# Asymptotic behavior of the principal eigenvalue of a linear second order
elliptic operator with large advection and general boundary conditions

Rui Peng† ,  Guanghui Zhang§,∗ and  Maolin Zhou‡

###### Abstract.

Consider the eigenvalue problem of a linear second order elliptic operator:

$-D\Delta\varphi-2\alpha\nabla
m(x)\cdot\nabla\varphi+V(x)\varphi=\lambda\varphi\ \ \hbox{ in }\Omega,$

complemented by the Dirichlet boundary condition or the following general
Robin boundary condition:

$\frac{\partial\varphi}{\partial n}+\beta(x)\varphi=0\ \ \hbox{ on
}\partial\Omega,$

where $\Omega\subset\mathbb{R}^{N}(N\geq 1)$ is a bounded smooth domain,
$n(x)$ is the unit exterior normal to $\partial\Omega$ at
$x\in\partial\Omega$, $D>0$ and $\alpha>0$ are, respectively, the diffusion
and advection coefficients, $m\in C^{2}(\overline{\Omega}),\,V\in
C(\overline{\Omega})$, $\beta\in C(\partial\Omega)$ are given functions, and
$\beta$ allows to be positive, sign-changing or negative.

In [41], the asymptotic behavior of the principal eigenvalue of the above
eigenvalue problem as $D\to 0$ or $D\to\infty$ was studied. In this paper,
when $N\geq 2$, under proper conditions on the advection function $m$, we
establish the asymptotic behavior of the principal eigenvalue as
$\alpha\to\infty$, and when $N=1$, we obtain a complete characterization for
such asymptotic behavior provided $m^{\prime}$ changes sign at most finitely
many times. Our results complement or improve those in [4, 7, 44] and also
partially answer some questions raised in [4].

###### Key words and phrases:

Second order elliptic operator; principal eigenvalue; large advection;
asymptotics.

###### 2010 Mathematics Subject Classification:

35P15, 35J20, 35P20.

†School of Mathematical Sciences, Zhejiang Normal University, Jinhua,
Zhejiang, 321004, China. (Email: pengrui $\b{}$<EMAIL_ADDRESS>
§School of Mathematics and Statistics, Huazhong University of Science and
Technology, Wuhan, 430074, China. (Email<EMAIL_ADDRESS>
∗Hubei Key Laboratory of Engineering Modeling and Scientific Computing,
Huazhong University of Science and Technology, Wuhan 430074, China.

‡Chern Institute of Mathematics and LPMC, Nankai University, Tianjin, 300071,
China. (Email<EMAIL_ADDRESS>
M. Zhou was supported by the National Key Research and Development Program of
China (2021YFA1002400), Nankai Zhide Foundation and NSF of China (No.
12271437, 11971498). R. Peng was supported by NSF of China (No. 12271486,
12171176). G. Zhang was supported by NSF of China (No. 12171176, 11971187).

##  1\. Introduction

In this paper, we continue our work on the following eigenvalue problem of a
linear second order elliptic operator with Robin boundary condition:

$\left\\{\begin{aligned} &-D\Delta\varphi-2\alpha\nabla
m(x)\cdot\nabla\varphi+V(x)\varphi=\lambda\varphi&&\text{in }\Omega,\\\
&\frac{\partial\varphi}{\partial n}+\beta(x)\varphi=0&&\text{on
}\partial\Omega,\end{aligned}\right.$ (1.1)

where $\Omega\subset\mathbb{R}^{N}(N\geq 1)$ is a bounded smooth domain,
$n(x)$ is the unit exterior normal to $\partial\Omega$ at
$x\in\partial\Omega$, $V\in C(\overline{\Omega})$, $\beta\in
C(\partial\Omega)$, the advection function $m\in C^{2}(\overline{\Omega})$,
and $D>0$ and $\alpha>0$ stand for, respectively, the diffusion and advection
coefficients. It is worth mentioning that the function $\beta$ on the boundary
condition allows to be positive, sign-changing or negative in our study.

We are also concerned with the eigenvalue problem with Dirichlet boundary
condition:

$\left\\{\begin{aligned} &-D\Delta\varphi-2\alpha\nabla
m(x)\cdot\nabla\varphi+V(x)\varphi=\lambda\varphi&&\text{in }\Omega,\\\
&\varphi=0&&\text{on }\partial\Omega\end{aligned}\right.$ (1.2)

with $m,\,V$ and $\Omega$ fulfilling the same smoothness assumptions as
before.

Given $D>0,\,\alpha,\,m$ and $V$, it is well known that (1.1) (and (1.2))
admits a smallest eigenvalue (also called as principal eigenvalue), denoted by
$\lambda_{1}(\alpha)$, which corresponds to a positive eigenfunction (called
as principal eigenfunction). The existence and uniqueness of the principal
eigenvalue of the eigenvalue problems (1.1) and (1.2) and its variational
characterization to be mentioned later are standard facts; see, for example,
[16, 23, 44]. The principal eigenvalue is a basic concept in the area of
partial differential equations, and it has found many important applications
in various kinds of nonlinear modelling problems arising from biology,
chemical reactions, material sciences, etc; one may refer to [1, 6, 7, 9, 10,
11, 16, 17, 18, 19, 25, 31, 32, 33, 38, 37, 42, 43, 46, 47] and the references
therein.

In [41], we investigated the asymptotic behavior of the principal eigenvalue
$\lambda_{1}(\alpha)$ of (1.1) and (1.2) as the diffusion coefficient $D$ goes
to $0$ or $\infty$. In this paper, we aim to explore the asymptotic behavior
of $\lambda_{1}(\alpha)$ as the advection coefficient $\alpha$ tends to
$\infty$. Let us first recall some existing research works on the asymptotic
behavior of the principal eigenvalue with respect to large advection closely
related to (1.1) and (1.2).

In [4], Berestycki, Hamel and Nadirashvili considered the following eigenvalue
problem subject to Neumnn boundary condition:

$\left\\{\begin{aligned} &-D\Delta\varphi-2\alpha
v(x)\cdot\nabla\varphi+V(x)\varphi=\lambda\varphi&&\text{in }\Omega,\\\
&\frac{\partial\varphi}{\partial n}=0&&\text{on
}\partial\Omega\end{aligned}\right.$ (1.3)

as well as the one subject to Dirichlet boundary condition:

$\left\\{\begin{aligned} &-D\Delta\varphi-2\alpha
v(x)\cdot\nabla\varphi+V(x)\varphi=\lambda\varphi&&\text{in }\Omega,\\\
&\varphi=0&&\text{on }\partial\Omega,\end{aligned}\right.$ (1.4)

where $v\in L^{\infty}(\Omega)$ is a divergence-free vector field; that is,

${\rm div}\,v=0\ \ \,\mbox{ in}\ \mathcal{D}^{\prime}(\Omega).$ (1.5)

Actually, [4] dealt with a bit more general elliptic operator and also the
associated space periodic eigenvalue problem. Note that if $v\in
C^{1}(\overline{\Omega})$ is a gradient field (i.e, $v=\nabla m$), then (1.3)
and (1.4) reduce, respectively, to (1.1) with $\beta=0$ and (1.2).

We now recall the following definition introduced in [4].

######  Definition 1.

(i) A function $u$ is said to be a first integral of the vector field $v$ if
$u\in H^{1}(\Omega)$ and $v\cdot\nabla u=0$ almost everywhere in $\Omega$.

(ii) Denote $\mathcal{I}$ (resp. $\mathcal{I}_{0}$) to be the set of all first
integrals of $v$ which belong to $H^{1}(\Omega)$ (resp. $H^{1}_{0}(\Omega)$).

Berestycki, Hamel and Nadirashvili derived the following important result for
both (1.3) and (1.4); see [4, Theorems 2.1 and 2.2].

######  Theorem 1.1.

Assume that (1.5) holds. The following assertions hold.

(i) Let $\lambda_{1}(\alpha)$ be the principal eigenvalue of (1.3) and assume
that $v\cdot n=0$ in $L_{loc}^{1}(\partial\Omega)$. Then $\lambda_{1}(\alpha)$
is bounded and

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\min_{u\in\mathcal{I}}\frac{\int_{\Omega}{D|\nabla
u|^{2}+V(x)u^{2}}}{\int_{\Omega}u^{2}}.$

(ii) Let $\lambda_{1}(\alpha)$ be the principal eigenvalue of (1.4). Then
$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ if and only if $v$ has no
first integral in $H^{1}_{0}(\Omega)$, and if $v$ has a first integral in
$H^{1}_{0}(\Omega)$, then

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\min_{u\in\mathcal{I}_{0}}\frac{\int_{\Omega}{D|\nabla
u|^{2}+V(x)u^{2}}}{\int_{\Omega}u^{2}}.$

As mentioned in [4, Reamrk 2.5], in Theorem 1.1(i), the extra assumption that
$v\cdot n=0$ in $L_{loc}^{1}(\partial\Omega)$ is necessary in order to
determine the limit $\lim_{\alpha\to\infty}\lambda_{1}(\alpha)$ of problem
(1.3). On the other hand, for problem (1.4), Devinatz, Ellis and Friedman
[13], Friedman [24] and Wentzell [45] derived several nice results on the
limiting behavior (even the asymptotic rate) of the principal eigenvalue as
$\alpha\to\infty$; see [13, 24, 45] for more details. Also pointed out by [4,
Reamrk 1.1], if the vector field $v$ is not divergence-free, Theorem 1.1(ii)
does not hold; indeed, [24] showed that if $V=0$, $v$ is continuous on
$\overline{\Omega}$ and $v\cdot n<0$ on $\partial\Omega$, then
$\lambda_{1}(\alpha)=O(\alpha e^{-c_{0}\alpha})$ for some $c_{0}>0$ as
$\alpha\to\infty$. Observe that Theorem 1.1(ii) claims that
$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)$ is $+\infty$ or a positive
constant when $V=0$ and $v$ is divergence-free.

Recently, in [7], as $\alpha\to\infty$, Chen and Lou studied the asymptotic
behavior of $\lambda_{1}(\alpha)$ of the following eigenvalue problem with a
gradient field and Neumann boundary condition (i.e., $v=\nabla m$ in (1.3)):

$\left\\{\begin{aligned} &-\Delta\varphi-2\alpha\nabla
m(x)\cdot\nabla\varphi+V(x)\varphi=\lambda\varphi&&\hbox{ in }\Omega,\\\
&\frac{\partial\varphi}{\partial n}=0&&\hbox{ on
}\partial\Omega.\end{aligned}\right.$ (1.6)

In the following, we shall introduce the main result in [7]. To do so, we need
to recall the definitions of critical points and local maxima for the
advection function $m$ used in [7].

######  Definition 2.

Assume that $m\in C^{2}(\overline{\Omega})$.

(i) A point $x\in\Omega$ is called an interior critical point of $m$ if
$|\nabla m(x)|=0$. An interior critical point $x$ is called non-degenerate if
$\text{det}(D^{2}m(x))\neq 0$.

(ii) A point $x\in\partial\Omega$ is called a boundary critical point of $m$
if $|\nabla m(x)|=|n(x)\cdot\nabla m(x)|$. A boundary critical point $x$ is
called non-degenerate if either $|\nabla m(x)|=0$,
$\text{det}\left(D^{2}m(x)\right)\neq 0$ or $|\nabla m(x)|\neq
0,\text{det}\left(D^{2}_{\partial\Omega}m(x)\right)\neq 0$.

(iii) A point $x\in\overline{\Omega}$ is call a local maximum of $m$ if
$m(x)\geq m(y)$ for every $y$ in a small neighborhood of $x$, and there exists
some sequence $\\{r_{j}\\}$ of positive numbers such that
$\lim_{j\to\infty}r_{j}=0$ and

$m(x)>\max_{y\in\overline{\Omega}\cap\partial B(x,r_{j})}m(y),\ \ \forall
j\in\mathbb{N}.$

From now on, let us denote

$\Sigma=\\{x\in\overline{\Omega}:\ x\ \,\mbox{is a local maximum of}\ m\\}.$

Chen and Lou established the following important result; see [7, Theorems
1.1].

######  Theorem 1.2.

Assume that all critical points of $m$ are non-degenerate. Let
$\lambda_{1}(\alpha)$ be the principal eigenvalue of (1.6). Then

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\min_{x\in\Sigma}V(x).$

We note that since all critical points of $m$ are non-degenerate, the set
$\Sigma$ consists of finitely many isolated points. We further point out that
in the setting of Theorem 1.2, as $m\in C^{2}(\overline{\Omega})$, (1.5)
becomes $\Delta m=0$ on $\overline{\Omega}$, and the additional assumption
that $v\cdot n=0$ in $L_{loc}^{1}(\partial\Omega)$ imposed in Theorem 1.1(i)
becomes $\frac{\partial m}{\partial n}=0$ on $\partial\Omega$. Thus, $m$ must
be a constant and the advection term disappears in (1.6). This shows that
Theorem 1.1(i) does not apply to problem (1.6) once $m\in
C^{2}(\overline{\Omega})$. However, Theorem 1.1(i) and (ii) can allow the
vector field $v$ to be degenerate so that it has a first integral; for
instance, if $v$ vanishes on an open subset $\Omega_{0}\subset\subset\Omega$,
clearly any nonzero function $u\in H_{0}^{1}(\Omega_{0})$, extended by $0$ in
$\Omega\backslash{\Omega_{0}}$, is a first integral of $v$.

It is also worth mentioning that Theorem 1.1 had been applied in [4] to
investigate the asymptotic behavior of the speeds of propagation of pulsating
travelling fronts, and [7] applied Theorem 1.2 to reveal new coexistence
phenomena of a Lotka-Volterra reaction-diffusion model for two competing
species, both with respect to the large advection coefficient.

More recently, in one space dimension (i.e., $\Omega$ is a finite open
interval), when the advection term $m$ admits nature kinds of degeneracy, the
first-named and third-named authors [44] studied an eigenvalue problem with a
general boundary condition (see problem (3.1) in Section 3). A rather
satisfactory understanding of the limiting behavior of the principle
eigenvalue and its eigenfunction was obtained in [44]. As far as time-periodic
parabolic eigenvalue problems are concerned, in the past decades there have
been several research works devoted to the study of the asymptotic behavior of
the principal eigenvalue, especially with respect to small or large frequency,
diffusion coefficient or advection coefficient in spatially or temporally
heterogeneous and even degenerate environments; see [2, 12, 28, 29, 30, 33,
34, 35, 36, 40, 41, 42], to mention a few.

We are now in a position to state the main results obtained in this paper.
Before going further, let us define the following three subsets of the local
maxima set $\Sigma$ of $m$:

$\Sigma_{1}=\Sigma\cap\Omega,\ \ \Sigma_{2}=\\{x\in\Sigma\cap\partial\Omega:\
\ \beta(x)=0\\},\ \ \Sigma_{3}=\\{x\in\Sigma\cap\partial\Omega:\ \
\beta(x)<0\\}.$

For the eigenvalue problem (1.1), we need the following assumption.

(A1): Assume that for every $x_{0}\in\Sigma_{2}$, there exists a neighborhood
$U$ of $x_{0}$ such that $\beta(x)=0$ on $\partial\Omega\cap U$.

Concerning the eigenvalue problem (1.1) with nonnegative $\beta$, our result
reads as follows.

######  Theorem 1.

Assume that (A1) holds, $\beta\geq 0$ and all critical points of $m$ are non-
degenerate. Let $\lambda_{1}(\alpha)$ be the principal eigenvalue of (1.1).
Then the following assertions hold.

(i) If $\Sigma_{1}\cup\Sigma_{2}\neq\emptyset$, we have

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\min_{x\in\Sigma_{1}\cup\Sigma_{2}}V(x).$

(ii) If $\Sigma_{1}\cup\Sigma_{2}=\emptyset$, we have

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty.$

######  Remark 1.

Regarding Theorem 1, we would like to make some comments as follows.

Notice that if $\beta=0$ on $\partial\Omega$, then (A1) holds automatically,
and Theorem 1 reduces to Theorem 1.2 obtained by Chen and Lou.

In [4, Page 478], Berestycki, Hamel and Nadirashvili proposed several
interesting open questions, and one of them was to ask whether one can find a
necessary and sufficient condition for the boundedness of the principal
eigenvalue $\lambda_{1}(\alpha)$ with respect to large $\alpha$ for similar
elliptic operators with Robin boundary condition. Our Theorem 1 above and
Theorem 3 below make some progress towards such a question.

For the Dirichlet eigenvalue problem (1.2), we have the following result.

######  Theorem 2.

Assume that all critical points of $m$ are non-degenerate. Let
$\lambda_{1}(\alpha)$ be the principal eigenvalue of (1.2). Then the following
assertions hold.

(i) If $\Sigma_{1}\neq\emptyset$, we have

$\lim_{\alpha\to\infty}{\lambda}(\alpha)=\min_{x\in\Sigma_{1}}V(x).$

(ii) If $\Sigma_{1}=\emptyset$, we have

$\lim_{\alpha\to\infty}{\lambda}(\alpha)=\infty.$

######  Remark 2.

Regarding Theorem 2, we want to make some comments as follows.

Under the assumption that all critical points of $m$ are non-degenerate,
Theorem 2 shows that $\lim_{\alpha\to\infty}{\lambda}(\alpha)=\infty$ if and
only if $m$ has no local maximum in the interior of $\Omega$; if $m$ has at
least one local maximum in the interior of $\Omega$, then
$\lim_{\alpha\to\infty}{\lambda}(\alpha)$ will be determined by Theorem 2(i).
Indeed, such assertions remain true as long as $\beta>0$ on $\partial\Omega$
by Theorem 1.

In dimension $N=2$, if $v$ is a $C^{1}(\overline{\Omega})$ satisfying (1.5)
and $v\cdot n=0$ on $\partial\Omega$, then it is not hard to see that $v$ has
a first integral, and therefore $\lim_{\alpha\to\infty}{\lambda}(\alpha)$ is
finite and is determined by Theorem 1.1(ii). Nevertheless, if $v\in
C^{1}(\overline{\Omega})$ is a gradient field, we take $v=\nabla m$ in (1.2)
so that (1.5) becomes $\Delta m=0$ on $\overline{\Omega}$, and the above-
mentioned condition $v\cdot n=0$ on $\partial\Omega$ becomes $\frac{\partial
m}{\partial n}=0$ on $\partial\Omega$. In such case, $m$ must be a constant so
that $v=0$.

In dimension $N=1$, that is, $\Omega$ is an open interval, say $\Omega=(0,1)$,
then $\Delta m=0$ on $\overline{\Omega}$ implies $m(x)=ax+b,\,\forall
x\in[0,1]$ for some constants $a,\,b$, and so $v(x)=a$ has a first integral in
$H_{0}^{1}(\Omega)$ if and only if $a=0$. It turns out that this result is a
very special case of problem (3.1) to be treated in Section 3 below. As
mentioned before, for any dimension $N\geq 1$, Theorem 1.1(i) and (ii) can
allow $v$ to be degenerate so that it has a first integral.

The discussion above shows that Theorem 1.1(ii) does not cover Theorem 2, and
vice versa.

If we allow $\beta$ to change its sign or be negative, generally one can not
expect $\lambda_{1}(\alpha)$ to be bounded from below with respect to all
large $\alpha$. In order to deal with such a case, let us denote

$\partial\Omega^{-}=\\{x\in\partial\Omega:\ \ \beta(x)<0\\},\ \ \
\Omega_{\delta}^{-}=\\{x\in\Omega:\ \
\text{dist}(x,\partial\Omega^{-})<\delta\\}.$

We further formulate the following assumption.

(A2): There exist a smooth vector field $\nu(x)$ and a small constant
$\delta>0$ such that $\nu(x)\cdot n(x)\geq\delta$ and $\nabla
m(x)\cdot\nu(x)\leq 0$ for all $x\in\Omega_{\delta}^{-}$.

For the general boundary function $\beta$, our result can be stated as
follows.

######  Theorem 3.

Assume that (A1) holds and all critical points of $m$ are non-degenerate. Let
$\lambda_{1}(\alpha)$ be the principal eigenvalue of (1.1). Then the following
assertions hold.

(i) If $\Sigma_{3}\neq\emptyset$, we have

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty.$

(ii) If $\Sigma_{3}=\emptyset$, $\Sigma_{1}\cup\Sigma_{2}\neq\emptyset$ and
(A2) holds, we have

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\min_{\Sigma_{1}\cup\Sigma_{2}}V(x).$

(iii) If $\Sigma_{1}\cup\Sigma_{2}\cup\Sigma_{3}=\emptyset$ and (A2) holds, we
have

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty.$

######  Remark 3.

Notice that $\lambda_{1}(\alpha)$ must be bounded from below as
$\alpha\to\infty$ provided (A2) holds; see Lemma 4 in Section 2. Moreover,
given $x_{0}\in\Sigma_{3}$, we have $\nabla m(x_{0})\cdot\nu(x_{0})\geq 0$ for
any vector field $\nu(x_{0})$ satisfying $\nu(x_{0})\cdot n(x_{0})>0$, which,
together with Theorem 3(i), implies that (A2) is almost a necessary condition
to guarantee the boundedness of $\lambda_{1}(\alpha)$ from below as
$\alpha\to\infty$. One may refer to Remark 8 in Section 3 for further
discussions in dimension $N=1$. On the other hand, $\lambda_{1}(\alpha)$ must
be bounded from above as $\alpha\to\infty$ as long as
$\Sigma_{1}\cup\Sigma_{2}\neq\emptyset$; see Lemma 1 below.

Let $d(x)=\mathrm{dist}(x,\partial\Omega)$. Then there exists a small constant
$d_{0}>0$ such that $d(x)$ is smooth in $\\{x\in\Omega:\ d(x)<d_{0}\\}$. We
now extend the definition of $n(x)$ by letting $n(x)=-\nabla d(x)$ for all
$x\in\Omega$ satisfying $d(x)\leq d_{0}$. Thus, for $x\in\partial\Omega$,
$n(x)$ is the unit outward normal to $\partial\Omega$.

In all the above stated theorems, the non-degeneracy assumption on all
possible critical points of $m$ in $\Omega$ can be relaxed by the following
weaker one:

(A3): Assume that $m$ has only finitely many critical points. Let
$x_{0}\in\overline{\Omega}\setminus\Sigma$ be a critical point of $m$. Assume
that the following hold.

(i) If $x_{0}\in\Omega$, then one of the following is satisfied.

(a). There exist an $N-1$ dimensional $C^{1}$-surface $\Gamma$ with
$x_{0}\in\Gamma$, and positive constants $r_{0},\delta$ and a smooth unit
vector field $\xi(x)$ such that $\xi(x)\cdot\nu(x)\geq\delta$, and for all
$x\in\Gamma\cap B(x_{0},r_{0})$,

$\xi(x)\cdot\nabla m(x)=0\ \text{on}\ \Gamma\cap B(x_{0},r_{0}),\ \
\xi(x)\cdot\nabla m(x)\geq 0\ \text{ in }D_{1},\ \ \xi(x)\cdot\nabla m(x)\leq
0\ \text{ in }D_{2},$

where $\nu(x)$ is the unit normal vector field of $\Gamma$, $D_{1}$ and
$D_{2}$ are the two connected components of $B(x_{0},r_{0})\setminus\Gamma$
such that $\nu(x)$ is outward to $D_{2}$.

(b). There exist a smooth unit vector field $\xi(x)$ and a constant $r_{0}>0$
such that $\xi(x)\cdot\nabla m(x)\geq 0$ for all $x\in
B(x_{0},r_{0})\cap\Omega$.

(ii) If $x_{0}\in\partial\Omega$, then either $x_{0}$ is non-degenerate or one
of the following is satisfied.

(a). There exist a smooth unit vector field $\xi(x)$ and positive constants
$r_{0}$ and $\delta$ such that

$\xi(x)\cdot n(x)\geq\delta\ \,\text{ on }\partial\Omega\cap B(x_{0},r_{0}),\
\ \nabla m(x)\cdot\xi(x)\leq 0\ \,\text{ in }\Omega\cap B(x_{0},r_{0}).$

(b). There exist a smooth unit vector field $\xi(x)$ and positive constants
$r_{0}$ and $\delta$ such that

$\xi(x)\cdot n(x)\leq 0\ \,\text{ on }\partial\Omega\cap B(x_{0},r_{0}),\ \
\nabla m(x)\cdot\xi(x)\leq 0\ \,\text{ in }\Omega\cap B(x_{0},r_{0}).$

(c). There exist an $N-1$ dimenstional $C^{1}$-surface $\Gamma$, a smooth unit
vector field $\xi(x)$ and positive constants $r_{0}$ and $\delta$ such that
$x_{0}\in\Gamma$, $|n(x_{0})\cdot\nu(x_{0})|\neq 1$,
$\xi(x)\cdot\nu(x)>\delta$ on $\Gamma\cap B(x_{0},r_{0})$, $\xi(x)\cdot\nabla
m(x)=0$ for all $x\in\Gamma\cap B(x_{0},r_{0})$, and

$\xi(x)\cdot n(x)\leq 0\ \,\text{ on }\partial\Omega\cap\overline{D}_{1},\ \
\nabla m(x)\cdot\nu(x)\geq 0\ \,\text{ in }D_{1},$

and

$\xi(x)\cdot n(x)\geq 0\ \,\text{ on }\partial\Omega\cap\overline{D}_{2},\ \
\nabla m(x)\cdot\nu(x)\leq 0\ \,\text{ in }D_{2},$

where $\nu(x)$ is the unit normal vector field of $\Gamma$, $D_{1}$ and
$D_{2}$ are the two connnected components of $\left(\Omega\cap
B(x_{0},r_{0})\right)\setminus\Gamma$ such that $n(x)$ is outward to $D_{2}$.

######  Remark 4.

We would like to provide some concrete examples so that (A3) holds.

(i) Suppose that $x_{0}\in\Omega\setminus\Sigma$ is a non-degenerate critical
point of $m$, then it is easy to check that $m$ satisfies (i)-(a) of (A3) at
$x_{0}$.

(ii) Suppose that $x_{0}\in\Omega$ is a critical point of $m$ and

$m(x)=m(x_{0})+c(e\cdot(x-x_{0}))^{3}+o(|x|^{3})$

in a small neighborhood of $x_{0}$ with some constant $c\neq 0$ and $e\in
S^{N-1}$, where $S^{N-1}$ is the unit sphere of $\mathbb{R}^{N}$. Then $m$
satisfies (i)-(b) of (A3) at $x_{0}$.

(iii) Suppose that $x_{0}\in\partial\Omega$ and

$m(x)=m(x_{0})+c_{1}n(x_{0})\cdot(x-x_{0})+c_{2}(\tilde{\nu}\cdot(x-x_{0}))^{3}+o(|x|^{3})$

in a small neighborhood of $x_{0}$ with some constants $c_{1}>0,c_{2}\neq 0$,
where $\tilde{\nu}$ is a tangent vector of $\Gamma$ at $x_{0}$. Then $m$
satisfies (ii)-(a) of (A3) at $x_{0}$.

(iv) Suppose that $x_{0}\in\partial\Omega$ and

$m(x)=m(x_{0})+c(n(x_{0})\cdot(x-x_{0}))^{3}+o(|x|^{3})$

in a small neighborhood of $x_{0}$ with some constant $c<0$. Then $m$
satisfies (ii)-(b) of (A3) at $x_{0}$.

(v) Suppose that $x_{0}\in\partial\Omega$, $\Omega$ is concave in a
neighborhood of $x_{0}$, and

$m(x)=m(x_{0})+c_{1}n(x_{0})\cdot(x-x_{0})+c_{2}(\tilde{\nu}\cdot(x-x_{0}))^{4}+o(|x|^{4})$

in a small neighborhood of $x_{0}$ with some constants $c_{1}\geq 0,c_{2}>0$,
where $\tilde{\nu}$ is a tangent vector of $\Gamma$ at $x_{0}$. Then $m$
satisfies (ii)-(c) of (A3) at $x_{0}$.

Theorem 1 and Theorem 3 can be improved as follows.

######  Theorem 4.

Assume that (A1) and (A3) hold. Let $\lambda_{1}(\alpha)$ be the principal
eigenvalue of (1.1). Then the following assertions hod.

(i) If $\Sigma_{3}\neq\emptyset$, we have

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty.$

(ii) If $\Sigma_{3}=\emptyset$, $\Sigma_{1}\cup\Sigma_{2}\neq\emptyset$ and
(A2) holds, we have

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\min_{\Sigma_{1}\cup\Sigma_{2}}V(x).$

(iii) If $\Sigma_{1}\cup\Sigma_{2}\cup\Sigma_{3}=\emptyset$ and (A2) holds, we
have

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty.$

We want to make the following remarks.

######  Remark 5.

(i) Notice that in the Neumann boundary problem (1.6), our assumptions (A1)
and (A2) hold automatically. In addition, we do not require the non-degeneracy
of local maxima of $m$. Thus, under the weaker condition (A3), the assertion
(ii) of Theorem 4 improves Theorem 1.2 obtained by Chen and Lou in [7].

(ii) Both Theorem 3(i) and Theorem 4(i) remain true only if the set
$\Sigma_{3}$ contains one local maximum in the sense of Definition 2(iii) and
all the critical points of $m$ allow to be degenerate; see Remark 7 below.

(iii) Theorem 2 remains valid provided the non-degeneracy assumption of all
critical points of $m$ is replaced by (A3).

######  Remark 6.

For any dimension $N\geq 1$, let $L_{i}\,(1\leq i\leq N)$ be given positive
numbers. Assume that $m\in C^{2}(\mathbb{R}^{N})$ and $V\in C(\mathbb{R}^{N})$
are $L$-periodic with respect to the spatial variable $x$ in the sense of [3,
4]; that is,
$w(x_{1}+L_{1},x_{2}+L_{2},\cdots,x_{n}+L_{N})=w(x_{1},x_{2},\cdots,x_{N})$
for all $x=(x_{1},x_{2},\cdots,x_{N})\in\mathbb{R}^{N}$ and $w\in\\{m,\,V\\}$.
Consider the following spatial-periodic eigenvalue problem:

$\left\\{\begin{aligned} &-\Delta\varphi-2\alpha\nabla
m(x)\cdot\nabla\varphi+V(x)\varphi=\lambda\varphi&&\hbox{ in
}\mathbb{R}^{N},\\\ &\varphi\ \mbox{is $L$-periodic in}\
x.\end{aligned}\right.$ (1.7)

Let $\Sigma$ be the local maxima set of $m$ in the sense of Definition 2(iii)
and denote by $\lambda_{1}(\alpha)$ the unique principal eigenvalue of (1.7).
Assume that $m$ has only finitely many critical points on
$\\{x=(x_{1},x_{2},\cdots,x_{N})\in\mathbb{R}^{N}:\ x_{i}\in(0,2L_{i}),\ 1\leq
i\leq N\\}$. For any critical point $x_{0}$ of $m$ with
$x_{0}\in\mathbb{R}^{N}\setminus\Sigma$, we further assume that either
(A3)(i)-(a) or (i)-(b) holds. Then, by the similar analysis to prove Theorem
4, we can conclude that

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\min_{\Sigma}V(x).$

This result is not covered by [4] and gives an explicit limit, compared to [4,
Theorem 2.2].

Let us now comment on the proofs of the main results of this paper. Though our
approach is in the spirit of [7], several essential difficulties arise, which
require delicate techniques to overcome. One of such difficulties is to
determine whether the following energy functional

$\int_{\Omega}\\{|\nabla w-\alpha w\nabla m|^{2}\\}+\int_{\partial\Omega}\beta
w^{2}$ (1.8)

is bounded or not for all large $\alpha$, where $w$ is the principal
eigenfunction corresponding to $\lambda_{1}(\alpha)$ with
$\int_{\Omega}w^{2}=1$ for all $\alpha\geq 0$. When $\beta=0$, it is clear to
see that (1.8) is bounded from both above and below. This is one of the key
points in [7] to derive Theorem 1.2. Nevertheless, such a property does not
hold for a general boundary function $\beta$.

In the case of $\beta\geq 0$, obviously (1.8) is bounded from below.
Furthermore, as long as $\Sigma_{1}\cup\Sigma_{2}\neq\emptyset$ holds, (1.8)
is also bounded from above (see Lemma 1). In fact, thanks to Lemma 1, in this
case, in order to prove Theorem 1(i), we just need to determine
$\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)$. This will be achieved by
analyzing the support of the weak limit $\mu$ of $w$, which is a probability
measure. The primary task is to show the support of $\mu$ satisfies
$\text{supp}(\mu)\subset\Sigma_{1}\cup\Sigma_{2}$, which involves some
nontrivial analysis. If $\Sigma_{1}\cup\Sigma_{2}=\emptyset$, we shall show
that (1.8) must be unbounded from above and in turn
$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ by resorting to a
contradiction argument so as to lead to $\text{supp}(\mu)=\emptyset$.

The case that $\beta$ may be negative somewhere on $\partial\Omega$ turns out
to be more challenging to handle since (1.8) may be unbounded from either
below or above. Among other ingredients in our argument, we shall establish a
trace-type inequality (see Lemma 3), which allows us to control the boundary
integral term in (1.8) by the first integral term of (1.8). As a result, under
proper assumptions on the advection function $m$, we can determine whether
(1.8) is bounded or unbounded; based on such information, $\text{supp}(\mu)$
can be therefore clarified.

All the aforementioned results apply to any dimension case (i.e., $N\geq 1$).
As far as the dimension $N=1$ is concerned, by allowing the advection function
$m$ to have natural kinds of degeneracy, we can give a complete description of
the limiting behavior of the principal eigenvalue of the associated eigenvalue
problem (i.e., (3.1) in Section 3); our results exhaust all parameter ranges
of the boundary conditions and thus complement those in [44]. See Section 3
for the precise details. We want to stress that the trace inequality given by
Lemma 3 is also crucial in yielding the results of the one-dimensional problem
(3.1). We also note that the principal eigenvalue of the periodic eigenvalue
problem (1.7) with $N=1$ and the degenerate function $m$ was treated in [44];
see Theorem 2.13 there.

The rest of our paper is organized as follows. In Section 2, we will give the
proofs of Theorem 1-4. In Section 3, we will consider the one-dimensional
problem and obtain a complete characterization of the asymptotic behavior of
the principal eigenvalue, and some typical examples will also be used to
demonstrate the impact of boundary conditions and the degeneracy of $m$.

##  2\. Proofs of Theorems 1-4

This section is devoted to the proofs of our main results: Theorems 1-4.
Without loss of generality, we always take $D=1$ from now on. Throughout this
paper, we use $S^{k-1}$ to denote the unit sphere of $\mathbb{R}^{k}$,
$\mathcal{H}^{k}$ to denote the $k$ dimensional Hausdorff measure, and $f^{+}$
and $f^{-}$, respectively, to denote the positive part and negative part of a
function $f$.

###  2.1. Two lemmas

For later purpose, we need to make the following transformation:

$w(x)=e^{\alpha m(x)}\varphi(x)$

and also normalize $w$ with $\int_{\Omega}w^{2}dx=1$. Then, in view of (1.1)
satisfied by $\varphi$, direct computation shows that $w$ solves

$\left\\{\begin{aligned} &-\Delta w+\left(\alpha^{2}|\nabla
m|^{2}+\alpha\Delta m+V-\lambda\right)w=0&&\hbox{ in }\Omega,\\\
&\frac{\partial w}{\partial n}-\alpha w\frac{\partial m}{\partial n}+\beta
w=0\ &&\hbox{ on }\partial\Omega,\\\
&\int_{\Omega}w^{2}=1.\end{aligned}\right.$ (2.1)

It is also well known that the principal eigenvalue $\lambda_{1}(\alpha)$ can
be variationally characterized by

$\begin{split}\lambda_{1}(\alpha)&=\inf_{\varphi\in
H^{1}(\Omega),\,\varphi\not=0}\frac{\int_{\Omega}e^{2\alpha
m}(|\nabla\varphi|^{2}+V\varphi^{2})dx+\int_{\partial\Omega}\beta e^{2\alpha
m}\varphi^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}e^{2\alpha m}\varphi^{2}dx}\\\
&=\inf_{w\in H^{1}(\Omega),\,\int_{\Omega}w^{2}dx=1}\int_{\Omega}|\nabla
w-\alpha w\nabla m|^{2}+Vw^{2}dx+\int_{\partial\Omega}\beta
w^{2}d\mathcal{H}^{N-1}.\end{split}$ (2.2)

One may refer to [7, 8, 41].

We first establish an estimate for the upper bound of $\lambda_{1}(\alpha)$
once $\Sigma_{1}\cup\Sigma_{2}\neq\emptyset$.

######  Lemma 1.

If $\Sigma_{1}\cup\Sigma_{2}\neq\emptyset$, then it holds that

$\limsup_{\alpha\to\infty}\lambda_{1}(\alpha)\leq\min_{x\in\Sigma_{1}\cup\Sigma_{2}}V(x).$

###### Proof.

Our analysis is similar to that of [7, Lemma 2.4]. Fix
$x_{0}\in\Sigma_{1}\cup\Sigma_{2}$. Then, by our Definition 2(iii), there
exists a sequence $\\{r_{j}\\}$ such that $\lim_{j\to\infty}r_{j}=0$ and

$m(x_{0})>\max_{\overline{\Omega}\cap\partial B(x_{0},r_{j})}m(x),\ \ \
\forall j\in\mathbb{N}.$

Now, we first choose a sequence $\\{s_{j}\\}$ such that

$0<s_{j}<r_{j},\ \
\min_{\overline{\Omega}\cap\overline{B(x_{0},s_{j})}}m(x):=m_{j}>\max_{\overline{\Omega}\cap\partial
B(x_{0},r_{j})}m(x),$

and then choose another sequence $\\{t_{j}\\}$ such that

$s_{j}<t_{j}<r_{j},\ \
\min_{\overline{\Omega}\cap\overline{B(x_{0},s_{j})}}m(x):=m_{j}>M_{j}:=\max_{\overline{\Omega}\cap\overline{B(x_{0},r_{j})}\setminus
B(x_{0},t_{j})}m(x).$

We define

$\varphi_{j}(x)=\left\\{\begin{aligned} &1&&\text{ if
}x\in\overline{B(x_{0},t_{j})},\\\ &(r_{j}-|x|)/(r_{j}-t_{j})&&\text{ if
}x\in\overline{B(x_{0},r_{j})}\setminus B(x_{0},t_{j}),\\\ &0&&\text{ if
}x\in\mathbb{R}^{N}\setminus B(x_{0},r_{j}).\end{aligned}\right.$

In the case that $x_{0}\in\Sigma_{2}$, according to the assumption (A1), there
exists $\delta>0$ such that $\beta(x)=0$ for all $x\in\partial\Omega\cap
B(x_{0},\delta)$. Thus, in either case of $x_{0}\in\Sigma_{1}$ or
$x_{0}\in\Sigma_{2}$, we may assume that, for sufficiently large $j$,
$\beta(x)\varphi_{j}(x)\equiv 0$ on $\partial\Omega$. Direct calculation
yields from (2.2) that

$\displaystyle\lambda_{1}(\alpha)$
$\displaystyle\leq\frac{\int_{\Omega}e^{2\alpha
m}|\nabla\varphi_{j}|^{2}dx}{\int_{\Omega}e^{2\alpha
m}\varphi_{j}^{2}dx}+\frac{\int_{\Omega}e^{2\alpha
m}V\varphi_{j}^{2}dx}{\int_{\Omega}e^{2\alpha
m}\varphi_{j}^{2}dx}+\frac{\int_{\partial\Omega}e^{2\alpha
m}\beta\varphi_{j}^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}e^{2\alpha
m}\varphi_{j}^{2}dx}$ $\displaystyle\leq\frac{e^{2\alpha
M_{j}}r_{j}^{N}}{|r_{j}-t_{j}|^{2}s_{j}^{N}e^{2\alpha
m_{j}}}+\max_{\overline{B(x_{0},r_{j})}}V(x)$

for sufficiently large $j$. Here we used the first variational
characterization of $\lambda_{1}(\alpha)$ in (2.2). Letting $\alpha\to\infty$
and then $j\to\infty$, we obtain that

$\limsup_{\alpha\to\infty}\lambda_{1}(\alpha)\leq V(x_{0}),$

which completes the proof. ∎

We next consider the case of $\Sigma_{3}\neq\emptyset$ and obtain the
following result.

######  Lemma 2.

If $\Sigma_{3}\neq\emptyset$, then it holds that

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty.$

###### Proof.

Let $x_{0}\in\Sigma_{3}$. Then there exists an $r_{0}>0$ such that

$\beta(x)\leq\frac{\beta(x_{0})}{2}:=-\delta$ for all $x\in\partial\Omega\cap
B(x_{0},r_{0})$. (2.3)

We choose the sequences $\\{r_{j}\\},\\{s_{j}\\}$ and $\\{t_{j}\\}$ and the
functions $\varphi_{j}$ the same as in the proof of Lemma 1. For any given
$\epsilon>0$, there exists a constant $M$ such that, for $\alpha,j>M$,

$\frac{\int_{\Omega}e^{2\alpha
m}|\nabla\varphi_{j}|^{2}dx}{\int_{\Omega}e^{2\alpha
m}\varphi_{j}^{2}dx}+\frac{\int_{\Omega}e^{2\alpha
m}V\varphi_{j}^{2}dx}{\int_{\Omega}e^{2\alpha m}\varphi_{j}^{2}dx}\leq
V(x_{0})+\epsilon.$ (2.4)

Without loss of generality, we may assume that $x_{0}=0$ and
$n(x_{0})=(0,\cdots,0,-1)$. Let $r=|x|$, $f_{j}(r)=\varphi_{j}(x)$. We define
$\Phi:\partial\Omega\cap B(0,r_{0})\to B^{\prime}(0^{\prime},r_{0})$,

$\Phi(x)=\frac{|x|}{|x^{\prime}|}x^{\prime},$

where $x=(x^{\prime},x_{N})$,
$B^{\prime}(0,r_{0})=\\{x^{\prime}\in\mathbb{R}^{N-1}:\ \
|x^{\prime}|<r_{0}\\}=B(0,r_{0})\cap\mathbb{R}^{N-1}$. Since $\partial\Omega$
is smooth, $\Phi$ is a diffeomorphism from $\partial\Omega\cap B(0,r_{0})$ to
$B^{\prime}(0^{\prime},r_{0})$ for sufficiently small $r_{0}$. Moreover, there
exists a constant $C$ such that

$C^{-1}\int_{B^{\prime}(0,r_{0})}|u(\Phi^{-1}(x))|dx^{\prime}\leq\int_{\partial\Omega\cap
B(0,r_{0})}|u(x)|d\mathcal{H}^{N-1}\leq
C\int_{B^{\prime}(0,r_{0})}|u(\Phi^{-1}(x))|dx^{\prime}$ (2.5)

for all $u\in L^{1}(\partial\Omega\cap B(0,r_{0}))$.

Notice that $\varphi_{j}(\Phi^{-1}(x))=\varphi_{j}(x)=f_{j}(r)$. We thus
deduce from (2.5) that

$\displaystyle\lim_{j\to\infty}\frac{\int_{\partial\Omega}\varphi_{j}^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}\varphi_{j}^{2}dx}$
$\displaystyle=\lim_{j\to\infty}\frac{\int_{\partial\Omega\cap
B({0},r_{0})}\varphi_{j}^{2}d\mathcal{H}^{N-1}}{\int_{\mathbb{R}^{N}}\varphi_{j}^{2}dx}$
$\displaystyle\geq\lim_{j\to\infty}\frac{1}{C}\frac{\int_{\mathbb{R}^{N-1}}\varphi_{j}^{2}dx}{\int_{\mathbb{R}^{N}}\varphi_{j}^{2}dx}$
$\displaystyle=\lim_{j\to\infty}\frac{1}{C}\frac{\int_{S^{N-2}}\int_{0}^{r_{j}}f_{j}(r)r^{N-2}drd\mathcal{H}^{N-2}}{\int_{S^{N-1}}\int_{0}^{r_{j}}f_{j}(r)r^{N-1}drd\mathcal{H}^{N-1}}$
$\displaystyle\geq\lim_{j\to\infty}\frac{1}{C}\frac{\mathcal{H}^{N-2}(S^{N-2})}{\mathcal{H}^{N-1}(S^{N-1})}\frac{1}{r_{j}}$
$\displaystyle=\infty.$

By fixing $\alpha>M$, due to (2.3), one can easily see that

$\lim_{j\to\infty}\frac{\int_{\partial\Omega}\beta e^{2\alpha
m}\varphi_{j}^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}e^{2\alpha
m}\varphi_{j}^{2}dx}\leq-\delta\frac{\min_{\overline{B({0},r_{j})}}e^{2\alpha
m}}{e^{2\alpha\max_{x\in\overline{\Omega}}|m({x})|}}\lim_{j\to\infty}\frac{\int_{\partial\Omega}\varphi_{j}^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}\varphi_{j}^{2}dx}=-\infty.$
(2.6)

Therefore, by (2.4) and (2.6), we have

$\displaystyle\lambda_{1}(\alpha)$
$\displaystyle\leq\frac{\int_{\Omega}e^{2\alpha
m}|\nabla\varphi_{j}|^{2}dx}{\int_{\Omega}e^{2\alpha
m}\varphi_{j}^{2}dx}+\frac{\int_{\Omega}e^{2\alpha
m}V\varphi_{j}^{2}dx}{\int_{\Omega}e^{2\alpha
m}\varphi_{j}^{2}dx}+\frac{\int_{\partial\Omega}e^{2\alpha
m}\beta\varphi_{j}^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}e^{2\alpha
m}\varphi_{j}^{2}dx}$ $\displaystyle\leq
V({0})+\epsilon-\delta\frac{\min_{\overline{B({0},r_{j})}}e^{2\alpha
m}}{e^{2\alpha\max_{x\in\overline{\Omega}}|m({x})|}}\frac{\int_{\partial\Omega}\varphi_{j}^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}\varphi_{j}^{2}dx}$
$\displaystyle\to-\infty,\ \ \ \mbox{as}\ j\to\infty.$

This completes the proof. ∎

######  Remark 7.

The proof of Lemma 2 does not use the degeneracy of critical points of $m$ in
$\Omega$; that is, Lemma 2 holds as long as $\Sigma_{3}$ contains an isolated
local maximum of $m$.

###  2.2. Proof of Theorem 1

In this subsection, we assume that $\beta\geq 0$ and aim to prove Theorem 1.

Before going further, let us introduce some notations to be used frequently
later. Let $\varphi_{\alpha}$ be the eigenfunctions corresponding to the
principal eigenvalue $\lambda_{1}(\alpha)$, and $w_{\alpha}=e^{\alpha
m}\varphi_{\alpha}$. Without loss of generality, we may assume that
$\int_{\Omega}w_{\alpha}^{2}dx=1$ for all $\alpha$.

For each $\alpha\in\mathbb{R}$, we extend $w_{\alpha}$ to all $\mathbb{R}^{N}$
by setting it to be zero on $\mathbb{R}^{N}\setminus\Omega$. We define a
measure $\mu_{\alpha}$ on $\mathbb{R}^{N}$ by

$\mu_{\alpha}(A)=\int_{A}w_{\alpha}^{2}dx.$

Then $\mu_{\alpha}$ is a Radon measure in $\mathbb{R}^{N}$ with
$\mathrm{supp}(\mu_{\alpha})\subset\overline{\Omega}$ and
$\mu_{\alpha}(\overline{\Omega})=1$, where $\mathrm{supp}(\mu_{\alpha})$
denotes the support of $\mu_{\alpha}$. By the weak compactness of Radon
measures, there exist a sequence $\\{\alpha_{k}\\}_{k\in\mathbb{N}}$ and a
Radon measure $\mu$ such that $\alpha_{k}\to\infty$ as $k\to\infty$ and
$\mu_{\alpha_{j}}$ weakly converges to $\mu$ in the sense of

$\lim_{k\to\infty}\int_{\Omega}w_{\alpha_{k}}^{2}(x)\zeta(x)dx=\int_{\overline{\Omega}}\zeta(x)d\mu,\
\ \ \forall\zeta\in C(\overline{\Omega}),$ (2.7)

fulfilling $\mathrm{supp}(\mu)\subset(\overline{\Omega})$ and
$\mu(\overline{\Omega})=1$. One may refer to [21].

To prove Theorems 1-4, as it will be seen later, the key ingredient of our
mathematical analysis is to gain a precise understanding of
$\mathrm{supp}(\mu)$. To the end, as in [7], we denote the set of non-critical
interior points of $m$ by

$\Omega_{1}=\\{x\in\Omega:\ \ |\nabla m(x)|\neq 0\\},$

and the set of non-degenerate interior critical points which are not local
maxima by

$\Omega_{2}=\\{x\in\Omega:\ \ |\nabla m(x)|=0,\ \exists e\in S^{N-1}\ \text{
such that}\ (e\cdot\nabla)^{2}m(x)>0\\},$

and the set of non-critical boundary points by

$\Omega_{3}=\left\\{x\in\partial\Omega:\ \ |\nabla m(x)|>\left|\frac{\partial
m}{\partial n}(x)\right|\right\\},$

and divide all boundary critical points which are not local maxima into the
following three categories:

$\displaystyle\Omega_{4}=\left\\{x\in\partial\Omega:\ \ |\nabla
m(x)|=-\frac{\partial m}{\partial n}(x)>0\right\\},$
$\displaystyle\Omega_{5}=\left\\{x\in\partial\Omega:\ \ |\nabla
m(x)|=\frac{\partial m}{\partial n}(x)>0,\ \exists e\in S^{N-1}\ \text{such
that}\ e\perp n,\ (e\cdot\nabla)^{2}m(x)>0\right\\},$
$\displaystyle\Omega_{6}=\left\\{x\in\partial\Omega:\ \ |\nabla m(x)|=0,\
\exists e\in S^{N-1}\ \text{such that}\ (e\cdot\nabla)^{2}m(x)>0\right\\}.$

With the above preparation, we are now ready to present the proof of Theorem
1.

###### Proof of Theorem 1.

We first prove (i). The upper bound estimate

$\limsup_{\alpha\to\infty}\lambda_{1}(\alpha)\leq\min_{\Sigma_{1}\cup\Sigma_{2}}V(x)$
(2.8)

follows directly from Lemma 1.

In what follows, we shall prove

$\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)\geq\min_{\Sigma_{1}\cup\Sigma_{2}}V(x).$
(2.9)

If we can show $\text{supp}(\mu)\subset\Sigma_{1}\cup\Sigma_{2}$, it then
follows that

$\begin{split}\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)&=\liminf_{\alpha\to\infty}\frac{\int_{\Omega}e^{2\alpha
m}|\nabla\varphi_{\alpha}|^{2}+e^{2\alpha
m}V\varphi_{\alpha}^{2}dx+\int_{\partial\Omega}\beta e^{2\alpha
m}\varphi_{\alpha}^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}e^{2\alpha
m}\varphi_{\alpha}^{2}dx}\\\
&\geq\liminf_{\alpha\to\infty}\frac{\int_{\Omega}e^{2\alpha
m}V\varphi_{\alpha}^{2}dx}{\int_{\Omega}e^{2\alpha
m}\varphi_{\alpha}^{2}dx}\\\
&=\liminf_{\alpha\to\infty}\frac{\int_{\Omega}Vw_{\alpha}^{2}dx}{\int_{\Omega}w_{\alpha}^{2}dx}\\\
&=\frac{\int_{\overline{\Omega}}Vd\mu}{\mu(\overline{\Omega})}\\\
&\geq\min_{\Sigma_{1}\cup\Sigma_{2}}V(x),\end{split}$ (2.10)

which verifies (2.9). Hence, a combination of (2.8) and (2.10) gives

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\min_{x\in\Sigma_{1}\cup\Sigma_{2}}V(x),$

as wanted.

Let $\Omega_{i}\ (1\leq i\leq 6)$ be defined as before, and further set

$\Sigma_{4}=\\{x\in\Sigma\cap\partial\Omega:\ \ \beta(x)>0\\}.$

In order to show $\text{supp}(\mu)\subset\Sigma_{1}\cup\Sigma_{2}$, it is not
hard to see that we just need to prove $\mu(\Sigma_{4})=0$ and
$\mu(\Omega_{i})=0\ (1\leq i\leq 6)$.

By means of (2.8) and $\beta\geq 0$, we observe that

$\int_{\Omega}|\nabla w_{\alpha}-\alpha w_{\alpha}\nabla m|^{2}dx\leq C$
(2.11)

for some positive constant $C$, which does not depend on $\alpha\geq 0$. Thus,
making use of (2.11), one can follow the same arguments as in [7, Lemmas
3.1-3.4 and Lemma 4.1] to claim that $\mu(\Omega_{i})=0$ for each $1\leq i\leq
6$. Thus, it remains to verify $\mu(\Sigma_{4})=0$.

In the sequel, we are going to prove that

$\mu(\Sigma_{4})=0.$ (2.12)

Note that $\Sigma_{4}$ is a discrete set. It is sufficient to show that
$\mu(\\{x_{0}\\})=0$ for each $x_{0}\in\Sigma_{4}$.

By (2.8) and $\int_{\Omega}w_{\alpha}^{2}dx=1$, it is clear that

$\int_{\Omega}|\nabla w_{\alpha}-\alpha w_{\alpha}\nabla
m|^{2}dx+\int_{\partial\Omega}\beta w_{\alpha}^{2}d\mathcal{H}^{N-1}\leq
C_{*}$ (2.13)

for some positive constant $C_{*}$, independent of $\alpha\geq 0$.

We may assume that $x_{0}=0$, and shall prove $\mu(\\{0\\})=0$ by a
contradiction argument. Supposing that $\mu(\\{0\\})>0$, we will show that
either

$\limsup_{\alpha\to\infty}\int_{\partial\Omega}\beta
w_{\alpha}^{2}d\mathcal{H}^{N-1}=\infty,$

or

$\limsup_{\alpha\to\infty}\int_{\Omega}|\nabla w_{\alpha}-\alpha
w_{\alpha}\nabla m|^{2}dx=\infty$

holds, which leads to a contradiction against the fact (2.13).

We assume that

$\limsup_{\alpha\to\infty}\int_{\partial\Omega}\beta
w_{\alpha}^{2}d\mathcal{H}^{N-1}<\infty,$ (2.14)

otherwise there is nothing to do.

In the following, we have to distinguish two different cases.

Case 1. $\partial_{n}m(0)=|\nabla m(0)|>0$ and $(e\cdot\nabla)^{2}m<0$ for all
$e\perp n(0)$, $e\in S^{N-1}$. In this case, by a rotation, we may assume that
$n(0)=(0,\cdots,0,-1)$. Near the point $0$, the boundary of $\Omega$ can be
expressed by

$\\{x=(x^{\prime},x_{N}=\psi(x^{\prime}))\\},\ \ \ \text{ where }\
\,x^{\prime}=(x_{1},\cdots,x_{N-1})$

with $\psi(x^{\prime})\in C^{2}(\mathbb{R}^{N-1})$ and
$\nabla_{x^{\prime}}\psi(0^{\prime})=0$. Locally the boundary can be flattened
by a simple diffeomorphisim

$x=X(z):=(z^{\prime},\psi(z^{\prime})+z_{N}),\ \
z=Z(x)=(x^{\prime},x_{N}-\psi(x^{\prime})).$

It is easy to see that

$\det\left(\frac{\partial X(z)}{\partial z}\right)=1,\ \ \
\det\left(\frac{\partial Z(x)}{\partial x}\right)=1.$

Let $\tilde{w}_{\alpha}(z)=w_{\alpha}(X(z))$,
$\tilde{\varphi}_{\alpha}(z)=\varphi_{\alpha}(X(z))$ and
$\tilde{m}(z)=m(X(z))$. We choose a small constant $r_{0}>0$ such that, for
$z\in B^{\prime}(0^{\prime},r_{0})$, $|\nabla\psi(z^{\prime})|<1$ and

$\beta(X(z))\geq\frac{\beta(0)}{2}:=\delta>0.$

Since

$\int_{X(B^{\prime}(0^{\prime},r_{0}))}w_{\alpha}^{2}d\mathcal{H}^{N-1}=\int_{B^{\prime}(0^{\prime},r_{0})}\tilde{w}_{\alpha}^{2}\left(1+|\nabla_{x^{\prime}}\psi|^{2}\right)^{1/2}dx^{\prime},$

we have

$\int_{B^{\prime}(0^{\prime},r_{0})}\tilde{w}_{\alpha}^{2}dx^{\prime}\leq\int_{X(B^{\prime}(0^{\prime},r_{0}))}w_{\alpha}^{2}d\mathcal{H}^{N-1}\leq\frac{1}{\delta}\int_{X(B^{\prime}(0^{\prime},r_{0}))}\beta
w_{\alpha}^{2}d\mathcal{H}^{N-1}<\infty$

due to (2.14). Thus there exists a positive constant $C$ independent of all
large $\alpha$, such that

$\int_{B^{\prime}(0^{\prime},r_{0})}\tilde{w}_{\alpha}^{2}dx^{\prime}\leq C.$
(2.15)

We choose $r>0$ sufficiently small such that $Z(B(0,r/2)\cap\Omega)\subset
B^{\prime}(0,r)\times(0,r)$. By the assumption that $\mu(\\{0\\})>0$, there
exists a sequence of $\\{\alpha\\}$, denoted by itself for convenience, such
that

$\int_{B(0,r/2)\cap\Omega}w_{\alpha}^{2}dx>\frac{\mu(\\{0\\})}{2},$

which yields

$\int_{B^{\prime}(0^{\prime},r)\times(0,r)}\tilde{w}_{\alpha}^{2}dz\geq\frac{\mu(\\{0\\})}{2}.$

Thus there exists $\tau\in(0,r)$ such that

$\int_{B^{\prime}(0^{\prime},r)\times\\{\tau\\}}\tilde{w}_{\alpha}^{2}dz^{\prime}\geq\frac{\mu(\\{0\\})}{2r}.$
(2.16)

In addition, because of $\partial_{n}m(0)=-\partial_{z_{N}}\tilde{m}(0)>0$, we
may choose $r\in(0,r_{0})$ small enough such that

$\partial_{z_{N}}\tilde{m}(z)<0\ \ \text{ in
}\,B^{\prime}(0^{\prime},r)\times[0,r].$ (2.17)

For any given $f$, let $\tilde{f}(z)=f(X(z))=f(x)$. Then we have

$\partial_{z_{k}}f(z)=\partial_{x_{k}}f(x)+\partial_{x_{N}}f(x)\partial_{x_{k}}\psi(x^{\prime})$,
for $k=1,2,\cdots,N-1,$

and $\partial_{Z_{N}}\tilde{f}(z)=\partial_{x_{N}}f(x)$. Thus

$\displaystyle|\nabla\tilde{f}(z)|$ $\displaystyle=|\nabla
f(x)+\partial_{x_{N}}f(x)(\nabla_{x^{\prime}}\psi(x^{\prime}),0)|$ (2.18)
$\displaystyle\leq|\nabla
f(x)|+|\partial_{x_{N}}f(x)(\nabla_{x^{\prime}}\psi(x^{\prime}),0)|$
$\displaystyle\leq 2|\nabla f(x)|,$

where
$\nabla_{x^{\prime}}\psi=(\partial_{x_{1}}\psi,\cdots,\partial_{x_{N-1}}\psi)$.

Using (2.17) and (2.18), some calculation shows that

$\displaystyle\int_{\Omega}|\nabla w_{\alpha}-\alpha w_{\alpha}\nabla
m|^{2}dx$ (2.19) $\displaystyle=\int_{\Omega}e^{2\alpha
m}|\nabla\varphi_{\alpha}|^{2}dx$
$\displaystyle\geq\frac{1}{2}\int_{B^{\prime}(0^{\prime},r)\times(0,r)}e^{2\alpha\tilde{m}}|\nabla\tilde{\varphi}_{\alpha}|^{2}dz$
$\displaystyle=\frac{1}{2}\int_{B^{\prime}(0^{\prime},r)\times(0,r)}|\nabla\tilde{w}_{\alpha}-\alpha\tilde{w}_{\alpha}\nabla\tilde{m}|^{2}dz$
$\displaystyle\geq\frac{1}{2}\int_{B^{\prime}(0^{\prime},r)\times(0,r)}|\partial_{z_{N}}\tilde{w}_{\alpha}-\alpha\tilde{w}_{\alpha}\partial_{z_{N}}\tilde{m}|^{2}\chi_{\\{\partial_{z_{N}}\tilde{w}_{\alpha}>0\\}}dz$
$\displaystyle\geq\frac{1}{2}\int_{B^{\prime}(0^{\prime},r)\times(0,r)}[(\partial_{z_{N}}\tilde{w}_{\alpha})^{+}]^{2}dz$
$\displaystyle=\frac{1}{2}\int_{B^{\prime}(0^{\prime},r)}\int_{0}^{r}[(\partial_{z_{N}}\tilde{w}_{\alpha})^{+}]^{2}dz_{N}dz^{\prime}$
$\displaystyle\geq\frac{1}{2}\int_{B^{\prime}(0^{\prime},r)}\left(\int_{0}^{r}1^{2}dz_{N}\right)^{-1}\left(\int_{0}^{r}(\partial_{z_{N}}\tilde{w}_{\alpha})^{+}dz_{N}\right)^{2}dz^{\prime}$
$\displaystyle=\frac{1}{2r}\int_{B^{\prime}(0^{\prime},r)}\left(\int_{0}^{r}(\partial_{z_{N}}\tilde{w}_{\alpha})^{+}dz_{N}\right)^{2}dz^{\prime}.$

Let $\mathcal{Q}=\\{x^{\prime}\in B^{\prime}(0^{\prime},r):\ \
\tilde{w}_{\alpha}(x^{\prime},\tau)\geq\tilde{w}_{\alpha}(x^{\prime},0)\\}$.
Then for each $x^{\prime}\in\mathcal{Q}$, we have

$\tilde{w}_{\alpha}(x^{\prime},\tau)-\tilde{w}_{\alpha}(x^{\prime},0)=\int_{0}^{\tau}\partial_{z_{N}}\tilde{w}_{\alpha}(x^{\prime},z_{N})dz_{N}\leq\int_{0}^{\tau}\left(\partial_{z_{N}}\tilde{w}_{\alpha}\right)^{+}(x^{\prime},z_{N})dz_{N}.$

Thus, it holds that

$\displaystyle\int_{\mathcal{Q}}\left(\int_{0}^{r}(\partial_{z_{N}}\tilde{w}_{\alpha})^{+}dz_{N}\right)^{2}dz^{\prime}$
$\displaystyle\geq\int_{\mathcal{Q}}\left(\tilde{w}_{\alpha}(z^{\prime},\tau)\right)^{2}-2\tilde{w}_{\alpha}(z^{\prime},\tau)\tilde{w}_{\alpha}(z^{\prime},0)+\left(\tilde{w}_{\alpha}(z^{\prime},0)\right)^{2}dz^{\prime}$
(2.20)
$\displaystyle\geq\int_{\mathcal{Q}}\frac{1}{2}\left(\tilde{w}_{\alpha}(z^{\prime},\tau)\right)^{2}-3\left(\tilde{w}_{\alpha}(z^{\prime},0)\right)^{2}dz^{\prime}$
$\displaystyle=\frac{1}{2}\int_{\mathcal{Q}}\left(\tilde{w}_{\alpha}(z^{\prime},\tau)\right)^{2}-6\left(\tilde{w}_{\alpha}(z^{\prime},0)\right)^{2}dz^{\prime}.$

On the other hand, for $x^{\prime}\in
B^{\prime}(0^{\prime},r)\setminus\mathcal{Q}$, due to

$\tilde{w}_{\alpha}(x^{\prime},\tau)<\tilde{w}_{\alpha}(x^{\prime},0),$

we have

$\int_{B^{\prime}(0^{\prime},r)\setminus\mathcal{Q}}\left(\tilde{w}_{\alpha}(z^{\prime},\tau)\right)^{2}-6\left(\tilde{w}_{\alpha}(z^{\prime},0)\right)^{2}dz\leq
0.$ (2.21)

Therefore, it follows from (2.20) and (2.21) that

$\displaystyle\int_{B^{\prime}(0^{\prime},r)}\left(\int_{0}^{r}(\partial_{z_{N}}\tilde{w}_{\alpha})^{+}dz_{N}\right)^{2}dz^{\prime}$
$\displaystyle\geq\int_{\mathcal{Q}}\left(\int_{0}^{r}(\partial_{z_{N}}\tilde{w}_{\alpha})^{+}dz_{N}\right)^{2}dz^{\prime}$
$\displaystyle\geq\frac{1}{2}\int_{\mathcal{Q}}\left(\tilde{w}_{\alpha}(z^{\prime},\tau)\right)^{2}-6\left(\tilde{w}_{\alpha}(z^{\prime},0)\right)^{2}dz^{\prime}$
$\displaystyle\geq\frac{1}{2}\int_{B^{\prime}(0^{\prime},r)}\left(\tilde{w}_{\alpha}(z^{\prime},\tau)\right)^{2}-6\left(\tilde{w}_{\alpha}(z^{\prime},0)\right)^{2}dz^{\prime}$
$\displaystyle=\frac{1}{2}\left(\int_{B^{\prime}(0,r)\times\\{\tau\\}}\tilde{w}_{\alpha}^{2}dz^{\prime}-6\int_{B^{\prime}(0,r)\times\\{0\\}}\tilde{w}_{\alpha}^{2}dz^{\prime}\right),$

and consequently, thanks to (2.15) and (2.16), we obtain from (2.19) that, for
any given $M<\infty$,

$\displaystyle\int_{\Omega}|\nabla w_{\alpha}-\alpha m\nabla
w_{\alpha}|^{2}dx$
$\displaystyle\geq\frac{1}{2r}\int_{B^{\prime}(0^{\prime},r)}\left(\int_{0}^{r}(\partial_{z_{N}}\tilde{w}_{\alpha})^{+}dz_{N}\right)^{2}dz^{\prime}$
$\displaystyle\geq\frac{1}{4r}\left(\frac{\mu(\\{0\\})}{2r}-6C\right)$
$\displaystyle\geq M,$

provided that $r$ is sufficiently small and $\alpha$ is sufficiently large,
which contradicts with (2.13). Thus we can conclude that $\mu(\\{0\\})=0$.

Case 2. $|\nabla m(0)|=0$ and $(e\cdot\nabla)^{2}m(0)<0$ for all $e\in
S^{N-1}$. We shall utilize a similar technique as in Case 1 to produce a
contradiction.

Since $|\nabla m(0)|=0$ and $\det(D^{2}m(0))\neq 0$, by [8, Lemma 3.1], there
exist a constant $r_{0}>0$ and a smooth vector field

$\tau(x):\ \ B(0,r_{0})\to\mathbb{R}^{N}$ such that $\tau\cdot n>0$ and
$\tau\cdot\nabla m=0$ on $\partial\Omega\cap B(0,r_{0})$.

We may assume that $|\tau|\equiv 1$. Since

$(\tau(0)\cdot\nabla)^{2}m(0)<0,$

there exists $r_{1}\in(0,r_{0})$ such that

$(\tau(x)\cdot\nabla)^{2}m(x)<0,\ \ \,\forall x\in\Omega\cap B(0,r_{1}),$

from which it then follows that

$\tau(x)\cdot\nabla m(x)<0,\ \ \ \forall x\in\Omega\cap B(0,r_{1}).$

Let

$\Omega_{t}=\\{x\in\Omega:\ \ \text{dist}(x,\partial\Omega)>t\\}.$

Since $\partial\Omega$ is smooth, $\partial\Omega_{t}$ is an
$(N-1)$-dimensional smooth manifold if $t$ is small enough. For
$x\in\partial\Omega_{t}$, we still use $n(x)$ to denote the unit outward
normal vector of $\Omega_{t}$ at $x$. Since $\tau(x)\cdot n(x)>0$ for
$x\in\partial\Omega\cap B(0,r_{0})$, there exist constants $\tau_{0}$ and
$r_{2}\in(0,r_{1})$ such that

$\inf_{x\in\Omega\cap B(0,r_{2})}\tau(x)\cdot n(x)\geq\tau_{0}>0.$

For any given $x\in\partial\Omega\cap B(0,r_{2})$, we denote $\xi(x,t)$ to be
the solution of the ordinary differential system

$\frac{d\xi(t)}{dt}=-\tau(\xi(t)),\ \ \ \xi(0)=x.$

For each $x\in\partial\Omega\cap B(0,r_{2})$ and $t\in(0,r_{2})$, there exists
$p(x,t)$ such that $\xi(x,p(x,t))\in\partial\Omega_{t}$. Moreover,

$\frac{\partial}{\partial t}p(x,t)=(\tau(x)\cdot
n(x))^{-1}\in[1,\tau_{0}^{-1}].$ (2.22)

Let $r_{3}\in(0,r_{2})$. For each $t\in[0,r_{3})$, we define

$\Gamma_{t}=\\{\xi(x,p(x,t)):\ \ x\in\partial\Omega\cap B(0,r_{3})\\}.$

Then $\Gamma_{t}\subset\partial\Omega_{t}$ is an $(N-1)$-dimensional smooth
manifold. We may choose $r_{3}$ small enough such that $\Gamma_{t}\subset
B(0,r_{2})$ for all $t\in(0,r_{3})$. Clearly, for sufficiently small
$r_{4}\in(0,r_{3})$,

$\overline{\Omega}\cap B(0,r_{4})\subset\cup_{t\in[0,r_{3}]}\Gamma_{t}.$
(2.23)

For each $t\in(0,r_{3})$,

$\Phi^{t}:\Gamma_{0}\to\Gamma_{t},\ \ \ \Phi^{t}(x)=\xi(x,p(x,t))$

is a diffeomorphism. There exists a constant $\sigma>0$ such that

$\sigma\mathcal{H}^{N-1}|_{\Gamma_{t}}\leq\Phi^{t}_{*}\left(\mathcal{H}^{N-1}|_{\Gamma_{0}}\right)\leq\sigma^{-1}\mathcal{H}^{N-1}|_{\Gamma_{t}}$
(2.24)

for all $t\in(0,r_{3})$. Here,
$\Phi^{t}_{*}\left(\mathcal{H}^{N-1}|_{\Gamma_{0}}\right)$ is the push-forward
measure of $\mathcal{H}^{N-1}$ restricted to $\Gamma_{0}$.

Using the formula of integration over level sets (see e.g. [21, Theorem
3.13]), we have

$\int_{\cup_{t\in[0,\delta]}\Gamma_{t}}f(x)dx=\int_{0}^{\delta}\int_{\Gamma_{t}}f(x)d\mathcal{H}^{N-1}dt,$
(2.25)

for all $f\in L^{1}(\Omega)$.

As in Case 1, towards a contradiction we suppose that $\mu(\\{0\\})>0$ and
(2.14) holds. Since $\beta(0)>0$, we may choose $r_{4}\in(0,r_{3})$ so small
that $\beta(x)\geq\frac{\beta(0)}{2}$ for all $x\in\partial\Omega\cap
B(0,r_{4})$. Thus, by (2.14), there exists a constant $C$, independent of all
$\alpha\geq 1$, such that

$\int_{\Gamma_{0}}w_{\alpha}^{2}d\mathcal{H}^{N-1}\leq C,\ \ \
\forall\alpha\geq 1.$ (2.26)

On the other hand, by virtue of $\mu(\\{0\\})>0$, we may assume

$\int_{B(0,r_{4})\cap\Omega}w_{\alpha}^{2}dx>\frac{\mu(\\{0\\})}{2}$

for all large $\alpha$. Combined with (2.23) and (2.25), this then gives

$\int_{\cup_{t\in[0,r_{4}]}\Gamma_{t}}w_{\alpha}^{2}dx=\int_{0}^{r_{4}}\int_{\Gamma_{t}}w_{\alpha}^{2}d\mathcal{H}^{N-1}dt\geq\int_{\Omega\cap
B(0,r_{4})}w_{\alpha}^{2}dx\geq\frac{\mu(\\{0\\})}{2}>0,$

which allows us to find $\delta\in(0,r_{4})$ such that

$\int_{\Gamma_{\delta}}w_{\alpha}^{2}d\mathcal{H}^{N-1}\geq\frac{\mu(\\{0\\})}{2r_{4}}.$
(2.27)

Similarly to Case 1, since $\tau\cdot\nabla m<0$ and $|\tau|\equiv 1$ in
$\Omega\cap B(0,r_{1})$, we have

$\displaystyle\int_{\Omega}|\nabla w_{\alpha}-\alpha w_{\alpha}\nabla
m|^{2}dx$ $\displaystyle\geq\int_{\cup_{t\in[0,\delta]}\Gamma_{t}}|\nabla
w_{\alpha}-\alpha w_{\alpha}\cdot\nabla m|^{2}dx$ (2.28)
$\displaystyle\geq\int_{\cup_{t\in[0,\delta]}\Gamma_{t}}|\tau\cdot\nabla
w_{\alpha}-\alpha w_{\alpha}\tau\cdot\nabla m|^{2}dx$
$\displaystyle\geq\int_{0}^{\delta}\int_{\Gamma_{t}}[(\tau\cdot\nabla
w_{\alpha})^{+}]^{2}d\mathcal{H}^{N-1}dt.$

On the other hand, it follows from (2.24) that

$\sigma\int_{\Gamma_{0}}|u(\Phi^{t}(x))|^{2}d\mathcal{H}^{N-1}\leq\int_{\Gamma_{t}}|u(x)|^{2}d\mathcal{H}^{N-1}\leq\sigma^{-1}\int_{\Gamma_{0}}|u(\Phi^{t}(x))|^{2}d\mathcal{H}^{N-1},$

for all $u\in W^{1,2}(\Omega)$. Thus, in light of (2.22) and (2.28), we obtain

$\displaystyle\int_{\Omega}|\nabla w_{\alpha}-\alpha w_{\alpha}\nabla
m|^{2}dx$
$\displaystyle\geq\sigma\int_{0}^{\delta}\int_{\Gamma_{0}}[(\tau\cdot\nabla
w_{\alpha})^{+}]^{2}(\xi(x,p(x,t)))d\mathcal{H}^{N-1}dt$ (2.29)
$\displaystyle=\sigma\int_{\Gamma_{0}}\int_{0}^{\delta}[(\tau\cdot\nabla
w_{\alpha})^{+}]^{2}(\xi(x,p(x,t)))dtd\mathcal{H}^{N-1}$
$\displaystyle=\sigma\int_{\Gamma_{0}}\int_{0}^{\delta}[(\partial_{t}w_{\alpha}(\xi(x,p(x,t)))^{+}\left(\partial_{t}p(x,t)\right)^{-1}]^{2}dtd\mathcal{H}^{N-1}$
$\displaystyle\geq\sigma\tau_{0}^{2}\int_{\Gamma_{0}}\int_{0}^{\delta}[(\partial_{t}w_{\alpha}(\xi(x,p(x,t))))^{+}]^{2}dtd\mathcal{H}^{N-1}$
$\displaystyle\geq\sigma\tau_{0}^{2}\int_{\Gamma_{0}}\left(\int_{0}^{\delta}1^{2}dt\right)^{-1}\left(\int_{0}^{\delta}[\partial_{t}w_{\alpha}(\xi(x,p(x,t)))]^{+}dt\right)^{2}d\mathcal{H}^{N-1}$
$\displaystyle\geq\frac{\sigma\tau_{0}^{2}}{\delta}\int_{\Gamma_{0}}\left(\int_{0}^{\delta}[\partial_{t}w_{\alpha}(\xi(x,p(x,t)))]^{+}dt\right)^{2}d\mathcal{H}^{N-1}.$

Denote $\mathcal{Q}=\\{x:\ \
x\in\Gamma_{0},\,w_{\alpha}(\xi(x,p(x,\delta)))\geq
w_{\alpha}(\xi(x,p(x,0)))\\}$. Similarly to Case 1, we have

$\displaystyle\int_{\Gamma_{0}}\left(\int_{0}^{\delta}[\partial_{t}w_{\alpha}(\xi(x,p(x,t)))]^{+}dt\right)^{2}d\mathcal{H}^{N-1}$
$\displaystyle\geq\int_{\mathcal{Q}}\left(\int_{0}^{\delta}[\partial_{t}w_{\alpha}(\xi(x,p(x,t)))]^{+}dt\right)^{2}d\mathcal{H}^{N-1}$
(2.30)
$\displaystyle\geq\int_{\mathcal{Q}}\left(w_{\alpha}(\xi(x,p(x,\delta))-w_{\alpha}(\xi(x,0)\right)^{2}d\mathcal{H}^{N-1}$
$\displaystyle\geq\int_{\mathcal{Q}}\frac{1}{2}(w_{\alpha}(\xi(x,p(x,\delta)))^{2}-3(w_{\alpha}(\xi(x,0))^{2}d\mathcal{H}^{N-1}$
$\displaystyle\geq\frac{1}{2}\int_{\Gamma_{0}}(w_{\alpha}(\xi(x,p(x,\delta)))^{2}-6(w_{\alpha}(\xi(x,0))^{2}d\mathcal{H}^{N-1}.$

As a result, using (2.26)-(2.30), we infer that

$\displaystyle\int_{\Omega}|\nabla w_{\alpha}-\alpha w_{\alpha}\nabla
m|^{2}dx$
$\displaystyle\geq\frac{\sigma\tau_{0}^{2}}{\delta}\int_{\Gamma_{0}}\left(\int_{0}^{\delta}[\partial_{t}w_{\alpha}(\xi(x,p(x,t)))]^{+}dt\right)^{2}d\mathcal{H}^{N-1}$
(2.31)
$\displaystyle\geq\frac{\sigma\tau_{0}^{2}}{2\delta}\int_{\Gamma_{0}}(w_{\alpha}(\xi(x,p(x,\delta)))^{2}-6(w_{\alpha}(\xi(x,0))^{2}d\mathcal{H}^{N-1}$
$\displaystyle\geq\frac{\sigma^{2}\tau_{0}^{2}}{2\delta}\int_{\Gamma_{\delta}}w_{\alpha}^{2}d\mathcal{H}^{N-1}-\frac{6\sigma\tau_{0}^{2}}{\delta}\int_{\Gamma_{0}}w_{\alpha}^{2}d\mathcal{H}^{N-1}$
$\displaystyle\geq\frac{\sigma^{2}\tau_{0}^{2}\mu(\\{0\\})}{4\delta
r_{4}}-\frac{6\sigma\tau_{0}^{2}C}{\delta}.$

For any given $r_{4}$, the inequality (2.31) holds for all large $\alpha$.
Thus we can choose $r_{4}$ sufficiently small such that

$\frac{\sigma^{2}\tau_{0}^{2}\mu(\\{0\\})}{4\delta
r_{4}}-\frac{6\sigma\tau_{0}^{2}C}{\delta}>C_{*},$

which contradicts with (2.13). Therefore we can assert that $\mu(\\{0\\})=0$
and so Theorem 1(i) holds.

We next verify Theorem 1(ii). We argue indirectly again by supposing that

$\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)\leq M,$ (2.32)

for some constant $M$. Then there exists a sequence of $\alpha$ which
converges to $\infty$, denoted by itself for notational convenience, such that
$\lambda_{1}(\alpha)\leq M+1$ for all $\alpha\geq 0$. This, together with
(2.32), implies that

$\int_{\Omega}|\nabla w_{\alpha}-\alpha w_{\alpha}\nabla
m|^{2}dx+\int_{\partial\Omega}\beta w_{\alpha}^{2}d\mathcal{H}^{N-1}\leq
M+1+\max_{x\in\overline{\Omega}}V(x)$

for all large $\alpha$. Similarly to the argument in (i), under the assumption
$\Sigma_{1}\cup\Sigma_{2}=\emptyset$, we are able to show that

$\text{supp}(\mu)\cap\overline{\Omega}=\emptyset,$

which leads to $\mu(\overline{\Omega})=0$, contradicting with
$\mu(\overline{\Omega})=1$. Therefore one must have

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty.$

This proves Theorem 1(ii). ∎

###  2.3. Proof of Theorem 2

In this subsection, we consider the eigenvalue problem (1.2). It is known that
the principal eigenvalue of (1.2) can be characterized by

$\begin{split}{\lambda}(\alpha)&=\inf_{\varphi\in
H_{0}^{1}(\Omega)}\frac{\int_{\Omega}e^{2\alpha
m}(|\nabla\varphi|^{2}+V\varphi^{2})dx}{\int_{\Omega}e^{2\alpha
m}\varphi^{2}dx}\\\ &=\inf_{w\in
H_{0}^{1}(\Omega),\,\int_{\Omega}w^{2}dx=1}\int_{\Omega}|\nabla w-\alpha
w\nabla m|^{2}+Vw^{2}dx.\end{split}$ (2.33)

Indeed, the second variational characterization in (2.33) is derived through
the substitution $\varphi=e^{-\alpha m}w$ in (1.2). Obviously, $w=e^{\alpha
m}\varphi$ solves

$\begin{cases}-\Delta w+\left(\alpha^{2}|\nabla m|^{2}+\alpha\Delta
m+V\right)w=\lambda_{1}(\alpha)w&\hbox{ in }\Omega,\\\ w=0&\hbox{ on
}\partial\Omega.\end{cases}$

We now present the proof of Theorem 2.

###### Proof of Theorem 2.

We consider the eigenvalue problem (1.1) with $\beta\equiv 1$, and denote the
corresponding principal eigenvalue by $\lambda_{1}(\alpha)$. Also for
notational clarity, we denote by $\tilde{\lambda}_{1}(\alpha)$ the principal
eigenvalue of the eigenvalue problem (1.2). Then, it is well known that

$\tilde{\lambda}_{1}(\alpha)\geq\lambda_{1}(\alpha),\ \ \ \forall\alpha>0.$

Observe that $\Sigma_{2}=\emptyset$ due to $\beta\equiv 1$. Thus, if
$\Sigma_{1}=\emptyset$, Theorem 2(ii) readily follows from Theorem 1(ii).

If $\Sigma_{1}\not=\emptyset$, we infer from Theorem 1(i) that

$\liminf_{\alpha\to\infty}\tilde{\lambda}_{1}(\alpha)\geq\liminf_{\alpha\to\infty}{\lambda_{1}}(\alpha)=\min_{x\in\Sigma_{1}}V(x).$
(2.34)

Let $x_{0}\in\Sigma_{1}$ and $\varphi_{j}(x)$ be the functions defined in the
proof of Lemma 1. Clearly $\varphi_{j}(x)=0$ on $\partial\Omega$. Substituting
$\varphi_{j}$ into the second expression of (2.33), we obtain, similarly to
the proof of Lemma 1, that

$\limsup_{\alpha\to\infty}\tilde{\lambda}_{1}(\alpha)\leq V(x_{0}).$

Since $x_{0}$ is arbitrary, we have

$\limsup_{\alpha\to\infty}\tilde{\lambda}_{1}(\alpha)\leq\min_{x\in\Sigma_{1}}V(x).$
(2.35)

Combining (2.34) and (2.35), we deduce Theorem 2(i). ∎

###  2.4. Proof of Theorem 3

In this subsection, we consider the general case that $\beta(x)$ may change
its sign or be negative.

Given $0<r_{0}<1$, for each
$x\in\partial\Omega^{-}_{r_{0}}=\\{x\in\partial\Omega:\ \
\mathrm{dist}(x,\partial\Omega^{-})<r_{0}\\}$, let $\eta(x,t)$ be the solution
of

$\frac{d\eta}{dt}(x,t)=\frac{-\nu(\eta(x,t))}{\nu(\eta(x,t))\cdot
n(\eta(x,t))},\ \ \ \eta(x,0)=x.$

For each $x\in\Omega^{-}_{r_{0}}$ and $t\in(0,r_{0})$,
$\eta(x,t)\in\partial\Omega_{t}^{-}$. Let us define

$\Gamma_{t}=\\{\eta(x,t):\ \ x\in\partial\Omega_{r_{0}}^{-}\\}.$

Then the mapping

$\Psi^{t}:\ \ \Gamma_{0}\to\Gamma_{t},\ \ \Psi^{t}(x)=\eta(x,t)$

is a diffeomorphism, and there exists a constant $\sigma>0$ such that

$\sigma\mathcal{H}^{N-1}|_{\Gamma_{t}}\leq\Psi^{t}_{*}\
\left(\mathcal{H}^{N-1}|_{\Gamma_{0}}\right)\leq\sigma^{-1}\mathcal{H}^{N-1}|_{\Gamma_{t}}$

for all $t\in(0,r_{0})$, where
$\Phi^{t}_{*}\left(\mathcal{H}^{N-1}|_{\Gamma_{0}}\right)$ is the push-forward
measure of $\mathcal{H}^{N-1}$ restricted to $\Gamma_{0}$. Thus, for all $f\in
W^{1,1}(\Omega)$ and $t\in(0,r_{0})$, it holds that

$\sigma\int_{\Gamma_{t}}|f(x)|d\mathcal{H}^{N-1}\leq\int_{\Gamma_{0}}|f(\eta(x,t))|d\mathcal{H}^{N-1}\leq\sigma^{-1}\int_{\Gamma_{t}}|f(x)|d\mathcal{H}^{N-1}.$
(2.36)

In the following, we prepare a trace inequality which will be used later.

######  Lemma 3.

Let $x_{0}\in\partial\Omega$, $r\in(0,r_{0})$ and $\zeta\in
C_{0}^{1}(B(x_{0},r))$. Assume that there exist a smooth unit vector field
$\nu(x)$ and positive constants $\delta$ and $r$ such that $\nu(x)\cdot
n(x)\geq\delta$ and $\nu(x)\cdot\nabla m(x)\leq 0$ for all $x\in\Omega\cap
B(x_{0},r)$. Then for any given $\epsilon>0$, there exists a constant
$C(\epsilon)$, which is independent of $\alpha\geq 0$, such that

$\int_{\partial\Omega\cap B(x_{0},r)}u^{2}\zeta^{2}e^{2\alpha
m}d\mathcal{H}^{N-1}\leq\epsilon\int_{\Omega\cap B(x_{0},r)}|\nabla
u|^{2}\zeta^{2}e^{2\alpha m}dx+C(\epsilon)\int_{\Omega\cap
B(x_{0},r)}u^{2}e^{2\alpha m}dx$ (2.37)

for all $u\in W^{1,2}(\Omega\cap B(x_{0},r))$.

###### Proof.

It suffices to prove (2.37) for any smooth function $u$.

For any given $u\in C^{\infty}(\overline{\Omega\cap B(x_{0},r)})$, in view of
(2.36), we have

$\displaystyle\int_{\partial\Omega\cap
B(x_{0},r)}u^{2}(x)\zeta^{2}(x)e^{2\alpha m(x)}d\mathcal{H}^{N-1}$
$\displaystyle=\int_{\partial\Omega\cap
B(x_{0},r)}\int_{0}^{r}-\frac{d}{dt}\left[u^{2}(\eta(x,t))\zeta^{2}(\eta(x,t))e^{2\alpha
m(\eta(x,t))}\right]dtd\mathcal{H}^{N-1}$
$\displaystyle=\int_{\partial\Omega\cap
B(x_{0},r)}\int_{0}^{r}-\frac{d\eta(x,t)}{dt}\cdot\left(2u\nabla
u\zeta^{2}e^{2\alpha m}+2u^{2}\zeta\nabla\zeta e^{2\alpha m}+2\alpha\nabla
mu^{2}\zeta^{2}e^{2\alpha m}\right)(\eta(x,t))dtd\mathcal{H}^{N-1}$
$\displaystyle=\int_{\partial\Omega\cap
B(x_{0},r)}\int_{0}^{r}\frac{1}{\nu(\eta(x,t))\cdot
n(\eta(x,t))}\nu(\eta(x,t))\cdot\left(2u\nabla u\zeta^{2}e^{2\alpha
m}+2u^{2}\zeta\nabla\zeta e^{2\alpha m}\right)(\eta(x,t))dtd\mathcal{H}^{N-1}$
$\displaystyle\ \ \ +2\alpha\int_{\partial\Omega\cap
B(x_{0},r)}\int_{0}^{r}\frac{\left(u^{2}\zeta^{2}e^{2\alpha
m}\right)(\eta(x,t))}{\nu(\eta(x,t))\cdot
n(\eta(x,t))}\nu(\eta(x,t))\cdot\nabla m(\eta(x,t))dtd\mathcal{H}^{N-1}$
$\displaystyle\leq\delta^{-1}\int_{\partial\Omega\cap
B(x_{0},r)}\int_{0}^{r}\left|\nu(x)\cdot\left(2u\nabla u\zeta^{2}e^{2\alpha
m}+2u^{2}\zeta\nabla\zeta e^{2\alpha
m}\right)(\eta(x,t))\right|dtd\mathcal{H}^{N-1}$
$\displaystyle\leq\delta^{-1}\int_{0}^{r}\int_{\partial\Omega\cap
B(x_{0},r)}\left(2|u||\nabla u|\zeta^{2}e^{2\alpha
m}+2u^{2}|\zeta||\nabla\zeta|e^{2\alpha
m}\right)(\eta(x,t))d\mathcal{H}^{N-1}dt$
$\displaystyle\leq\delta^{-1}\sigma^{-1}\int_{0}^{r}\int_{\Gamma_{t}\cap
B(x_{0},r)}\left(2|u||\nabla u|\zeta^{2}e^{2\alpha
m}+2u^{2}|\zeta||\nabla\zeta|e^{2\alpha m}\right)(x)d\mathcal{H}^{N-1}dt$
$\displaystyle=\delta^{-1}\sigma^{-1}\int_{\Omega\cap B(x_{0},r)}2|u||\nabla
u|\zeta^{2}e^{2\alpha m}+2u^{2}|\zeta||\nabla\zeta|e^{2\alpha m}dx$
$\displaystyle\leq\epsilon\int_{\Omega\cap B(x_{0},r)}|\nabla
u|^{2}\zeta^{2}e^{2\alpha m}dx+\frac{1}{\epsilon\delta\sigma}\int_{\Omega\cap
B(x_{0},r)}u^{2}\zeta^{2}e^{2\alpha m}dx$ $\displaystyle\ \ \
+2\delta^{-1}\sigma^{-1}\max_{\overline{\Omega\cap
B(x_{0},r)}}|\zeta||\nabla\zeta|\int_{\Omega\cap B(x_{0},r)}u^{2}e^{2\alpha
m}dx$ $\displaystyle\leq\epsilon\int_{\Omega\cap B(x_{0},r)}|\nabla
u|^{2}\zeta^{2}e^{2\alpha m}dx+C(\epsilon)\int_{\Omega\cap
B(x_{0},r)}u^{2}e^{2\alpha m}dx.$

This completes the proof. ∎

With the help of Lemma 2.37, we are able to obtain the following boundedness
of $\\{\lambda_{1}(\alpha)\\}$ from below as $\alpha\to\infty$.

######  Lemma 4.

Assume that there exist a smooth unit vector field $\nu(x)$ and positive
constants $\delta$ and $r_{0}$ such that $\nu(x)\cdot n(x)\geq\delta$ and
$\nu(x)\cdot\nabla m(x)\leq 0$ for all $x\in\Omega_{r_{0}}^{-}$. Then we have

$\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)>-\infty.$

###### Proof.

Let $r\in(0,\delta)$. We choose a set $\\{x_{k}\in\partial\Omega^{-}:\ 1\leq
k\leq K\\}$ consisting of finitely many points and a sequence of smooth
functions $\\{\zeta_{k}:\ 1\leq k\leq K\\}$ such that

$\partial\Omega^{-}\subset\bigcup_{k=1}^{K}B(x_{k},r)$, ${\zeta_{k}}=0$ in
$\mathbb{R}^{N}\setminus B(x_{k},r)$, $\sum_{k=1}^{K}\zeta_{k}^{2}\leq 1$

and

$\sum_{k=1}^{K}\zeta_{k}^{2}(x)=1$ for all $x\in\partial\Omega^{-}$.

For any $\epsilon>0$, by Lemma 3, there exist constants $C_{k}(\epsilon)$ such
that, for all $\varphi\in W^{1,2}(\Omega)$,

$\int_{\partial\Omega\cap B(x_{k},r)}\varphi^{2}\zeta_{k}^{2}e^{2\alpha
m}d\mathcal{H}^{N-1}\leq\epsilon\int_{\Omega\cap
B(x_{k},r)}|\nabla\varphi|^{2}\zeta_{k}^{2}e^{2\alpha
m}dx+C_{k}(\epsilon)\int_{\Omega\cap B(x_{k},r)}\varphi^{2}e^{2\alpha m}dx.$

Thus, it holds that

$\displaystyle\int_{\partial\Omega^{-}}\varphi^{2}e^{2\alpha
m}d\mathcal{H}^{N-1}$
$\displaystyle\leq\int_{\partial\Omega^{-}}\sum_{k=1}^{K}\varphi^{2}\zeta_{k}^{2}e^{2\alpha
m}d\mathcal{H}^{N-1}$ (2.38)
$\displaystyle\leq\sum_{k=1}^{K}\left(\epsilon\int_{\Omega}|\nabla\varphi|^{2}\zeta_{k}^{2}e^{2\alpha
m}dx+C_{k}(\epsilon)\int_{\Omega}\varphi^{2}e^{2\alpha m}dx\right)$
$\displaystyle\leq\epsilon\int_{\Omega}|\nabla\varphi|^{2}e^{2\alpha
m}dx+C(\epsilon)\int_{\Omega}\varphi^{2}e^{2\alpha m}dx$

for some constant $C(\epsilon)$.

Let $\beta^{*}=\sup_{\partial\Omega^{-}}|\beta(x)|$ and choose a small
$\epsilon$ such that $\epsilon\beta^{*}\leq 1$. Then for all $\varphi\in
W^{1,2}(\Omega)$, it holds that

$\displaystyle\frac{\int_{\Omega}e^{2\alpha
m}(|\nabla\varphi|^{2}+V\varphi^{2})dx+\int_{\partial\Omega}\beta e^{2\alpha
m}\varphi^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}e^{2\alpha m}\varphi^{2}dx}$
$\displaystyle\geq\frac{\int_{\Omega}e^{2\alpha
m}(|\nabla\varphi|^{2}+V\varphi^{2})dx-\beta^{*}\left(\epsilon\int_{\Omega}|\nabla\varphi|^{2}e^{2\alpha
m}dx+C(\epsilon)\int_{\Omega}e^{2\alpha
m}\varphi^{2}dx\right)}{\int_{\Omega}e^{2\alpha m}\varphi^{2}}$
$\displaystyle\geq\frac{\int_{\Omega}Ve^{2\alpha
m}\varphi^{2}-\beta^{*}C(\epsilon)e^{2\alpha
m}\varphi^{2}dx}{\int_{\Omega}e^{2\alpha m}\varphi^{2}dx}$
$\displaystyle\geq\min_{\overline{\Omega}}V(x)-\beta^{*}C(\epsilon).$

Therefore, we have

$\lambda_{1}(\alpha)=\inf_{\varphi\in
W^{1,2}(\Omega)}\frac{\int_{\Omega}e^{2\alpha
m}(|\nabla\varphi|^{2}+V\varphi^{2})dx+\int_{\partial\Omega}\beta e^{2\alpha
m}\varphi^{2}d\mathcal{H}^{N-1}}{\int_{\Omega}e^{2\alpha
m}\varphi^{2}dx}\geq\min_{\overline{\Omega}}V(x)-\beta^{*}C(\epsilon)$

for all $\alpha>0$, and consequently,

$\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)\geq\min_{\overline{\Omega}}V(x)-\beta^{*}C(\epsilon).$

This completes the proof. ∎

Now we are ready to prove Theorem 3.

###### Proof of Theorem 3.

The assertion (i) follows directly from Lemma 2.

In the sequel, we will prove the assertion (ii) under conditions (A1) and
(A2). By Lemma 4, we have

$\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)>-\infty.$

Since $\Sigma_{1}\cup\Sigma_{2}\neq\emptyset$, due to Lemma 1, we see that

$\limsup_{\alpha\to\infty}\lambda_{1}(\alpha)\leq\min_{\Sigma_{1}\cup\Sigma_{2}}V(x).$
(2.39)

It remains to show that

$\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)\geq\min_{\Sigma_{1}\cup\Sigma_{2}}V(x).$

As in Subsection 2.2, let us take $\varphi_{\alpha}$ to be the eigenfunctions
corresponding to the principal eigenvalues $\lambda_{1}(\alpha)$ and set
$w_{\alpha}=e^{\alpha m}\varphi_{\alpha}$ with
$\int_{\Omega}w_{\alpha}^{2}dx=1$. We may assume that $w_{\alpha}^{2}$ weakly
converges to a Radon measure $\mu$ in the sense of (2.7) with
$\mu(\overline{\Omega})=1$. Noticing that, by (2.39), for all large $\alpha$,

$\lambda_{1}(\alpha)=\int_{\Omega}(|\nabla\varphi_{\alpha}|^{2}+V\varphi_{\alpha}^{2})e^{2\alpha
m}dx+\int_{\partial\Omega}\beta\varphi_{\alpha}^{2}e^{2\alpha
m}d\mathcal{H}^{N-1}\leq\min_{\Sigma_{1}\cup\Sigma_{2}}V(x)+1.$ (2.40)

On the other hand, similarly as to derive (2.38), one can use Lemma 3 to claim

$\int_{\partial\Omega}\beta^{-}\varphi_{\alpha}^{2}e^{2\alpha
m}d\mathcal{H}^{N-1}\leq\frac{1}{2}\int_{\Omega}|\nabla\varphi_{\alpha}|^{2}e^{2\alpha
m}dx+C$ (2.41)

for some constant $C(\epsilon)$ independent of $\alpha$. Hence, by (2.40) and
(2.41), one can find a constant $M$, independent of all $\alpha\geq 0$, such
that

$\int_{\Omega}|\nabla\varphi_{\alpha}|^{2}e^{2\alpha
m}dx+\int_{\partial\Omega}\beta^{+}\varphi_{\alpha}^{2}e^{2\alpha
m}d\mathcal{H}^{N-1}\leq M,$

leading to

$\int_{\Omega}|\nabla w_{\alpha}-\alpha w_{\alpha}\nabla m|^{2}dx\leq M.$
(2.42)

As before, it follows from [7] that the inequality (2.42) implies that
$\mu(\Omega_{i})=0$ for $i=1,\cdots,6$.

Furthermore, in view of (2.42), we can use the same argument as in the proof
of Theorem 1 to claim that $\mu(\Sigma_{4})=0$. As a result, we have

$\text{supp}(\mu)\subset\Sigma_{1}\cup\Sigma_{2}.$ (2.43)

Let $\delta>0$ such that
$\Omega_{\delta}^{-}\cap(\Sigma_{1}\cup\Sigma_{2})=\emptyset$. We choose
$r\in(0,\delta/2)$ sufficiently small, the set
$\\{x_{k}\in\partial\Omega^{-}:\ 1\leq k\leq K\\}$ and a sequence of smooth
functions $\\{\zeta_{k}:\ 1\leq k\leq K\\}$ such that

$\partial\Omega^{-}\subset\cup_{k=1}^{K}B(x_{k},r)$, ${\zeta_{k}}=0$ in
$\mathbb{R}^{N}\setminus B(x_{k},r)$, $\sum_{k=1}^{K}\zeta_{k}^{2}\leq 1$

and

$\sum_{k=1}^{K}\zeta_{k}^{2}(x)=1$ for all $x\in\partial\Omega^{-}$.

Let $\eta\in C_{0}^{\infty}(\mathbb{R}^{N})$ be a nonnegative function such
that

$\eta\equiv 0$ in $\Omega\setminus\Omega_{\delta}^{-}$ and $\eta(x)\equiv 1$
in $\Omega_{r}^{-}$.

For any $\epsilon>0$, by Lemma 3, there exist constants $C_{k}(\epsilon)$ such
that, for all $\alpha\geq 0$,

$\displaystyle\int_{\partial\Omega\cap
B(x_{k},r)}\varphi_{\alpha}^{2}\zeta_{k}^{2}e^{2\alpha m}d\mathcal{H}^{N-1}$
$\displaystyle\leq\epsilon\int_{\Omega\cap
B(x_{k},r)}|\nabla\varphi_{\alpha}|^{2}\zeta_{k}^{2}e^{2\alpha
m}dx+C_{k}(\epsilon)\int_{\Omega\cap B(x_{k},r)}\varphi_{\alpha}^{2}e^{2\alpha
m}dx$ $\displaystyle\leq\epsilon\int_{\Omega\cap
B(x_{k},r)}|\nabla\varphi_{\alpha}|^{2}\zeta_{k}^{2}e^{2\alpha
m}dx+C_{k}(\epsilon)\int_{\Omega}\eta\varphi_{\alpha}^{2}e^{2\alpha m}dx.$

Thus, we obtain

$\displaystyle\int_{\partial\Omega^{-}}\varphi_{\alpha}^{2}e^{2\alpha
m}d\mathcal{H}^{N-1}$
$\displaystyle\leq\int_{\partial\Omega^{-}}\sum_{k=1}^{K}\varphi_{\alpha}^{2}\zeta_{k}^{2}e^{2\alpha
m}d\mathcal{H}^{N-1}$ (2.44)
$\displaystyle\leq\sum_{k=1}^{K}\left(\epsilon\int_{\Omega}|\nabla\varphi_{\alpha}|^{2}\zeta_{k}^{2}e^{2\alpha
m}dx+C_{k}(\epsilon)\int_{\Omega}\eta\varphi_{\alpha}^{2}e^{2\alpha
m}dx\right)$
$\displaystyle\leq\epsilon\int_{\Omega}|\nabla\varphi_{\alpha}|^{2}e^{2\alpha
m}dx+C(\epsilon)\int_{\Omega}\eta\varphi_{\alpha}^{2}e^{2\alpha m}dx$

for some constant $C(\epsilon)$ which is independent of $\alpha$.

By means of (2.44), we have

$\displaystyle\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)$
$\displaystyle=\lim_{\alpha\to\infty}\int_{\Omega}(e^{2\alpha
m}|\nabla\varphi_{\alpha}|^{2}+Ve^{2\alpha
m}\varphi_{\alpha}^{2})dx+\int_{\partial\Omega}\beta e^{2\alpha
m}\varphi_{\alpha}^{2}d\mathcal{H}^{N-1}$ (2.45)
$\displaystyle\geq\lim_{\alpha\to\infty}\int_{\Omega}(e^{2\alpha
m}|\nabla\varphi_{\alpha}|^{2}+Ve^{2\alpha
m}\varphi_{\alpha}^{2})dx+\int_{\partial\Omega^{-}}\beta e^{2\alpha
m}\varphi_{\alpha}^{2}d\mathcal{H}^{N-1}$
$\displaystyle\geq\lim_{\alpha\to\infty}\left(1-\beta^{*}\epsilon\right)\int_{\Omega}|\nabla\varphi_{\alpha}|^{2}e^{2\alpha
m}dx+\int_{\Omega}Ve^{2\alpha
m}\varphi_{\alpha}^{2}dx-\beta^{*}C(\epsilon)\int_{\Omega}\eta e^{2\alpha
m}\varphi_{\alpha}^{2}dx,$

where $\beta^{*}=\sup_{x\in\partial\Omega^{-}}\beta^{-}(x)$. We choose
$\epsilon$ sufficiently small such that $1-\beta^{*}\epsilon\geq 0$. Note that

$\mathrm{supp}(\eta)\cap\mathrm{supp}(\mathrm{\mu})\subset\mathrm{supp}({\eta})\cap(\Sigma_{1}\cup\Sigma_{2})=\emptyset$

due to (2.43) and the choice of $\eta$. It then follows from the weak
convergence of $\mu_{\alpha}$ (see (2.7)) that

$\lim_{\alpha\to\infty}\int_{\Omega}\eta\varphi_{\alpha}^{2}e^{2\alpha
m}dx=\lim_{\alpha\to\infty}\int_{\mathbb{R}^{N}}\eta
d\mu_{\alpha}=\int_{\mathbb{R}^{N}}\eta d\mu=0.$

Therefore, because of (2.45), we can deduce

$\liminf_{\alpha\to\infty}\lambda_{1}(\alpha)\geq\lim_{\alpha\to\infty}\int_{\Omega}V\varphi_{\alpha}^{2}e^{2\alpha
m}dx\geq\min_{x\in\Sigma_{1}\cup\Sigma_{2}}V(x).$

This verifies the assertion (ii).

We finally prove (iii) under the conditions (A1), (A2) and
$\Sigma_{1}\cup\Sigma_{2}\cup\Sigma_{3}=\emptyset$. Arguing indirectly, we
suppose that there exists a sequence of $\alpha$, still denoted by itself,
such that

$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)<\infty,$

which implies (2.42) holds. Then, the same analysis as in the proof of the
assertion (ii) yields $\mu(\Omega_{i})=0$ for $i=1,\cdots,6$ and
$\mu(\Sigma_{4})=0$. Thus, $\text{supp}(\mu)\subset\Sigma_{1}\cup\Sigma_{2}$,
which, in turn, gives $\mu(\overline{\Omega})=0$ thanks to
$\Sigma_{2}\cup\Sigma_{3}=\emptyset$. This contradicts with
$\mu(\overline{\Omega})=1$. Consequently, we must have
$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$. The proof is now
completed. ∎

###  2.5. Proof of Theorem 4

This subsection is devoted to the proof of Theorem 4.

###### Proof of Theorem 4.

Theorem 4(i) follows directly from Lemma 2.

Denote $\Omega_{i}\ (i=1,3)$ as before, and let
$\\{\alpha_{k}\\}_{k\in\mathbb{N}}$ be a sequence such that
$\lim_{k\to\infty}\alpha_{k}=\infty$ and $\mu_{\alpha_{k}}$ weakly converges
to a Radon measure $\mu$ with $\mathrm{supp}(\mu)\subset\overline{\Omega}$ and
$\mu(\overline{\Omega})=1$. It follows from the same argument as in [7] that
$\mu(\Omega_{i})=0$ for $i=1,3$.

Define

$\tilde{\Omega}_{2}=\\{x\in\Omega\setminus\Sigma:\ \nabla m(x)=0\\},$
$\tilde{\Omega}_{4}=\\{x\in\partial\Omega\setminus\Sigma:\ \nabla m(x)=0\
\text{ or }\ |\nabla m(x)|=-\frac{\partial}{\partial n}m(x)>0\\}$

and

$\Sigma_{4}=\\{x\in\Sigma\cap\partial\Omega:\ \ \beta(x)>0\\}.$

If $\Sigma_{3}=\emptyset$ and $\Sigma_{1}\cup\Sigma_{2}\neq\emptyset$,
similarly to the proof of Theorem 3, we have

$\int_{\Omega}|\nabla w_{\alpha_{k}}-\alpha w_{\alpha_{k}}\nabla m|^{2}dx<C$
(2.46)

for some constant $C$, independent of $k$. Then Lemma 5 below tells us that
$\mu(\tilde{\Omega}_{2})=0$ and $\mu(\tilde{\Omega}_{4})=0$, which implies
$\mathrm{supp}(\mu)\subset\cup_{i=1}^{4}\Sigma_{i}$. Now, the same arguments
as in Theorem 1 and Theorem 3 can be used to deduce Theorem 4(ii) and (iii). ∎

######  Lemma 5.

Let $x_{0}\in{\overline{\Omega}}\setminus\Sigma$ be a critical point of $m$.
Assume that (A3) and (2.46) hold. Then we have $\mu(\\{x_{0}\\})=0$.

###### Proof.

Case 1. $x_{0}\in\Omega$. We assume that (a)-(i) of (A3) holds; the proof for
case (b) is similar. Let $\Gamma,\,\nu,\,\xi,\,r_{0},\,\delta,\,D_{1},\,D_{2}$
be given as in (a)-(i) of (A3). For $x\in B(x_{0},r_{0})$, we define the
signed distance function

$\tilde{d}(x)=\begin{cases}\inf_{y\in\Gamma}|x-y|,&\text{ if }x\in D_{1},\\\
-\inf_{y\in\Gamma}|x-y|,&\text{ if }x\in D_{2}.\end{cases}$

We may assume that $r_{0}$ is sufficiently small such that $\tilde{d}(x)$ is
$C^{1}$ and $\xi(x)\cdot\nu(x)\geq\delta/2$ for all $x\in B(x_{0},r_{0})$,
where $\tilde{\nu}(x)=\nabla\tilde{d}(x)$ and $\tilde{\nu}(x)=\nu(x)$ for
$x\in\Gamma\cap B(x_{0},r_{0})$. Let $\eta(x,t)$ be the solution of

$\frac{d}{dt}\eta(x,t)=\frac{\xi(\eta(x,t))}{\xi(\eta(x,t))\cdot\tilde{\nu}(\eta(x,t))},\
\ \eta(x,0)=x.$

Then $d(\eta(x,t))=t$ for all $x\in\Gamma$ and $t\in(-r_{0},r_{0})$. Given
$r_{1}\in(0,r_{0})$, let us denote

$\Gamma_{t}=\\{\eta(x,t):\ \ x\in\Gamma\cap B(x_{0},r_{1})\\}.$

We may further choose $r_{1}$ sufficiently small such that $\Gamma_{t}\subset
B(x_{0},r_{0})$ for all $t\in(-r_{1},r_{1})$. Then for each
$t\in(-r_{1},r_{1})$,

$\Phi^{t}:\ \ \Gamma_{0}\to\Gamma_{t},\ \ \ \Phi^{t}(x)=\eta(x,t)$

is a differomorphism. Hence, there exists a positive constant $\sigma$, which
is independent of $t$, such that

$\sigma\mathcal{H}^{N-1}|_{\Gamma_{t}}\leq\Phi^{t}_{*}(\mathcal{H}^{N-1}|_{\Gamma_{0}})\leq\sigma^{-1}\mathcal{H}^{N-1}|_{\Gamma_{t}}.$

Suppose towards a contradiction that $\mu(\\{x_{0}\\})>0$. Then, for any given
$\epsilon>0$,

$\int_{-\epsilon r_{1}}^{\epsilon
r_{1}}\int_{\Gamma_{t}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}dt=\int_{\cup_{t\in(-\epsilon
r_{1},\epsilon
r_{1})}\Gamma_{t}}w_{\alpha_{k}}^{2}dx\geq\frac{\mu(\\{x_{0}\\})}{2}$

for all large $k$. Thus, there exist constants $s_{k}\in(-\epsilon
r_{1},\epsilon r_{1})$ such that

$\int_{\Gamma_{s_{k}}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}\geq\frac{\mu(\\{x_{0}\\})}{4\epsilon
r_{1}},\ \ \mbox{for all large}\ k.$ (2.47)

Without loss of generality, we may assume that $s_{k}\in[0,\epsilon r_{1})$.

On the other hand, it follows from the same argument as in [7, Lemma 3.1] that
$\mu(\Omega_{1})=0$. Since all critical points are isolated, we may assume
that $\inf_{\cup_{t\in(r_{1}-\epsilon r_{1},r_{1})}\Gamma_{t}}|\nabla
m(x)|>0.$ This, together with the definitions of $\Omega_{1}$ and the weak
convergence of $w_{\alpha_{k}}^{2}\to\mu$, yields that

$\lim_{k\to\infty}\int_{\cup_{t\in(r_{1}-\epsilon
r_{1},r_{1})}\Gamma_{t}}w_{\alpha_{k}}^{2}dx\leq\mu(\Omega_{1})=0,$

from which it follows, for all large $k$, that

$\int_{r_{1}-\epsilon
r_{1}}^{r_{1}}\int_{\Gamma_{t}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}dt=\int_{\cup_{t\in(r_{1}-\epsilon
r_{1},r_{1})}\Gamma_{t}}w_{\alpha_{k}}^{2}dx<\epsilon\mu(\\{x_{0}\\}).$

Thus, we have

$\int_{\Gamma_{t_{k}}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}<\frac{\mu(\\{x_{0}\\})}{r_{1}}$
(2.48)

for some constant $t_{k}\in(r_{1}-\epsilon r_{1},r_{1})$ and all large $k$.

By the assumption, we have, for all $t\in(0,r_{1})$, $\xi(x)\cdot\nabla
m(x)\geq 0$ on $\Gamma_{t}$. As a consequence, by means of (2.46), (2.47), and
(2.48), we obtain

$\displaystyle\frac{\sigma\mu(\\{x_{0}\\})}{4\epsilon
r_{1}}-\frac{\mu(\\{x_{0}\\})}{\sigma r_{1}}$
$\displaystyle\leq\sigma\int_{\Gamma_{s_{k}}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}-\sigma^{-1}\int_{\Gamma_{t_{k}}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}$
$\displaystyle\leq\int_{\Gamma_{0}}w_{\alpha_{k}}^{2}(\eta(x,s_{k}))d\mathcal{H}^{N-1}-\int_{\Gamma_{0}}w_{\alpha_{k}}^{2}(\eta(x,t_{k}))d\mathcal{H}^{N-1}$
$\displaystyle=\int_{\Gamma_{0}}\int_{s_{k}}^{t_{k}}-\partial_{t}w_{\alpha_{k}}^{2}(\eta(x,t))dtd\mathcal{H}^{N-1}$
$\displaystyle=\int_{\Gamma_{0}}\int_{s_{k}}^{t_{k}}-2w_{\alpha_{k}}(\eta(x,t))\nabla
w_{\alpha_{k}}(\eta(x,t))\cdot\frac{\xi(\eta(x,t))}{\xi(\eta(x,t))\cdot\tilde{\nu}(\eta(x,t))}dtd\mathcal{H}^{N-1}$
$\displaystyle\leq\frac{2}{\delta}\int_{\Gamma_{0}}\int_{s_{k}}^{t_{k}}2\left(\nabla
w_{\alpha_{k}}(\eta(x,t))\cdot\xi(\eta(x,t))\right)^{-}w_{\alpha_{k}}(\eta(x,t))dtd\mathcal{H}^{N-1}$
$\displaystyle\leq\frac{2}{\delta}\int_{\Gamma_{0}}\int_{s_{k}}^{t_{k}}\left[\left(\nabla
w_{\alpha_{k}}(\eta(x,t))\cdot\xi(\eta(x,t))\right)^{-}\right]^{2}+w_{\alpha_{k}}^{2}(\eta(x,t))dtd\mathcal{H}^{N-1}$
$\displaystyle=\frac{2}{\delta}\int_{s_{k}}^{t_{k}}\int_{\Gamma_{0}}\left[\left(\nabla
w_{\alpha_{k}}(\eta(x,t))\cdot\xi(\eta(x,t))\right)^{-}\right]^{2}+w_{\alpha_{k}}^{2}(\eta(x,t))d\mathcal{H}^{N-1}dt$
$\displaystyle\leq\frac{2}{\delta\sigma}\int_{s_{k}}^{t_{k}}\int_{\Gamma_{t}}\left[\left(\nabla
w_{\alpha_{k}}(x)\cdot\xi(x)\right)^{-}\right]^{2}+w_{\alpha_{k}}^{2}(x)d\mathcal{H}^{N-1}dt$
$\displaystyle\leq\frac{2}{\delta\sigma}\int_{\cup_{t\in[s_{k},t_{k}]}\Gamma_{t}}\left[\left(\nabla
w_{\alpha_{k}}(x)\cdot\xi(x)\right)^{-}\right]^{2}+w_{\alpha_{k}}^{2}(x)dx$
$\displaystyle\leq\frac{2}{\delta\sigma}\int_{\left(\cup_{t\in[s_{k},t_{k}]}\Gamma_{t}\right)\cap\\{x:\
\xi(x)\cdot\nabla m(x)\geq 0\\}}\left[\nabla
w_{\alpha_{k}}\cdot\xi-\alpha_{k}w_{\alpha_{k}}\xi\cdot\nabla
m\right]^{2}dx+\int_{\Omega}w_{\alpha_{k}}^{2}dx$
$\displaystyle\leq\frac{2}{\delta\sigma}\int_{\Omega}\left[\nabla
w_{\alpha_{k}}\cdot\xi-\alpha_{k}w_{\alpha_{k}}\xi\cdot\nabla
m\right]^{2}+w_{\alpha_{k}}^{2}dx$
$\displaystyle\leq\frac{2}{\delta\sigma}\int_{\Omega}\left|\nabla
w_{\alpha_{k}}-\alpha_{k}w_{\alpha_{k}}\nabla
m\right|^{2}+w_{\alpha_{k}}^{2}dx$
$\displaystyle\leq\frac{2(C+1)}{\delta\sigma},$

a contradiction provided $\epsilon$ is small enough, and in turn
$\mu(\\{x_{0}\\})=0$ holds.

Case 2. $x_{0}\in\partial\Omega$. We assume that (c)-(ii) in (A3) holds; the
other cases can be handled in a similar manner. Let
$\Gamma,\,\nu,\,\xi,\,r_{0},\,\delta,\,D_{1},\,D_{2}$ be given as in (c)-(ii)
of (A3). Also define the signed distance function $\tilde{d}(x)$,
$\tilde{\nu}(x)$ and $\eta(x,t)$ as in Case 1 such that
$\xi(x)\cdot\nu(x)\geq\delta/2$ for all $x\in B(x_{0},r_{0})$. Thus
$\tilde{d}(\eta(x,t))=t$ for all $x\in\Gamma$ and $t\in(-r_{0},r_{0})$. For
$r_{1}\in(0,r_{0})$, we also set $\Gamma_{t}=\\{\eta(x,t):\ \ x\in\Gamma\cap
B(x_{0},r_{1})\\}$. We may choose $r_{1}$ to be so small that
$\Gamma_{t}\subset B(x_{0},r_{0})$ for all $t\in(-r_{1},r_{1})$. Given
$s,t\in(-r_{1},r_{1})$, then $\Phi^{s,t}:\ \Gamma_{s}\to\Gamma_{t},\
\,\Phi^{s,t}(x)=\eta(x,t-s)$ is a differomorphism, and there exists a constant
$\sigma>0$, independent of $s$ and $t$, such that

$\sigma\mathcal{H}^{N-1}|_{\Gamma_{t}}\leq\Phi^{s,t}_{*}(\mathcal{H}^{N-1}|_{\Gamma_{s}})\leq\sigma^{-1}\mathcal{H}^{N-1}|_{\Gamma_{t}}.$

Let $\tilde{\Gamma}_{t}=\Gamma_{t}\cap\Omega$. Since $\xi(x)\cdot n(x)\leq 0$
on $\partial\Omega\cap\overline{D}_{1}$ and $\xi(x)\cdot n(x)\geq 0$ on
$\partial\Omega\cap\overline{D}_{2}$, we have

$\Phi^{s,t}(\tilde{\Gamma}_{s})\subset\tilde{\Gamma}_{t},\ \ \forall 0\leq
s\leq t\leq r_{1}\ \ \mbox{and}\ \
\Phi^{s,t}(\tilde{\Gamma}_{s})\subset\tilde{\Gamma}_{t},\ \ \forall 0\geq
s\geq t\geq-r_{1}.$

As before, we suppose that $\mu(\\{x_{0}\\})>0$. Then, for any given
$\epsilon>0$,

$\int_{-\epsilon r_{1}}^{\epsilon
r_{1}}\int_{\tilde{\Gamma}_{t}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}dt=\int_{\cup_{t\in(-\epsilon
r_{1},\epsilon
r_{1})}\tilde{\Gamma}_{t}}w_{\alpha_{k}}^{2}dx\geq\frac{\mu(\\{x_{0}\\})}{2}$

for all large $k$, which gives rise to

$\int_{\tilde{\Gamma}_{s_{k}}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}\geq\frac{\mu(\\{x_{0}\\})}{4\epsilon
r_{1}}$ (2.49)

for $s_{k}\in(-\epsilon r_{1},\epsilon r_{1})$ for all large $k$. We may
assume that $s_{k}\in[0,\epsilon r_{1})$.

By a similar reasoning as to derive (2.48), for all large $k$, it follows that

$\int_{\tilde{\Gamma}_{t_{k}}}w_{\alpha_{k}}^{2}dx<\frac{\mu(\\{x_{0}\\})}{r_{1}}$
(2.50)

for some constant $t_{k}\in(r_{1}-\epsilon r_{1},r_{1})$ and all large $k$. In
addition, we may assume that for all $t\in(0,r_{1})$, $\xi(x)\cdot\nabla
m(x)\geq 0$ on $\Gamma_{t}$.

Now, using (2.46), (2.49), (2.50) and the fact that
$\Phi^{t_{k},s_{k}}(\tilde{\Gamma}_{t_{k}})\subset\tilde{\Gamma}_{s_{k}}$,
similarly to Case 1, we have

$\displaystyle\frac{\mu(\\{x_{0}\\})}{4\epsilon
r_{1}}-\frac{\mu(\\{x_{0}\\})}{\sigma r_{1}}$
$\displaystyle\leq\int_{\tilde{\Gamma}_{s_{k}}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}-\sigma^{-1}\int_{\tilde{\Gamma}_{t_{k}}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}$
$\displaystyle\leq\int_{\tilde{\Gamma}_{s_{k}}}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}-\sigma^{-1}\int_{\Phi^{t_{k},s_{k}}(\tilde{\Gamma}_{t_{k}})}w_{\alpha_{k}}^{2}d\mathcal{H}^{N-1}$
$\displaystyle\leq\int_{\tilde{\Gamma}_{s_{k}}}w_{\alpha_{k}}^{2}(x)d\mathcal{H}^{N-1}-\int_{\tilde{\Gamma}_{s_{k}}}w_{\alpha_{k}}^{2}(\eta(x,t_{k}-s_{k}))d\mathcal{H}^{N-1}$
$\displaystyle=\int_{\tilde{\Gamma}_{s_{k}}}\int_{0}^{t_{k}-s_{k}}-\partial_{t}w_{\alpha_{k}}^{2}(\eta(x,t))dtd\mathcal{H}^{N-1}$
$\displaystyle=\int_{\tilde{\Gamma}_{s_{k}}}\int_{0}^{t_{k}-s_{k}}-2w_{\alpha_{k}}(\eta(x,t))\nabla
w_{\alpha_{k}}(\eta(x,t))\cdot\frac{\xi(\eta(x,t))}{\xi(\eta(x,t))\cdot\tilde{\nu}(\eta(x,t))}dtd\mathcal{H}^{N-1}$
$\displaystyle\leq\frac{2}{\delta}\int_{\tilde{\Gamma}_{s_{k}}}\int_{0}^{t_{k}-s_{k}}2\left(\nabla
w_{\alpha_{k}}(\eta(x,t))\cdot\xi(\eta(x,t))\right)^{-}w_{\alpha_{k}}(\eta(x,t))dtd\mathcal{H}^{N-1}$
$\displaystyle\leq\frac{2}{\delta}\int_{\tilde{\Gamma}_{s_{k}}}\int_{0}^{t_{k}-s_{k}}\left[\left(\nabla
w_{\alpha_{k}}(\eta(x,t))\cdot\xi(\eta(x,t))\right)^{-}\right]^{2}+w_{\alpha_{k}}^{2}(\eta(x,t))dtd\mathcal{H}^{N-1}$
$\displaystyle=\frac{2}{\delta}\int_{0}^{t_{k}-s_{k}}\int_{\tilde{\Gamma}_{s_{k}}}\left[\left(\nabla
w_{\alpha_{k}}(\eta(x,t))\cdot\xi(\eta(x,t))\right)^{-}\right]^{2}+w_{\alpha_{k}}^{2}(\eta(x,t))d\mathcal{H}^{N-1}dt$
$\displaystyle\leq\frac{2}{\delta\sigma}\int_{s_{k}}^{t_{k}}\int_{\tilde{\Gamma}_{t}}\left[\left(\nabla
w_{\alpha_{k}}(x)\cdot\xi(x)\right)^{-}\right]^{2}+w_{\alpha_{k}}^{2}(x)d\mathcal{H}^{N-1}dt$
$\displaystyle\leq\frac{2}{\delta\sigma}\int_{\left(\cup_{t\in[s_{k},t_{k}]}\tilde{\Gamma}_{t}\right)\cap\\{x:\
\xi(x)\cdot\nabla m(x)\geq 0\\}}\left[\nabla
w_{\alpha_{k}}\cdot\xi-\alpha_{k}w_{\alpha_{k}}\xi\cdot\nabla
m\right]^{2}dx+\int_{\Omega}w_{\alpha_{k}}^{2}dx$
$\displaystyle\leq\frac{2}{\delta\sigma}\int_{\Omega}\left[\nabla
w_{\alpha_{k}}\cdot\xi-\alpha_{k}w_{\alpha_{k}}\xi\cdot\nabla
m\right]^{2}+w_{\alpha_{k}}^{2}dx$
$\displaystyle\leq\frac{2(C+1)}{\delta\sigma},$

which is impossible if we send $\epsilon$ to be small enough. Therefore, we
have proved $\mu(\\{x_{0}\\})=0$. ∎

##  3\. One dimensional problem

In this section, we consider the eigenvalue problem with a general boundary
condition in one space dimension:

$\left\\{\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle-\varphi^{\prime\prime}(x)-2\alpha
m^{\prime}(x)\varphi^{\prime}(x)+V(x)\varphi(x)=\lambda\varphi(x),&0<x<1,\\\
\displaystyle-\hbar_{1}\varphi^{\prime}(0)+\ell_{1}\varphi(0)=\hbar_{2}\varphi^{\prime}(1)+\ell_{2}\varphi(1)=0,\end{array}\right.$
(3.1)

where $m\in C^{2}([0,1]),\,V\in C([0,1])$, and the constants
$\hbar_{i},\,\ell_{i}\,(i=1,2)$ satisfy $|\hbar_{i}|+|\ell_{i}|>0\,(i=1,2)$.

As in [44], in this section, unless otherwise specified, we always assume that
(A4): $m$ is not constant and $m^{\prime}(x)$ changes sign at most finitely
many times on $[0,1]$. Hence, $m$ can have various natural kinds of
degeneracy.

Before stating the main results of this section, we need to classify the set
of points of local maximum points of $m$ and introduce some notations as in
[44].

######  Definition 3.

(i) A point of local maximum of $m$ is a point $x\in[0,1]$ such that there is
a small $\epsilon_{0}>0$ such that $m(x)\geq m(y)$ in
$(x-\epsilon_{0},x+\epsilon_{0})\cap[0,1]$, and such $x$ is said to be an
interior point of local maximum if $x\in(0,1)$;

(ii) An isolated point of local maximum of $m$ is a point $x^{I}\in[0,1]$ such
that there is a small $\epsilon_{0}>0$ such that $m(x^{I})>m(x)$ in
$(x^{I}-\epsilon_{0},x^{I}+\epsilon_{0})\cap[0,1]\setminus\\{x^{I}\\}$;

(iii) A segment of local maximum of $m$ is a closed interval
$[a,b]\subset[0,1]$ such that there is a small $\epsilon_{0}>0$ such that $m$
is constant on the closed interval $[a,b]$, $m$ is monotone (non-increasing or
non-decreasing) on $[a-\epsilon_{0},a]\cap[0,1]$ and
$[b,b+\epsilon_{0}]\cap[0,1]$, and for any small $\epsilon>0$, $\\{x\in[0,1]:\
m^{\prime}(x)\not=0\\}\cap(a-\epsilon,a)\not=\emptyset$ if $0<a$ and
$\\{x\in[0,1]:\ m^{\prime}(x)\not=0\\}\cap(b,b+\epsilon)\not=\emptyset$ if
$b<1$.

Note that the definition of an isolated point of local maximum in Definition 3
is exactly that in Definition 2(iii) under the assumption (A4). Moreover, if
(A4) holds, clearly $m$ admits at most finitely many isolated points of local
maximum. For later purpose, we will use the same notations

$[a^{i},b^{j}]\subset[0,1]$ with $0\leq a<b\leq 1$,
$i,j\in\\{\mathcal{N},\,\mathcal{D}\\}$

as in [44] to distinguish all possible segments of local maximum of $m$; see
(1)-(5) in page 2527 of [44]. Here, the capital letters $I$ and $D$ represent
increasing and decreasing, respectively. For example, $[a^{{I}},b^{{D}}]$
means that $m$ increases in a left neighbourhood of $a^{{I}}$ and decreases in
a right neighbourhood of $b^{{D}}$ while $m$ is constant on
$[a^{{I}},b^{{D}}]$. Under the assumption (A4), $m$ admits at most finitely
many segments $[a^{{I}},b^{{D}}]$ and $[a^{{D}},b^{{I}}]$ of local maximum,
and it has at most countably many segments $[a^{{I}},b^{{I}}]$ and
$[a^{{D}},b^{{D}}]$ of local maximum.

As in [44], we have to introduce some further notations as follows. Given a
closed interval $[a,b]\subset[0,1]$ with $0<a<b<1$ and
$i,j\in\\{\mathcal{N},\,\mathcal{D}\\}$, we denote by
$\lambda_{1}^{{ij}}(a,b)$ the principal eigenvalue of the elliptic eigenvalue
problem:

$-\varphi^{\prime\prime}(x)+V(x)\varphi(x)=\lambda\varphi(x),\ a<x<b$

with the Neumann boundary condition (Dirichlet boundary condition,
respectively) at the left boundary point $a$ if $i=\mathcal{N}$ (if
$i=\mathcal{D}$, respectively), and Neumann boundary condition (Dirichlet
boundary condition, respectively) at the right boundary point $b$ if
$j=\mathcal{N}$ (if $j=\mathcal{D}$, respectively).

Given $[0,b]\subset[0,1]$ with $0<b<1$ and
$i\in\\{\mathcal{N},\,\mathcal{D}\\}$, denote by
$\lambda_{1}^{{\mathcal{R}i}}(0,b)$ the principal eigenvalue of

$-\varphi^{\prime\prime}(x)+V(x)\varphi(x)=\lambda\varphi(x),\ 0<x<b;\ \ \
-\hbar_{1}\varphi^{\prime}(0)+\ell_{1}\varphi(0)=0$

with the Neumann boundary condition (Dirichlet boundary condition,
respectively ) at $b$ if $i=\mathcal{N}$ (if $i=\mathcal{D}$, respectively).

Given $[a,1]\subset[0,1]$ with $0<a<1$ and
$i\in\\{\mathcal{N},\,\mathcal{D}\\}$, denote by
$\lambda_{1}^{{i\mathcal{R}}}(a,1)$ the principal eigenvalue of

$-\varphi^{\prime\prime}(x)+V(x)\varphi(x)=\lambda\varphi(x),\ a<x<1;\ \ \
\hbar_{2}\varphi^{\prime}(1)+\ell_{2}\varphi(1)=0$

with the Neumann boundary condition (Dirichlet boundary condition,
respectively ) at $a$ if $i=\mathcal{N}$ (if $i=\mathcal{D}$, respectively).

In view of (A4), the set $\mathcal{M}$ of local maxima of $m$ consists of

$\mathcal{M}=\cup_{i=1}^{9}\mathcal{M}_{i},$

where

$\left.\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\mathcal{M}_{1}=\\{x^{I}_{i}\\}_{i=1}^{h_{1}},\ \ \ \
\mathcal{M}_{2}=\\{[a_{i}^{{I}},b_{i}^{{I}}]\\}_{i=1}^{h_{2}},\ \ \ \
\mathcal{M}_{3}=\\{[a_{i}^{{I}},b_{i}^{{D}}]\\}_{i=1}^{h_{3}},\ \ \ \
\mathcal{M}_{4}=\\{[a_{i}^{{D}},b_{i}^{{I}}]\\}_{i=1}^{h_{4}},\\\ \vskip 6.0pt
plus 2.0pt minus
2.0pt\displaystyle\mathcal{M}_{5}=\\{[a_{i}^{{D}},b_{i}^{{D}}]\\}_{i=1}^{h_{5}},\
\ \mathcal{M}_{6}\subset\\{[0,a^{{I}}]\\},\ \
\mathcal{M}_{7}\subset\\{[0,a^{{D}}]\\},\ \
\mathcal{M}_{8}\subset\\{[a^{{I}},1]\\},\ \
\mathcal{M}_{9}\subset\\{[a^{{D}},1]\\}.\end{array}\right.$

Note that $h_{1},\,h_{3},\,h_{4}$ are finite integers while $h_{2},\,h_{5}$
may be finite integers or infinity. Though some $\mathcal{M}_{i}$ may be
empty, obviously $\mathcal{M}\not=\emptyset$.

For simplicity, let us also set

$\left.\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\mathfrak{L}&=&\displaystyle\min\Big{\\{}\min\\{\lambda_{1}^{\mathcal{ND}}(a^{I}_{i},b^{I}_{i}):\
[a^{I}_{i},b^{I}_{i}]\in\mathcal{M}_{2}\\},\ \
\min\\{\lambda_{1}^{\mathcal{NN}}(a^{I}_{i},b^{D}_{i}):\
[a^{I}_{i},b^{D}_{i}]\in\mathcal{M}_{3}\\},\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt&&\ \ \ \ \ \ \ \
\displaystyle\min\\{\lambda_{1}^{\mathcal{DD}}(a^{D}_{i},b^{I}_{i}):\
[a^{D}_{i},b^{I}_{i}]\in\mathcal{M}_{4}\\},\ \
\min\\{\lambda_{1}^{\mathcal{DN}}(a^{D}_{i},b^{D}_{i}):\
[a^{D}_{i},b^{D}_{i}]\in\mathcal{M}_{5}\\}\Big{\\}}.\end{array}\right.$

In [44], the situation

$\hbar_{1}\ell_{1}\geq 0,\ \ \hbar_{2}\ell_{2}\geq 0$ (3.2)

was studied. In this case, it is clear that
$\lambda_{1}(\alpha)\geq\min_{x\in[0,1]}V(x)$ for all $\alpha$. The main
result for problem (3.1) obtained in [44] reads as follows.

######  Theorem 3.1.

Assume that (A4) and (3.2) hold. Then we have the following assertions.

  * (i)

$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ if and only if

(i-1) $\mathcal{M}=\mathcal{M}_{1}\subset\\{0,1\\}$ when $\ell_{1}\not=0$ and
$\ell_{2}\not=0$.

(i-2) $\mathcal{M}=\mathcal{M}_{1}=\\{0\\}$ when $\ell_{1}\not=0$ and
$\ell_{2}=0$.

(i-3) $\mathcal{M}=\mathcal{M}_{1}=\\{1\\}$ when $\ell_{1}=0$ and
$\ell_{2}\not=0$.

  * (ii)

Otherwise, $-\infty<\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)<\infty$,
and

$\left.\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)&=&\vskip
6.0pt plus 2.0pt minus 2.0pt\displaystyle\min\Big{\\{}\mathfrak{L},\ \
\min\\{V(x):\ x\in\mathcal{M}_{1}\setminus\\{0,\,1\\}\\},\ \
\lambda_{1}^{\mathcal{RD}}(0,a^{I})\ (\mbox{if}\
[0,a^{I}]\in\mathcal{M}_{6}),\\\ &&\ \ \ \ \ \ \ \vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lambda_{1}^{\mathcal{RN}}(0,a^{D})\ (\mbox{if}\
[0,a^{D}]\in\mathcal{M}_{7}),\ \ \lambda_{1}^{\mathcal{NR}}(a^{I},1)\
(\mbox{if}\ [a^{I},1]\in\mathcal{M}_{8}),\\\ &&\ \ \ \ \ \ \ \vskip 6.0pt plus
2.0pt minus 2.0pt\displaystyle\lambda_{1}^{\mathcal{DR}}(a^{D},1)\ (\mbox{if}\
[a^{D},1]\in\mathcal{M}_{9}),\ \ V(0)\ (\mbox{if\ $0\in\mathcal{M}_{1}$ and
$\ell_{1}=0$}),\\\ &&\ \ \ \ \ \ \ \displaystyle V(1)\ (\mbox{if\
$1\in\mathcal{M}_{1}$ and $\ell_{2}=0$})\,\Big{\\}}.\end{array}\right.$

In the sequel, we will consider all the remaining cases. Our first result
concerns the case of

$\hbar_{1}\ell_{1}<0,\ \ \ \hbar_{2}\ell_{2}\geq 0.$ (3.3)

Indeed, we have the following conclusion.

######  Theorem 5.

Assume that (A4) and (3.3) hold. Then we have the following assertions.

  * (i)

$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ if and only if
$\mathcal{M}=\mathcal{M}_{1}=\\{1\\}$ when $\ell_{2}\not=0$.

  * (ii)

$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty$ if and only if
$\mathcal{M}=\mathcal{M}_{1}=\\{0\\}$.

  * (iii)

Otherwise, $-\infty<\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)<\infty$
and

$\left.\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)&=&\vskip
6.0pt plus 2.0pt minus 2.0pt\displaystyle\min\Big{\\{}\mathfrak{L},\ \
\min\\{V(x):\ x\in\mathcal{M}_{1}\setminus\\{0,\,1\\}\\},\ \
\lambda_{1}^{\mathcal{RD}}(0,a^{I})\ (\mbox{if}\
[0,a^{I}]\in\mathcal{M}_{6}),\\\ &&\ \ \ \ \ \ \ \vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lambda_{1}^{\mathcal{RN}}(0,a^{D})\ (\mbox{if}\
[0,a^{D}]\in\mathcal{M}_{7}),\ \ \lambda_{1}^{\mathcal{NR}}(a^{I},1)\
(\mbox{if}\ [a^{I},1]\in\mathcal{M}_{8}),\\\ &&\ \ \ \ \ \ \ \vskip 6.0pt plus
2.0pt minus 2.0pt\displaystyle\lambda_{1}^{\mathcal{DR}}(a^{D},1)\ (\mbox{if}\
[a^{D},1]\in\mathcal{M}_{9}),\ V(1)\ (\mbox{if\ $1\in\mathcal{M}_{1}$ and
$\ell_{2}=0$})\,\Big{\\}}.\end{array}\right.$

Parallel to case (3.3), for the case

$\hbar_{1}\ell_{1}\geq 0,\ \ \ \hbar_{2}\ell_{2}<0,$ (3.4)

we can state the following result.

######  Theorem 6.

Assume that (A4) and (3.4) hold. Then we have the following assertions.

  * (i)

$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ if and only if
$\mathcal{M}=\mathcal{M}_{1}=\\{0\\}$ when $\ell_{1}\not=0$.

  * (ii)

$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty$ if and only if
$\mathcal{M}=\mathcal{M}_{1}=\\{1\\}$.

  * (iii)

Otherwise, $-\infty<\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)<\infty$
and

$\left.\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)&=&\vskip
6.0pt plus 2.0pt minus 2.0pt\displaystyle\min\Big{\\{}\mathfrak{L},\ \
\min\\{V(x):\ x\in\mathcal{M}_{1}\setminus\\{0,\,1\\}\\},\ \
\lambda_{1}^{\mathcal{RD}}(0,a^{I})\ (\mbox{if}\
[0,a^{I}]\in\mathcal{M}_{6}),\\\ &&\ \ \ \ \ \ \ \vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lambda_{1}^{\mathcal{RN}}(0,a^{D})\ (\mbox{if}\
[0,a^{D}]\in\mathcal{M}_{7}),\ \ \lambda_{1}^{\mathcal{NR}}(a^{I},1)\
(\mbox{if}\ [a^{I},1]\in\mathcal{M}_{8}),\\\ &&\ \ \ \ \ \ \ \vskip 6.0pt plus
2.0pt minus 2.0pt\displaystyle\lambda_{1}^{\mathcal{DR}}(a^{D},1)\ (\mbox{if}\
[a^{D},1]\in\mathcal{M}_{9}),\ V(0)\ (\mbox{if\ $0\in\mathcal{M}_{1}$ and
$\ell_{1}=0$})\,\Big{\\}}.\end{array}\right.$

Finally, we consider the case

$\hbar_{1}\ell_{1}<0,\ \ \ \hbar_{2}\ell_{2}<0,$ (3.5)

and can obtain the following result.

######  Theorem 7.

Assume that (A4) and (3.5) hold. Then we have the following assertions.

  * (i)

$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty$ if and only if
$\mathcal{M}=\mathcal{M}_{1}\subset\\{0,1\\}$.

  * (ii)

Otherwise, $-\infty<\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)<\infty$
and

$\left.\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)&=&\vskip
6.0pt plus 2.0pt minus 2.0pt\displaystyle\min\Big{\\{}\mathfrak{L},\ \
\min\\{V(x):\ x\in\mathcal{M}_{1}\setminus\\{0,\,1\\}\\},\ \
\lambda_{1}^{\mathcal{RD}}(0,a^{I})\ (\mbox{if}\
[0,a^{I}]\in\mathcal{M}_{6}),\\\ &&\ \ \ \ \ \ \ \vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lambda_{1}^{\mathcal{RN}}(0,a^{D})\ (\mbox{if}\
[0,a^{D}]\in\mathcal{M}_{7}),\ \ \lambda_{1}^{\mathcal{NR}}(a^{I},1)\
(\mbox{if}\ [a^{I},1]\in\mathcal{M}_{8}),\\\ &&\ \ \ \ \ \ \ \vskip 6.0pt plus
2.0pt minus 2.0pt\displaystyle\lambda_{1}^{\mathcal{DR}}(a^{D},1)\ (\mbox{if}\
[a^{D},1]\in\mathcal{M}_{9})\,\Big{\\}}.\end{array}\right.$

######  Remark 8.

Regarding Theorem 3.1 and Theorems 5-7, we would like to make the following
comments.

  * (i)

Once $\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)$ is finite, it can be
given by Theorem 3.1(i), Theorem 5(iii), Theorem 6(iii) and Theorem 7(ii),
which exhaust all the possibilities of the parameters
$\hbar_{i},\,\ell_{i}\,(i=1,2)$. Moreover, we observe that if $0$ or $1$ is an
isolated point of local maximum of $m$,
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)$, when finite, is affected
by such a boundary point only if the Neumann boundary condition is prescribed
there.

  * (ii)

The limit $\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty$ only if
either $\hbar_{1}\ell_{1}<0$ or $\hbar_{2}\ell_{2}<0$. Precisely speaking,
when $\hbar_{1}\ell_{1}<0$,
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty$ if and only if $0$
is an isolated local maximum of $m$ (see Theorem 5(ii)); when
$\hbar_{2}\ell_{2}<0$,
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty$ if and only if $1$
is an isolated local maximum of $m$ (see Theorem 6(ii)); when
$\hbar_{1}\ell_{1}<0$ and $\hbar_{2}\ell_{2}<0$,
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty$ if and only if $1$
or $1$ is an isolated local maximum of $m$ (see Theorem 7(i)). It should be
stressed here that all these assertions remain valid without the assumption
(A4).

  * (iii)

Whether $\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ or not
depends on the sign $\hbar_{1}\ell_{1},\,\hbar_{2}\ell_{2}$ and the
monotonicity of $m$. More precisely, we can conclude the following statements
from Theorem 3.1 and Theorems 5-7.

    * (iii-1)

If $\hbar_{1}\ell_{1}\geq 0,\,\hbar_{2}\ell_{2}\geq 0$ and
$\ell_{1},\,\ell_{2}>0$, then
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ if and only if
either $m$ increases or decreases on $[0,1]$, or $m$ decreases on $[0,\theta]$
while increases on $[\theta,1]$ for some $\theta\in(0,1)$.

    * (iii-2)

If either $\hbar_{1}\ell_{1}<0,\,\hbar_{2}\ell_{2}\geq 0$, $\ell_{2}\not=0$ or
$\ell_{1}=0,\,\ell_{2}>0$, then
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ if and only if $m$
increases on $[0,1]$.

    * (iii-3)

If either $\hbar_{1}\ell_{1}\geq 0,\,\hbar_{2}\ell_{2}<0$, $\ell_{1}\not=0$ or
$\ell_{1}>0,\,\ell_{2}=0$, then
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ if and only if $m$
decreases on $[0,1]$.

Before presenting the proof of Theorems 5-7, we would like to demonstrate,
through several concrete examples, the impact of the boundary critical points
of the advection function $m$ and the boundary conditions on the asymptotic
behavior of the principal eigenvalue $\lambda_{1}(\alpha)$. We will mainly
focus on the situation where the degeneracy of $m$ occurs near the left
boundary point $x=0$ since similar behaviors happen near the right boundary
point $x=1$. For the impact of interior critical points of $m$ on the
asymptotic behavior of $\lambda_{1}(\alpha)$, one may refer to [44] for some
typical examples.

To emphasize the dependence of $\lim_{\alpha\to\infty}\lambda_{1}(\alpha)$ on
the sign of the boundary parameters $\hbar_{i},\,\ell_{i}\,(i=1,2)$ as well as
for convenience, instead of $\lim_{\alpha\to\infty}\lambda_{1}(\alpha)$, below
we will use $\lambda_{1}^{a_{i},a_{j}}(\infty)\,(i=1,2)$ with $a_{i}\,(i=1,2)$
given by

$a_{i}=+\ \text{ if }\,\hbar_{i}\ell_{i}>0\ \text{or}\ \hbar_{i}=0;\ \
a_{i}=0\ \text{ if }\ \ell_{i}=0;\ \ a_{i}=-\ \text{ if }\
\hbar_{i}\ell_{i}<0.$

Example 1: $m$ is constant on $[0,x_{1}]$ and increases on $[x_{1},1]$; see
Figure 1(a). Let $x_{1}$ shrink to the boundary point $0$ so that $m$
increases on $[0,1]$; see Figure 1(b).

Figure 1(a) Figure 1(b) Figure 2(a) Figure 2(b)

According to the results of this section, if $m$ is given as in Figure 1(a),
we have the following conclusions:

$\left.\begin{array}[]{llll}\lambda_{1}^{+,+}=\lambda_{1}^{\mathcal{RD}}(0,x_{1});&\lambda_{1}^{+,0}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(1)\\};&\lambda_{1}^{0,0}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(1)\\};\\\
\lambda_{1}^{0,-}=-\infty;&\lambda_{1}^{-,0}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(1)\\};&\lambda_{1}^{-,-}=-\infty.\end{array}\right.$

It should be pointed out that the values $\lambda_{1}^{\mathcal{RD}}(0,x_{1})$
appearing above are different because they depend on the boundary conditions
at $x=0$. Hereafter, we shall use the same notation whenever no confusion is
caused.

If $m$ is given as in Figure 1(b), we have

$\lambda_{1}^{+,+}=\infty;\ \
\lambda_{1}^{+,0}=\lambda_{1}^{0,0}=\lambda_{1}^{-,0}=V(1);\ \
\lambda_{1}^{0,-}=\lambda_{1}^{-,-}=-\infty.$

Notice that when the point $x_{1}$ shrinks to the boundary point $0$, all the
values $\lambda_{1}^{\mathcal{RD}}(0,x_{1})$ converge to $\infty$. Thus, the
asymptotic behavior of $\lim_{\alpha\to\infty}\lambda_{1}(\alpha)$ changes
continuously when $m$ changes continuously from Figure 1(a) to Figure 1(b).

Example 2: $m$ is constant on $[0,x_{1}]$, increases on $[x_{1},x_{2}]$ and
decreases on $[x_{2},1]$; see Figure 2(a). Let $x_{1}$ shrink to the boundary
point $0$ so that $m$ increases on $[0,x_{1}]$ and decreases on $[x_{2},1]$;
see Figure 2(b). If $m$ is given as in Figure 2(a), we have the following
conclusions:

$\left.\begin{array}[]{llll}\lambda_{1}^{+,+}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(x_{2})\\};&\lambda_{1}^{+,0}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(x_{2})\\};&\lambda_{1}^{0,0}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(x_{2})\\};\\\
\lambda_{1}^{0,-}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(x_{2})\\};&\lambda_{1}^{-,0}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(x_{2})\\};&\lambda_{1}^{-,-}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(x_{2})\\}.\end{array}\right.$

If $m$ is given as in Figure 2(b), we have

$\lambda_{1}^{+,+}=\lambda_{1}^{+,0}=\lambda_{1}^{0,0}=\lambda_{1}^{-,0}=\lambda_{1}^{0,-}=\lambda_{1}^{-,-}=V(x_{2}).$

Notice that when the point $x_{1}$ shrinks to the boundary point $0$, all the
values $\lambda_{1}^{\mathcal{RD}}(0,x_{1})$ converge to $\infty$. Again, the
asymptotic behavior of $\lim_{\alpha\to\infty}\lambda_{1}(\alpha)$ changes
continuously when $m$ changes continuously from Figure 2(a) to Figure 2(b).

Example 3: $m$ is constant on $[0,x_{1}]$ and decreases on $[x_{1},1]$; see
Figure 3(a). Let $x_{1}$ shrink to the boundary point $0$ so that $m$
decreases on $[0,1]$; see Figure 3(b).

Figure 3(a) Figure 3(b) Figure 4(a) Figure 4(b)

If $m$ is given as in Figure 3(a), we have

$\left.\begin{array}[]{llll}\lambda_{1}^{+,+}=\lambda_{1}^{\mathcal{RD}}(0,x_{1});&\
\ \ \lambda_{1}^{+,0}=\lambda_{1}^{\mathcal{RD}}(0,x_{1});&\ \ \
\lambda_{1}^{0,0}=\lambda_{1}^{\mathcal{RD}}(0,x_{1});\\\
\lambda_{1}^{0,-}=\lambda_{1}^{\mathcal{RD}}(0,x_{1});&\ \ \
\lambda_{1}^{-,0}=\lambda_{1}^{\mathcal{RD}}(0,x_{1});&\ \ \
\lambda_{1}^{-,-}=\lambda_{1}^{\mathcal{RD}}(0,x_{1}).\end{array}\right.$
(3.6)

If $m$ is given as in Figure 3(b), we have

$\lambda_{1}^{+,+}=\lambda_{1}^{+,0}=\infty;\ \
\lambda_{1}^{0,0}=\lambda_{1}^{0,-}=V(0);\ \
\lambda_{1}^{-,0}=\lambda_{1}^{-,-}=-\infty.$

Note that in (3.6), as $x_{1}$ shrinks to the boundary point $0$, it is easily
checked that
$\lambda_{1}^{+,+}=\lambda_{1}^{\mathcal{RD}}(0,x_{1})\to\infty,\,\lambda_{1}^{+,0}=\lambda_{1}^{\mathcal{RD}}(0,x_{1})\to\infty$,
$\lambda_{1}^{0,0}=\lambda_{1}^{\mathcal{RD}}(0,x_{1})\to V(0)$,
$\lambda_{1}^{0,-}=\lambda_{1}^{\mathcal{RD}}(0,x_{1})\to V(0)$, while
$\lambda_{1}^{-,0}=\lambda_{1}^{\mathcal{RD}}(0,x_{1})\to-\infty,\,\lambda_{1}^{-,-}=\lambda_{1}^{\mathcal{RD}}(0,x_{1})\to-\infty$.
Thus, in a certain sense, the asymptotic behavior of
$\lim_{\alpha\to\infty}\lambda_{1}(\alpha)$ changes continuously when $m$
changes continuously from Figure 3(a) to Figure 3(b).

Example 4: $m$ is constant on $[0,x_{1}]$, decreases on $[x_{1},x_{2}]$ for
some $x_{2}\in(x_{1},1)$ and increases on $[x_{2},1]$; see Figure 4(a). Let
$x_{1}$ shrink to the boundary point $0$ so that $m$ decreases on $[0,x_{2}]$
and increases on $[x_{2},1]$; see Figure 4(b). If $m$ is given as in Figure
4(a), we have

$\left.\begin{array}[]{llll}\lambda_{1}^{+,+}=\lambda_{1}^{\mathcal{RD}}(0,x_{1});&\lambda_{1}^{+,0}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(1)\\};&\lambda_{1}^{0,0}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(1)\\};\\\
\lambda_{1}^{0,-}=-\infty;&\lambda_{1}^{-,0}=\min\\{\lambda_{1}^{\mathcal{RD}}(0,x_{1}),\,V(1)\\};&\lambda_{1}^{-,-}=-\infty.\end{array}\right.$

If $m$ is given as in Figure 4(b), we have

$\lambda_{1}^{+,+}=\infty;\ \
\lambda_{1}^{+,0}=\lambda_{1}^{0,0}=\lambda_{1}^{-,0}=V(1);\ \
\lambda_{1}^{0,-}=\lambda_{1}^{-,-}=-\infty.$

Once again, it is easily seen that in a certain sense, the asymptotic behavior
of $\lim_{\alpha\to\infty}\lambda_{1}(\alpha)$ changes continuously when $m$
changes continuously from Figure 4(a) to Figure 4(b).

In what follows, we only give the proof of Theorem 5 since Theorem 6 and
Theorem 7 can be verified similarly.

###### Proof of Theorem 5.

As before, through the transformation $w(x)=e^{\alpha m(x)}\varphi(x)$,
problem (3.1) reduces to the following one:

$\left\\{\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle-w^{\prime\prime}(x)+[\alpha^{2}(m^{\prime}(x))^{2}+\alpha
m^{\prime\prime}(x)+V(x)]w(x)=\lambda_{1}(\alpha)w(x),\ 0<x<1,\\\
\displaystyle(\alpha\hbar_{1}+\ell_{1})w(0)-\hbar_{1}w^{\prime}(0)=(-\alpha\hbar_{2}+\ell_{2})w(1)+\hbar_{2}w^{\prime}(1)=0.\end{array}\right.$
(3.7)

For each $\alpha$, when $\hbar_{2}\not=0$, $\lambda_{1}(\alpha)$ has the
variational characterization:

$\left.\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lambda_{1}(\alpha)&=&\displaystyle\min_{w\in
H^{1}(0,1),\,\int_{0}^{1}w^{2}dx=1}\Big{\\{}\int_{0}^{1}[(w^{\prime}-\alpha
wm^{\prime})^{2}+Vw^{2}]dx+{{\ell_{1}}\over{\hbar_{1}}}w^{2}(0)+{{\ell_{2}}\over{\hbar_{2}}}w^{2}(1)\Big{\\}},\end{array}\right.$
(3.8)

and when $\hbar_{2}=0$, $\lambda_{1}(\alpha)$ admits the variational
characterization:

$\left.\begin{array}[]{lll}\vskip 6.0pt plus 2.0pt minus
2.0pt\displaystyle\lambda_{1}(\alpha)&=&\displaystyle\min_{w\in
H^{1}_{*}(0,1),\,\int_{0}^{1}w^{2}dx=1}\int_{0}^{1}[(w^{\prime}-\alpha
wm^{\prime})^{2}+Vw^{2}]dx+{{\ell_{1}}\over{\hbar_{1}}}w^{2}(0),\end{array}\right.$
(3.9)

where $H^{1}_{*}(0,1)=\\{g\in H^{1}(0,1):\,g(1)=0\\}$.

As before, we shall use $w_{\alpha}$ with $\int_{0}^{1}w_{\alpha}^{2}(x)dx=1$
to denote the principal eigenfunction of (3.7), and assume that, as
$\alpha\to\infty$, $w_{\alpha}^{2}$ converges weakly to a unique Radon measure
$\mu$ with $\mu([0,1])=1$.

We first assume that $\ell_{2}\not=0$ and
$\mathcal{M}=\mathcal{M}_{1}=\\{1\\}$, and claim
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$. Obviously, the
condition $\mathcal{M}=\mathcal{M}_{1}=\\{1\\}$ implies that the function
advection $m$ must be increasing on $[0,1]$, and so $m^{\prime}(x)\geq 0$ for
all $x\in[0,1]$.

To show $\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$, we proceed
indirectly and suppose that there exists a sequence of $\alpha$, labelled by
itself for convenience, such that $\lambda_{1}(\alpha)\leq C$ for some
constant $C>0$ and $\alpha\geq 0$. Hereafter, the positive constant $C$ may
change from place to place but does not depend on $\alpha\geq 0$.

As a result, it follows from (3.3), (3.8) and (3.9) that

$\int_{0}^{1}[(w_{\alpha}^{\prime}-\alpha
w_{\alpha}m^{\prime})^{2}+Vw_{\alpha}^{2}]dx+{{\ell_{1}}\over{\hbar_{1}}}w_{\alpha}^{2}(0)\leq
C,\ \ \forall\alpha\geq 0.$ (3.10)

On the other hand, from the proof of Theorem 3 (see (2.41)), we obtain

$-{{\ell_{1}}\over{\hbar_{1}}}w_{\alpha}^{2}(0)\leq\frac{1}{2}\int_{0}^{1}[(w_{\alpha}^{\prime}-\alpha
w_{\alpha}m^{\prime})^{2}dx+C,\ \ \forall\alpha\geq 0.$ (3.11)

Thus, because of $\ell_{1}\hbar_{1}<0$, we infer that

$-{{\ell_{1}}\over{\hbar_{1}}}w_{\alpha}^{2}(0)\leq C,\ \
\int_{0}^{1}[(w_{\alpha}^{\prime}-\alpha w_{\alpha}m^{\prime})^{2}dx\leq C,\ \
\forall\alpha\geq 0.$ (3.12)

Combined with (3.12) and the fact that $m$ is increasing on $[0,1]$, one can
use the same analysis as in the proofs of [44, Lemma 2.6 and Lemma 2.8] to
conclude that $\mu([0,1))=1$. Hence, $\mu(\\{1\\})=1$, which, together with
$\int_{0}^{1}w_{\alpha}^{2}(x)dx=1$ for all $\alpha$, enables us to conclude
that

$\limsup_{\alpha\to\infty}\|w_{\alpha}(\cdot)\|_{L^{\infty}((0,1))}=\infty.$

By virtue of this and (3.12), an argument similar to that of [44, Lemma 2.7]
can be employed to lead to a contradiction. Therefore,
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=\infty$ must hold.

We then claim $\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty$ under
the assumption $\mathcal{M}=\mathcal{M}_{1}=\\{0\\}$. Clearly,
$\ell_{1}\hbar_{1}<0$, $\mathcal{M}=\mathcal{M}_{1}=\\{0\\}$ and (3.3) imply
$\Sigma_{3}=\\{0\\}$. Thus, Lemma 2 gives
$\lim\limits_{\alpha\to\infty}\lambda_{1}(\alpha)=-\infty$, as desired.

If $\ell_{2}=0$, it is easily seen from (3.8) that
$\lambda_{1}(\alpha)\leq\max_{x\in[0,1]}V(x)$ for all $\alpha\geq 0$.

If $\ell_{2}\not=0$ and $1$ is not an isolated local maximum of $m$, then by
(A4), either $m$ is constant on $[1-a_{0},1]$ or $m$ is decreasing on
$[1-a_{0},1]$ for some small $a_{0}>0$. In view of $\ell_{1}\hbar_{1}<0$, the
analysis of [44, Lemma 2.1] can be used to show that
$\limsup\limits_{\alpha\to\infty}\lambda_{1}(\alpha)<\infty$. This shows that
$\lambda_{1}(\alpha)\leq C$ for all $\alpha\geq 0$ when either $\ell_{2}=0$ or
$1$ is not an isolated maximum of $m$. Thus, using (3.10) and (3.11), we see
that (3.12) holds.

On the other hand, if $\mathcal{M}\not=\\{0\\}$, then by (A4), either $m$ is
increasing on $[0,b_{0}]$ or $m$ is constant on $[0,b_{0}]$ for some small
$b_{0}>0$. In each case, by Lemma 4, we know that
$\liminf\limits_{\alpha\to\infty}\lambda_{1}(\alpha)>-\infty$.

Consequently, if either $\ell_{2}=0$ or $1$ is not an isolated maximum of $m$
and $\mathcal{M}\not=\\{0\\}$, from the above argument we can conclude that

$-\infty<\liminf\limits_{\alpha\to\infty}\lambda_{1}(\alpha)\leq\limsup\limits_{\alpha\to\infty}\lambda_{1}(\alpha)<\infty,$

and (3.12) holds. Then, one can safely follow the argument of [44, Theorem
1.2] to establish Theorem 5(iii). This completes the proof. ∎

Data availability Data sharing not applicable to this article as no datasets
were generated or analysed during the current study.

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