), 284-317.

[19] Zhu, M. Uniqueness results through a priori estimates, Part II: Dirichlet problem. J. Math. Anal. Appl. 2000, 248 (1), 156-172.