A SINGULARLY PERTURBED LINEAR EIGENVALUE PROBLEM IN C^{1} DOMAINS

Yuan Lou Department of Mathematics The Ohio state University Columbus, OH 43210

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

Abstract

Some non-existence result is established for bounded solutions of a Neumann problem on the upper half space. Based on this non-existence result, precise asymptotic behavior is given for the principal eigenvalue of some linear eigenvalue problem in bounded C^{1} domains, and this answers a question that appeared in [3].

1 Introduction

For any γ > 0, set

Λ(γ) = sup_{u∈H^{1}(Ω)\\{0\}} \frac{γ ∫_{∂Ω} u^{2} − ∫_{Ω} |∇u|^{2}}{∫_{Ω} u^{2}} ,  (1)

where Ω is a bounded domain in ℝ^{n} with boundary ∂Ω. It is straightforward to

show that the supremum of (1) is attained by some positive function u_{γ} ∈ H^{1}(Ω),

which is a weak solution of

Δu = Λ(γ)u in Ω,  \frac{∂u}{∂ν} = γu on ∂Ω,  (2)

where ν is the outward unit normal vector on ∂Ω; ν exists a.e. for Lipschitz domains. The goal of this paper is to understand the asymptotic behavior of Λ(γ) as γ → ∞ when ∂Ω ∈ C^{1}. Since Λ(γ) → ∞ when γ → ∞, (2) can be viewed as a singularly perturbed linear eigenvalue problem.

The asymptotic behavior of Λ(γ) was first studied by Lacey, Ockendon and Sabina in [3], where they investigated some reaction-diffusion model in which distributed nonlinear absorption mechanisms compete with nonlinear boundary sources. In order to describe the long time behaviors of solutions to this reaction-diffusion model, it is important to understand the asymptotic behavior of Λ(γ) as γ → ∞ (see [3] and the references therein). Among other things, Lacey, Ockendon and Sabina showed in [3] that

lim_{γ→∞} \frac{Λ(γ)}{γ^{2}} = 1  (3)

if ∂Ω is C^{2} and is differentially equivalent to the unit sphere. On the other hand, when Ω is a planar domain and ∂Ω is piecewise C^{1}, they proved that

lim inf_{γ→∞} \frac{Λ(γ)}{γ^{2}} ≥ cosec^{2}α ≥ 1,

where α is the smallest interior semiangle on ∂Ω. These considerations indicate that the asymptotic behavior of Λ(γ) is strongly affected by the smoothness of the boundary. In this connection, we prove

Theorem 1.1 (3) holds for any bounded C^{1} domain.

Remark 1.1 Similar result can be established for the problem

Δu = Λ(γ)u in Ω, \frac{∂u}{∂ν} = γb(x)u on ∂Ω.

More precisely, if b(x) ∈ C(∂Ω) is positive somewhere, then

lim_{r→∞} \frac{Λ(γ)}{γ^{2}} = max_{∂Ω}(b_{+})^{2}.

In the following we briefly sketch our approach. Since (2) is a singularly perturbed problem, it is natural to “blow up” u_{γ}, the solution of (2), near its maximum which must be attained on ∂Ω (via the Maximum Principle). That is, straightening out the boundary and rescaling u_{γ} suitably, by passing to the limit we are led to the following Neumann problem on the upper half space.

Δu = au in ℝ_{+}^{n},  \frac{∂u}{∂x_{n}} = −u on ∂ℝ_{+}^{n},  (4)

where a is the limit of Λ(γ)/γ^{2} (subject to a subsequence) as γ → ∞. By adequate choice of test function in (1), one can show that a ≥ 1. On the other hand, using some similar ideas as the sliding method developed in [1], we are able to show

Theorem 1.2 If a > 1, (4) has no bounded non-trivial solutions.

By some non-degeneracy result in Section 3, the solution of (4) obtained via the blowup process is indeed non-trivial. Hence, Theorem 1.2 ensures that a = 1, which in turn yields Theorem 1.1.

Remark 1.2 It turns out that Theorem 1.2 is sharp: for every a ≤ 1, (4) has bounded non-trivial solutions of the form w(x′)e^{−x_{n}}, where x = (x′, x_{n}) and w is a solution of

Δw = (a − 1)w in R^{n−1};  (5)

Theorem 1.2 also fails without the boundedness condition: (4) has positive (unbounded) solutions of the form w(x′)e^{−x_{n}}, where w is a positive solution of (5) for a > 1. We refer to [2] for the classification of positive solutions to (5).

The plan of this paper is as follows: Theorem 1.2 is established in Section 2. In Section 3, we first derive the relation between lim_{γ→∞} Λ(γ)/γ^{2} and (4), and then use it to complete the proof of Theorem 1.1.

2 The Proof of Theorem 1.2

We prove Theorem 1.2 in this section. Our idea is to construct some super-solution of (2), by employing some similar ideas as the sliding method of Berestycki, Caffarelli and Nirenberg (see, e.g., [1]).

Throughout this section, we assume that u(x) is a bounded solution to (4) and a > 0; Without loss of generality, we may assume that sup_{ℝ_{+}^{n}} u > 0.

Lemma 2.1 sup_{ℝ_{+}^{n}} u = sup_{∂ℝ_{+}^{n}} u.

Proof. Let {x^{j}}_{j=1}^{∞} be a sequence of points with u(x^{j}) → sup_{ℝ_{+}^{n}} u. Denote x_{n}^{j} the last component of x^{j}. We first show that x_{n}^{j} → 0 as j → ∞. If not, then there is a δ > 0, such that x_{n}^{j} > δ (after passing to some subsequence). We consider u^{j}(x) = u(x + x^{j}) for |x| ≤ δ. Notice that u^{j} is bounded in |x| ≤ δ. By standard elliptic estimates, we know that after passing to some subsequence, u^{j} → u_{0} in C^{2}(B_{δ/2}(0)), where u_{0} satisfies

Δu_{0} = au_{0} in B_{δ/2}(0),

and u_{0} assumes its positive maximum at the origin. This is clearly impossible.

Again, standard elliptic estimates yield that |∇u| ≤ C for some positive constant C. Therefore, sup_{ℝ_{+}^{n}} u = sup_{∂ℝ_{+}^{n}} u.

We normalize u(x) so that sup_{ℝ_{+}^{n}} u = 1. By Lemma 2.1,

sup_{∂ℝ_{+}^{n}} u = 1.  (6)

Define

Ω_{h} = {(x′, x_{n}) | x′ ∈ R^{n−1}, 0 < x_{n} < h},
Ω_{h,r} = {(x′, x_{n}) | |x′| < r, 0 < x_{n} < h},
Γ_{1,h,r} = ∂Ω_{h,r} ∩ {x_{n} > 0},
Γ_{2,h,r} = ∂Ω_{h,r} ∩ {x_{n} = 0}.

We first state a lemma concerning the sub- and super-solution method. Consider

−Δu = f(x, u) in Ω_{h,r},
u = g(x)  on Γ_{1,h,r},
\frac{∂u}{∂ν} = h(x, u)  on Γ_{2,h,r}.

(7)

where f(x, u), h(x, u) are Carathéodory functions, g(x) is continuous.

A function u ∈ H^{1}(Ω_{h,r}) is called a sub-solution to (7) if u ≤ g(x) on Γ_{1,h,r} and

∫_{Ω_{h,r}} [∇u∇η − f(x, u)η] − ∫_{Γ_{2,h,r}} h(x, u)η ≤ 0,

for all η ∈ C^{∞}(Ω_{h,r}), η ≥ 0, and η = 0 on Γ_{1,h,r}. Similarly u ∈ H^{1}(Ω_{h,r}) is called a super-solution to (7) if the above inequalities are reversed. We refer the proof of the following result to [5].

Lemma 2.2 Suppose that \overline{u}, \underline{u} ∈ H^{1}(Ω_{h,r}) are both bounded, \overline{u} ≥ \underline{u}, and they are super-solution, sub-solution of (7), respectively. Then there is a solution u ∈ H^{1}(Ω_{h,r}) to (7) such that \underline{u} ≤ u ≤ \overline{u} holds a.e. in Ω_{h,r}.

We now apply Lemma 2.2 to construct a super-solution of (4).

Lemma 2.3 For any fixed h, r > 0, there is a unique solution to

Δψ = aψ  in Ω_{h,r},
ψ = 1  on Γ_{1,h,r},
\frac{∂ψ}{∂ν} = 1  on Γ_{2,h,r}.

(8)

Proof. It is easy to check that \underline{u} = 0 is a sub-solution and \overline{u} = 1 + h − x_{n} is a super-solution to (8). The existence follows from Lemma 2.2, whereas the uniqueness follows from the Maximum Principle.

Lemma 2.4 Let u(x) be a solution to (4) and ψ_{h,r} be the unique solution to (8). Then ψ_{h,r} ≥ u in Ω_{h,r}.

Proof. Set w = ψ_{h,r} − u. Then w satisfies

Δw = aw  in Ω_{h,r},
w ≥ 0  on Γ_{1,h,r},
\frac{∂w}{∂ν} = 1 − u on Γ_{2,h,r}.

If w(x_{0}) = min_{\overline{Ω}_{h,r}} w(x) < 0, then x_{0} ∈ Γ_{2,h,r} \ \{|x′| = r\}. By the Hopf Boundary Lemma, we know that ∂w/∂ν(x_{0}) < 0, which contradicts

∂w/∂ν(x_{0}) = 1 − u(x_{0}) ≥ 0.

This proves Lemma 2.4.

Now consider

Δψ = aψ  in Ω_{h,r},
ψ = 1  on Γ_{1,h,r},
\frac{∂ψ}{∂ν} = 0  on Γ_{2,h,r}.

(9)

It is easy to see that \underline{u} = 0 and \overline{u} = 1 are sub- and super-solutions to (9), respectively. Thus there is a unique solution to (9). We denote it by ψ_{1,h,r}. Decompose ψ_{h,r} as

ψ_{h,r} = ψ_{1,h,r} + ψ_{2,h,r},

where ψ_{2,h,r} is the unique solution to

Δψ = aψ  in Ω_{h,r},
ψ = 0  on Γ_{1,h,r},
\frac{∂ψ}{∂ν} = 1  on Γ_{2,h,r}.

It again follows from the Maximum Principle that

0 < ψ_{1,h,r} < 1, 0 < ψ_{2,h,r} < 1 + h, ∀x ∈ Ω_{h,r}.

Furthermore, we can show Lemma 2.5

a) ψ_{1,h,r} is non-increasing in r. For any fixed h > 0, as r → ∞, ψ_{1,h,r} converges monotonically to ψ_{1,h}, where ψ_{1,h}(x) is a function of x_{n} alone and satisfies

Δψ = aψ  in Ω_{h},
ψ = 1  on {x_{n} = h},
\frac{∂ψ}{∂ν} = 0  on \{x_{n} = 0\}.

(10)

b) ψ_{2,h,r} is nondecreasing in r. For any fixed h > 0, as r → ∞, ψ_{2,h,r} converges monotonically to ψ_{2,h}, where ψ_{2,h}(x) is a function of x_{n} alone and satisfies

Δψ = aψ  in Ω_{h},
ψ = 0  on {x_{n} = h},
\frac{∂ψ}{∂ν} = 1  on \{x_{n} = 0\}.

(11)

Proof. We only give the proof of part (a) since part (b) can be established in the same spirit. For any r′ > r, set w = ψ_{1,h,r′} − ψ_{1,h,r} in Ω_{h,r}. Then w satisfies

Δw = aw  in Ω_{h,r},
w ≤ 0  on Γ_{1,h,r},
\frac{∂w}{∂ν} = 0  on Γ_{2,h,r}.

(12)

It follows from the Maximum Principle and the Hopf Boundary Lemma that w ≤ 0 in \overline{Ω}_{h,r}. This proves the monotonicity of ψ_{1,h,r}.

Therefore, for any fixed h > 0, as r → ∞, ψ_{1,h,r} monotonically converges to ψ_{1,h} in ℝ_{+}^{n} and ψ_{1,h} satisfies (10).

We still need to show that ψ_{1,h} is a function of x_{n} only. For any P ∈ ℝ^{n−1} and r′, r > 0 with r′ − r > |P |, consider the difference w_{1} = ψ_{1,h,r′}(x′ + P, x_{n}) − ψ_{1,h,r}(x′, x_{n}) in Ω_{h,r}. It is easy to see that w_{1} satisfies (12), thus w_{1} ≤ 0 in Ω_{h,r}. That is,

ψ_{1,h,r′}(x′ + P, x_{n}) ≤ ψ_{1,h,r}(x′, x_{n}), ∀x ∈ Ω_{h,r}, ∀P ∈ R^{n−1}.

Sending r, r′ → ∞, we have

ψ_{1,h}(x′ + P, x_{n}) ≤ ψ_{1,h}(x), ∀x ∈ Ω_{h}.

Hence ψ_{1,h}(x) = ψ_{1,h}(x_{n}).

Proof of Theorem 1.2

Denote ψ_{h} = ψ_{1,h} + ψ_{2,h}. Then by Lemmas 2.4 and 2.5, ψ_{h}(x) ≥ u(x) in Ω_{h} and ψ_{h} satisfies

\begin{matrix}ψ′′ − aψ = 0 in (0, h), \\ ψ′(0) = −1, ψ(h) = 1. \end{matrix}

Direct calculation shows that

ψ_{h}(x_{n}) = c_{1}e^{\sqrt{a}x_{n}} + c_{2}e^{−\sqrt{a}x_{n}},

where

c_{1} = \frac{1 − \frac{1}{\sqrt{a}}e^{−\sqrt{a}h}}{e^{\sqrt{a}h} + e^{−\sqrt{a}h}} ,

c_{2} = \frac{1 + \frac{1}{\sqrt{a}}e^{\sqrt{a}h}}{e^{\sqrt{a}h} + e^{−\sqrt{a}h}} .

Sending h → ∞, we have ψ_{h}(x_{n}) → \frac{1}{\sqrt{a}}e^{−\sqrt{a}x_{n}}. Thus by Lemma 2.4,

u(x) ≤ \frac{1}{\sqrt{a}}e^{−\sqrt{a}x_{n}}, ∀x ∈ ℝ_{+}^{n}.

It follows from (6) that

1 = sup_{∂ℝ_{+}^{n}} u(x) ≤ \frac{1}{\sqrt{a}},

i.e., a ≤ 1. This proves Theorem 1.2.

3 Asymptotic Behaviors of Eigenvalues

We prove Theorem 1.1 in this section. For every piecewise smooth domain Ω, it was proved in [3] that

\underline{lim}_{γ→∞} \frac{Λ(γ)}{γ^{2}} ≥ 1.

To prove Theorem 1.1, we need to show that when ∂Ω is C^{1},

\overline{lim}_{γ→∞} \frac{Λ(γ)}{γ^{2}} ≤ 1.  (13)

Proof of Theorem 1.1

For any γ > 1, let u_{γ} be a positive solution of (2) and u_{γ} attains its maximum at x_{γ}. By the Maximum Principle, we know that x_{γ} ∈ ∂Ω. After normalization we can assume that max\overline{Ω} u_{γ} = 1 and x_{γ} → 0 ∈ ∂Ω. Further, we can assume that there is a C^{1} function ϕ such that ∂Ω ∩ B_{2}(0) can be represented by x_{n} = ϕ(x′) for |x′| ≤ 2 with ϕ(0) = 0 and ∂ϕ/∂x_{i}(0) = 0 for i = 1, ..., n − 1.

For any η ∈ C_{0}^{∞}(B_{2}(0)), u_{γ} satisfies

∫_{Ω} ∇u_{γ} · ∇η + Λ(γ) ∫_{Ω} u_{γ}η − γ ∫_{∂Ω} u_{γ}η = 0.  (14)

Now we flatten ∂Ω near the origin. Let y = Φ(x) : Ω ∩ B_{2}(0) → Ω_{Φ} ≡ Φ(Ω ∩ B_{2}(0)), be such that

\begin{matrix}Φ_{i}(x) = x_{i}, i = 1, 2, ..., n − 1, \\ Φ_{n}(x) = x_{n} − ϕ(x′). \end{matrix}

Denote the inverse of y = Φ(x) by x = Ψ(y). Then (14) can be rewritten as ∑_{k,l=1}^{n} ∫_{Ω_{Φ}} \frac{∂u_{γ}}{∂y_{k}} \frac{∂η}{∂y_{l}} \frac{∂Φ_{k}}{∂x_{i}} (Ψ(y))\frac{∂Φ_{l}}{∂x_{i}} (Ψ(y))|DΨ|dy + Λ(γ) ∫_{Ω_{Φ}} u_{γ}η|DΨ|dy

−γ ∫_{∂Ω_{Φ}} u_{γ}η\sqrt{1 + |∇ϕ(y′)|^{2}}dy′ = 0,

where |DΨ| is the derterminant of DΨ. Notice that |∇ϕ| = o(1) as x′ → 0. Thus DΨ → I as |y| → 0, where I is the n × n identity matrix. We now consider two different cases.

Case 1:

\overline{lim}_{γ→∞} \frac{Λ(γ)}{γ^{2}} = a < +∞.

Without loss of generality, we may assume that

\frac{γ ∫_{∂Ω} u_{γ}^{2} − ∫_{Ω} |∇u_{γ}|^{2}}{γ^{2} ∫_{Ω} u_{γ}^{2}} = sup_{u∈H^{1}(Ω)\\{0\}} \frac{γ ∫_{∂Ω} u^{2} − ∫_{Ω} |∇u|^{2}}{γ^{2} ∫_{Ω} u^{2}} → a,

and u_{γ}(x_{γ}) = max_{Ω} u_{γ}(x) = 1, x_{γ} → 0. We let z = γ(y − y_{γ}), where y_{γ} = (x′_{γ}, 0), and set v_{γ}(z) = u_{γ}(y). Then for any R > 0 and η with compact support in B_{2R}, as γ becomes sufficiently large, v_{γ} satisfies

\begin{matrix}∑_{k,l=1}^{n} ∫_{B_{2R}^{+}} \frac{∂v_{γ}}{∂z_{k}} \frac{∂η}{∂z_{l}} \frac{∂Φ_{k}}{∂x_{i}} (Ψ(y_{γ} + \frac{z}{γ}))\frac{∂Φ_{l}}{∂x_{i}} (Ψ(y_{γ} + \frac{z}{γ}))|DΨ|(y_{γ} + \frac{z}{γ})dz \\ +\frac{Λ(γ)}{γ^{2}} ∫_{B_{2R}^{+}} v_{γ}η|DΨ|(y_{γ} + \frac{z}{γ})dz − ∫_{z_{n}=0} v_{γ}η\sqrt{1 + |∇ϕ(y′_{γ} + z′/γ)|^{2}}dz′ = 0. \end{matrix} (15)

Since for z ∈ B_{2R}^{+}, y_{γ} + z/γ → 0 as γ → ∞, we know that for sufficiently large γ,

( ∑_{i=1}^{n} \frac{∂Φ_{k}}{∂x_{i}} (Ψ(y_{γ} + \frac{z}{γ}))\frac{∂Φ_{l}}{∂x_{i}} (Ψ(y_{γ} + \frac{z}{γ})))_{kl} > \frac{I}{2}.

Let η_{R} be a cutoff function satisfying η_{R} = 1 in B_{R} with compact support in B_{2R} and |∇η_{R}| ≤ C. Choosing η = u · η_{R} in (15), we have,

∥∇v_{γ}∥_{L^{2}(B_{R})} ≤ C||v_{γ}||_{L^{∞}} = C.

Therefore, after passing to a subsequence, v_{γ} → v_{0} weakly in H_{loc}^{1}(ℝ_{+}^{n}) as γ → ∞, where v_{0} ∈ H_{loc}^{1}(ℝ_{+}^{n}) satisfies

\begin{matrix}Δv_{0} = av_{0} & in ℝ_{+}^{n}, \\ \frac{∂v_{0}}{∂ν} = v_{0} & on ∂ℝ_{+}^{n}, \\ 0 ≤ v_{0} ≤ 1 & in ℝ_{+}^{n}. \end{matrix}

To show that v_{0} is non-trivial, we claim that there is a constant C > 0 such that

1 = ||v_{γ}||_{L^{∞}(B_{1}^{+})} ≤ C(||v_{γ}||_{L^{2}(∂ℝ_{+}^{n}∩\{|x′|<2\})} + ||v_{γ}||_{L^{\frac{2n}{n−1}} (B_{2}^{+}(0))}).  (16)

Since the embeddings from W^{1,2}(B_{2}^{+}(0)) to L^{\frac{2n}{n−1}} (B_{2}^{+}(0)) and L^{2}(∂ℝ_{+}^{n}∩\{|x′| < 2\}) are both compact, we know that

1 ≤ C(||v_{0}||_{L^{2}(∂ℝ_{+}^{n}∩\{|x′|<1\})} + ||v_{0}||_{L^{\frac{2n}{n−1}} (B_{1}^{+}(0))}),

from which it follows that v_{0} ≠ 0. From Theorem 1.2 we see that a ≤ 1. Hence it suffices to establish (16).

Inequality (16) can be obtained via Moser iteration. Though it seems to be a standard result (see, e.g., [4]), we include a proof here for the completeness. Direct calculation shows that

\begin{aligned}∫_{B_{2}^{+}} ∇v_{γ} · ∇(v_{γ}^{k}ξ^{2}) = & \frac{4k}{(k+1)^{2}} ∫_{B_{2}^{+}} |∇(vγ^{\frac{k+1}{2}} ξ)|^{2} + \frac{k−1}{(k+1)^{2}} ∫_{B_{2}^{+}} v_{γ}^{k+1}Δξ^{2} \\ & −\frac{4k}{(k+1)^{2}} ∫_{B_{2}^{+}} v_{γ}^{k+1}|∇ξ|^{2} − \frac{k−1}{(k+1)^{2}} ∫_{\{x_{n}=0, |x′|≤2\}} v_{γ}^{k+1}∇ξ^{2} · ν. \end{aligned}

Choosing η = v_{γ}^{k}ξ^{2} in (15) with k > 1 and ξ having compact support in B_{2}^{+}, we have

\begin{matrix}∫_{B_{2}^{+}} |∇(vγ^{\frac{k+1}{2}} ξ)|^{2} \\ ≤ \frac{k−1}{4k} ∫_{B_{2}^{+}} v_{γ}^{k+1}Δξ^{2} + ∫_{B_{2}^{+}} v_{γ}^{k+1}|∇ξ|^{2} + \frac{k−1}{4k} ∫_{\{x_{n}=0, |x′|≤2\}} v_{γ}^{k+1}∇ξ^{2} · ν \\ +\frac{(k+1)^{2}C}{4k} ∫_{\{x_{n}=0, |x′|≤2\}} v_{γ}^{k+1}ξ + \frac{(k+1)^{2}C}{4k} ∫_{B_{2}^{+}} v_{γ}^{k+1}ξ. \end{matrix}  (17)

Let

r_{i} = 1 + \frac{1}{2^{i−1}} , i = 1, 2, ...,

and choose ξ_{i} satisfying

ξ_{i} = 1, |x| ≤ r_{i+1};
ξ_{i} = 0, |x| > r_{i};
|∇ξ_{i}| ≤ 2 · 2^{i}, |∇^{2}ξ_{i}| ≤ 4 · 4^{i}.

Replacing ξ by ξ_{i} in (17) and using the Sobolev inequality and the trace inequality, we have, for n ≥ 3,

\begin{matrix}( ∫_{B_{r_{i}}^{+}} (vγ^{\frac{k+1}{2}} ξ_{i})^{\frac{2n}{n−2}} )^{\frac{n−2}{n}} + ∫_{∂B_{r_{i}}^{+}∩\{x_{n}=0\}} (vγ^{\frac{k+1}{2}} ξ_{i})^{\frac{2(n−1)}{n−2}} \\ ≤ C · (4^{i} + \frac{(k+1)^{2}}{4k}) ∫_{B_{r_{i}}^{+}} v_{γ}^{k+1} + C · (4^{i} + \frac{(k+1)^{2}}{4k}) ∫_{∂B_{r_{i}}^{+}∩\{x_{n}=0\}} v_{γ}^{k+1}, \end{matrix}

where C is some universal constant. By

∫_{B_{r_{i}}^{+}} v_{γ}^{k+1} ≤ C( ∫_{B_{r_{i}}^{+}} v_{γ}^{(k+1)· \frac{n}{n−1}} )^{\frac{n−1}{n}} ,

we arrive at

\begin{matrix}( ∫_{B_{r_{i}}^{+}} (vγ^{\frac{k+1}{2}} ξ_{i})^{\frac{2n}{n−2}} )^{\frac{n−2}{n}} + ∫_{∂B_{r_{i}}^{+}∩\{x_{n}=0\}} (vγ^{\frac{k+1}{2}} ξ_{i})^{\frac{2(n−1)}{n−2}} \\ ≤ C · (4^{i} + \frac{(k+1)^{2}}{4k})( ∫_{B_{r_{i}}^{+}} v_{γ}^{(k+1)· \frac{n}{n−1}} )^{\frac{n−1}{n}} + C · (4^{i} + \frac{(k+1)^{2}}{4k}) ∫_{∂B_{r_{i}}^{+}∩\{x_{n}=0\}} v_{γ}^{k+1}. \end{matrix} (18)

Define β = (n−1)/(n−2), q_{0} = 2, q_{i+1} = βq_{i} and p_{i} = nq_{i}/(n−1) for i = 0, 1, ...; choosing k = q_{i} − 1 in (18), we have

\begin{matrix}||v_{γ}||_{L^{p_{i+1}}(B_{r_{i+1}}^{+})}^{q_{i}} + ||v_{γ}||_{L^{q_{i+1}}(Γ_{i+1})}^{q_{i}} \\ ≤ C · (4^{i} + \frac{q_{i}^{2}}{q_{i}−1}) · (||v_{γ}||_{L^{p_{i}}(B_{r_{i}}^{+})}^{q_{i}} + ||v_{γ}||_{L^{q_{i}}(Γ_{i})}^{q_{i}}), \end{matrix}

where Γ_{i} = ∂B_{r_{i}}^{+} ∩ {x_{n} = 0} for i = 0, 1, ...; since β > 1, (a^{β} + b^{β})^{1/β} ≤ a + b. It follows that

\begin{matrix}(||v_{γ}||_{L^{p_{i+1}}(B_{r_{i+1}}^{+})}^{q_{i+1}} + ||v_{γ}||_{L^{q_{i+1}}(Γ_{i+1})}^{q_{i+1}})^{1/q_{i+1}} \\ ≤ (C · (4^{i} + \frac{q_{i}^{2}}{q_{i}−1}))^{1/q_{i}}(||v_{γ}||_{L^{p_{i}}(B_{r_{i}}^{+})}^{q_{i}} + ||v_{γ}||_{L^{q_{i}}(Γ_{i})}^{q_{i}})^{1/q_{i}}. \end{matrix}  (19)

Since q_{i} = 2β^{i}, it is easy to see that

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

Thus

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

It follows that

||v_{γ}||_{L^{p_{i+1}}(B_{r_{i+1}}^{+})} ≤ C(||v_{γ}||_{L^{2}(∂ℝ_{+}^{n}∩\{|x′|<2\})} + ||v_{γ}||_{L^{\frac{2n}{n−1}} (B_{2}^{+}(0))}).

Sending i → ∞, we obtain (16). For n = 2, we can obtain (16) in the same spirit. We thereby complete the proof of Theorem 1.1 in Case 1.

Case 2:

\overline{lim}_{γ→∞} \frac{Λ(γ)}{γ^{2}} = ∞.

We will also rule out this possibility. Let u_{γ} be the sequence of positive functions such that

\frac{γ ∫_{∂Ω} u_{γ}^{2} − ∫_{Ω} |∇u_{γ}|^{2}}{γ^{2} ∫_{Ω} u_{γ}^{2}} = sup_{u∈H^{1}(Ω)\\{0\}} \frac{γ ∫_{∂Ω} u^{2} − ∫_{Ω} |∇u|^{2}}{γ^{2} ∫_{Ω} u^{2}} = a(γ) → ∞

as γ → ∞, and u_{γ}(x_{γ}) = max\overline{Ω} u(x) = 1. Define z = \sqrt{a(γ)}γ(y − y_{γ}) and v_{γ}(z) = u_{γ}(y). Then for any R > 0 and η with compact support in B_{2R}^{+}, as γ becomes sufficiently large, v_{γ} satisfies

\begin{matrix}∑_{k,l=1}^{n} ∫_{B_{2R}^{+}} \frac{∂v_{γ}}{∂z_{k}} \frac{∂η}{∂z_{l}} \frac{∂Φ_{k}}{∂x_{i}} (Ψ(y_{γ} + \frac{z}{γ}))\frac{∂Φ_{l}}{∂x_{i}} (Ψ(y_{γ} + \frac{z}{γ}))|DΨ|(y_{γ} + \frac{z}{γ})dz \\ +\frac{Λ(γ)}{a(γ)γ^{2}} ∫_{B_{2R}^{+}} v_{γ}η|DΨ|(y_{γ} + \frac{z}{γ})dz − \frac{1}{\sqrt{a(γ)}} ∫_{z_{n}=0} v_{γ}η\sqrt{1 + |∇ϕ(yγ^{′} + z′/γ)|^{2}}dz′ = 0. \end{matrix}

Similarly as in Case 1, we can show that v_{γ} → v_{0} weakly in H_{loc}^{1}(B_{R}^{+}), where 0 ≤ v_{0} ≤ 1, v_{0} ≠ 0, and v_{0} ∈ H_{loc}^{1}(ℝ_{+}^{n}) is a weak solution of

\begin{matrix}Δv_{0} = v_{0} & in ℝ_{+}^{n}, \\ \frac{∂v_{0}}{∂ν} = 0 & on ∂ℝ_{+}^{n}. \end{matrix}  (20)

Using even reflection, from (20) we know that there is a positive bounded function satisfying

Δv_{0} = v_{0} in ℝ^{n},

which does not exist. This finishes the proof of Theorem 1.1.

References

[1] Berestycki, H., Caffarelli, L. and Nirenberg, L., Symmetry for elliptic equations in a half space, in “Boundary value problems for partial differential equations and applications”, volume dedicated to E. Magenes, J.L. Lions et al., ed., Masson, Paris(1993), 27-42.

[2] Caffarelli, L. and Littman, W., Representation formulas for solutions to Δu − u = 0 in R^{n}, Studies in partial differential equations, 249-263, MAA Stud. Math., 23, Math. Assoc. America, Washington, D.C., 1982.

[3] Lacey, A., Ockendon, J. and Sabina, J., Multidimensional reaction diffusion equations with nonlinear boundary conditions, SIAM J. Appl. math. 58(1998), no. 5, 1622-1647.

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

[5] Struwe, M., Variational Methods: applications to nonlinear partial differential equations and Hamiltonian systems, Springer-Verlag, 2000.