\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 64, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/64\hfil Asymptotic shape of solutions]
{Asymptotic shape of solutions to
the perturbed simple pendulum problems}

\author[T. Shibata\hfil EJDE-2007/64\hfilneg]
{Tetsutaro Shibata}

\address{Tetsutaro Shibata \newline
Department of Applied Mathematics, Graduate School of Engineering,
Hiroshima University, Higashi-Hiroshima, 739-8527, Japan}
\email{shibata@amath.hiroshima-u.ac.jp}

\thanks{Submitted May 12, 2006. Published May 9, 2007.}
\subjclass[2000]{35J60}
\keywords{Asymptotic formulas; $L^q$-norm; simple pendulum}

\begin{abstract}
 We consider the positive solution of
 the perturbed simple pendulum problem
 $$
 u''(r) + \frac{N-1}{r}u'(r) - g(u(t)) + \lambda \sin u(r) = 0,
 $$
 with $0 < r < R$, $ u'(0) = u(R) = 0$. To understand well
 the shape of the solution $u_\lambda$ when $\lambda \gg 1$,
 we establish the leading and second terms of $\Vert u_\lambda\Vert_q$
 ($1 \le q < \infty$) with the estimate of third term as
 $\lambda \to \infty$. We also obtain the asymptotic formula for
 $u_\lambda'(R)$ as $\lambda \to \infty$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\section{Introduction}


We consider the perturbed simple pendulum problem
\begin{gather}
u''(r) + \frac{N-1}{r}u'(r) - g(u(t)) + \lambda \sin u(r) = 0,
\quad  0 < r < R, \label{e1.1} \\
u(r) > 0, \quad 0 \le r < R, \label{e1.2} \\
u'(0) = u(R) = 0, \label{e1.3}
\end{gather}
where $N \ge 2$, $R > 0$ is a constant and $\lambda > 0$ is a parameter.
We assume the following conditions:

\begin{itemize}
\item[(A1)] $g \in C^{m,\gamma}(\mathbb{R})$ ($m \ge 1, 0 < \gamma < 1$)
and $g(u) > 0$ for $u > 0$.

\item[(A2)] $g(0) = g'(0) = 0$.

\item[(A3)] $g(u)/u$ is strictly increasing for $0 < u < \pi$.

\end{itemize}

A typical example of $g(u)$ is $g(u) = \vert u\vert^{m-1}u$
$(m > 1)$ and $g(u)$ is regarded as the nonlinear
self-interaction term of the simple pendulum equation. The
following properties (P1) and (P2) are well-known and easy to show
(cf. \cite{c,f,h1}).

\begin{itemize}
\item[(P1)] For a given $\lambda \in \mathbb{R}$,
\eqref{e1.1}--\eqref{e1.3} has a unique solution
$u_\lambda \in C^3([0, R])$
if and only if
$\lambda > \lambda_1$, where $\lambda_1 > 0$ is the first eigenvalue
of $-\Delta$ in $B_R = \{\vert x \vert < R\} \subset \mathbb{R}^N$
with Dirichlet zero boundary condition.

\item[(P2)] $\Vert u_\lambda \Vert_\infty < \pi$
and $u_\lambda \to \pi$ uniformly on any compact interval in $[0, R)$ as
$\lambda \to \infty$.

\end{itemize}
Therefore, we see from (P2) that $u_\lambda$ is almost flat inside
$[0, R)$. The purpose of this paper is
to understand well the asymptotic behavior
of $u_\lambda$ as $\lambda \to \infty$ not from a local point of view, but
from a viewpoint of total shape of $u_\lambda$ in $[0, R]$. To this end,
we establish the precise asymptotic formula for $\Vert u_\lambda\Vert_q$
($1 \le q < \infty$) as $\lambda \to \infty$. Here $\Vert u\Vert_q
:= \vert S^{N-1}\vert\int_0^R r^{N-1}\vert u(r)\vert^q dr$ and
$\vert S^{N-1}\vert$ is the measure of $S^{N-1} = \{\vert x\vert = R\}$.

Singularly perturbed equations have been investigated by many authors.
We refer to \cite{h,o1,o2,s,s1}
 and the references therein.
In particular, one of the main concern in this area is to
investigate asymptotic shapes of the corresponding solutions.

As for the pointwise behavior of the solution
$u_\lambda$ of \eqref{e1.1}--\eqref{e1.3} as $\lambda \to \infty$,
there are some known results.
Let us consider the case $N = 1$ in the interval $(-R, R)$ and
$g \equiv 0$. We denote by $u_{0,\lambda}$ the unique solution
associated with given $\lambda \gg 1$. Then it is known
(cf. \cite{s0}) that as $\lambda \to \infty$
\begin{equation} \label{e1.4}
\Vert u_{0,\lambda}\Vert_\infty = \pi - 8(1 + o(1))
e^{-\sqrt{\lambda}(1 + o(1))R}.
\end{equation}
We remark that the second term in the righthand side of the
equation decays \emph{exponentially} as $\lambda \to \infty$.
Furthermore, when $N \ge 2$, $g(u) \not \equiv 0$ and satisfies
(A.1)--(A.3), the following asymptotic formula has been obtained
in \cite{s3}.

\begin{theorem}[\cite{s3}] \label{thm1}
 Let an arbitrary $0 \le r < R$ be fixed. Then
the following asymptotic formula for the solution $u_\lambda$
of \eqref{e1.1}--\eqref{e1.3} holds
as $\lambda \to \infty$.
\begin{equation} \label{e1.5}
u_\lambda(r) = \pi
- \sum_{k = 1}^{m+1}\frac{b_k}{\lambda^k}
+ o\big(\frac{1}{\lambda^{m+1}}\big),
\end{equation}
where $b_1 = g(\pi)$ and $b_k$ ($k = 2, 3, \dots, m+1$) are constants
determined by
$$
\{g^{(j)}(\pi)\}_{j=0}^{k-1}.
$$
\end{theorem}

In particular, we see from \eqref{e1.5} that
\begin{equation} \label{e1.6}
\Vert u_\lambda\Vert_\infty = \pi
- \sum_{k = 1}^{m+1}\frac{b_k}{\lambda^k}
+ o\big(\frac{1}{\lambda^{m+1}}\big).
\end{equation}

Theorem \ref{thm1} gives us the precise pointwise
information about $u_\lambda$ inside $[0, R)$
as $\lambda \to \infty$.
However, if we consider the asymptotic behavior of
$\Vert u_\lambda\Vert_q$ as $\lambda \to \infty$
($1 \le q < \infty$) for the better understanding of the
\emph{total shape} of $u_\lambda$ in $[0, R]$, then it is natural that
$\Vert u_\lambda\Vert_q$ is affected by both
the interior behavior of $u_\lambda$ and the behavior near
the boundary. Therefore, it is
expected that the asymptotic formula for
$\Vert u_\lambda\Vert_q$ ($1 \le q < \infty$) is different from \eqref{e1.6}.

Now we state our main results. Let $G(u) := \int_0^u g(s)ds$.


\begin{theorem} \label{thm2}
 Let $1 \le q < \infty$ be fixed. Then for
any fixed $0 < \delta < 1/4$, the
following asymptotic formula holds
as $\lambda \to \infty$:
\begin{equation} \label{e1.7}
\Vert u_\lambda\Vert_q = \vert B_R\vert^{1/q}
\Big(\pi - \frac{NC_0}{\pi^{q-1}qR}\lambda^{-1/2}
+ O\big(\lambda^{-(1/2+\delta)}\big)\Big),
\end{equation}
where $\vert B_R\vert$ is the volume of $B_R$ and
$$
C_0 = \int_0^\pi \frac{\pi^q - \theta^q}
{\sqrt{2(1 + \cos\theta)}}d\theta.
$$
\end{theorem}


\begin{theorem} \label{thm3}
 The following asymptotic formula holds
as $\lambda \to \infty$:
\begin{equation} \label{e1.8}
\vert u_\lambda'(R)\vert = 2\sqrt{\lambda} - \frac{2(N-1)}{R} + o(1).
\end{equation}
\end{theorem}

We see from Theorems \ref{thm1}, \ref{thm2}, \ref{thm3}
 that the second term of
$\Vert u_\lambda\Vert_q$ as $\lambda \to \infty$
is mainly affected by the slope of the boundary layer $u_\lambda'(R)$.
It should also be mentioned that for the case $N =1$,
the exact third term of
$\Vert u_\lambda\Vert_1$
has been obtained in \cite{s2}.

We briefly explain the difficulty to treat the case $N \ge 2$.
To prove Theorem \ref{thm2}, we calculate
$\Vert u_\lambda\Vert_q$ which is affected both the behavior of $u_\lambda$
inside $[0, R)$ and near the boundary. Moreover, \eqref{e1.1}
contains the term $(N-1)u_\lambda'(r)/r$, which is quite difficult to
treat and does not appear when $N = 1$. Therefore, the calculation to
obtain the remainder estimate in \eqref{e1.7} is quite delicate and complicated.
This is the reason why we need the restriction $0 < \delta < 1/4$.

\section{Proof of Theorem \ref{thm2}}

In what follows, $C$ denotes various positive constants independent of
$\lambda \gg 1$.
We begin with the fundamental properties of $u_\lambda$.
It is well known that
\begin{equation} \label{e2.1}
u_\lambda(0) = \Vert u_\lambda\Vert_\infty, \quad
u_\lambda'(r) < 0 \quad (0 < r \le R).
\end{equation}
Multiply \eqref{e1.1} by $u_\lambda'$. Then for
$r \in [0, R]$,
$$
\big\{u_\lambda''(r) + \frac{N-1}{r}u_\lambda'(r) +
\lambda\sin u_\lambda(r)-g(u_\lambda(r))\big\}u_\lambda'(r) = 0.
$$
By \eqref{e2.1}, this implies that for
$r \in [0, R]$,
\begin{equation} \label{e2.2}
\begin{aligned}
& \frac12u_\lambda'(r)^2 + \int_0^r \frac{N-1}{s}u_\lambda'(s)^2ds
- \lambda \cos u_\lambda(r) - G(u_\lambda(r))
\equiv \quad \mbox{constant}
\\
&= -\lambda\cos\Vert u_\lambda\Vert_\infty - G(\Vert u_\lambda\Vert_\infty)
\quad (\mbox{put $r = 0$})
\\
&= \frac12u_\lambda'(R)^2 + \int_0^R
\frac{N-1}{s}u_\lambda'(s)^2ds - \lambda
\quad (\mbox{put $r = R$}).
\end{aligned}
\end{equation}
Let $M_\lambda := \inf\{\theta > 0: \lambda\sin \theta = g(\theta)\}$.
It is clear that $M_\lambda < \pi$ and
$\lambda\sin\theta > g(\theta)$ for $0 < \theta < M_\lambda$.
We know from \cite{c} that
$\Vert u_\lambda \Vert_\infty < M_\lambda$.
Therefore, for $0 \le r \le R$, we have
\begin{equation} \label{e2.3}
\lambda\sin u_\lambda(r) > g(u_\lambda(r)).
\end{equation}
In particular,
\begin{equation} \label{e2.4}
\xi_\lambda := \lambda\sin \Vert u_\lambda\Vert_\infty
- g(\Vert u_\lambda \Vert_\infty) > 0.
\end{equation}
Furthermore, for $[0, R]$, we put
\begin{gather} \label{e2.5}
I_\lambda(r) :=
\lambda(\cos u_\lambda(r) - \cos \Vert u_\lambda\Vert_\infty)
+ G(u_\lambda(r)) - G(\Vert u_\lambda\Vert_\infty),
\\
II_\lambda(r) := \int_0^r \frac{N-1}{s}u_\lambda'(s)^2ds. \label{e2.6}
\end{gather}
Then for $r \in [0, R]$, by \eqref{e2.2}, we obtain
\begin{equation} \label{e2.7}
\frac12u_\lambda'(r)^2 = I_\lambda(r) - II_\lambda(r).
\end{equation}
We explain the basic idea of the proof of Theorem \ref{thm2}.
The main part of the proof of Theorem \ref{thm2} is to show the
following Proposition \ref{prop2.1}.


\begin{proposition} \label{prop2.1}
Let an arbitrary $0 < \delta < 1/4$
be fixed. Then for $\lambda \gg 1$
\begin{equation} \label{e2.8}
\vert B_R \vert\Vert u_\lambda\Vert_\infty^q
- \Vert u_\lambda\Vert_q^q =
\frac{N\vert B_R\vert C_0}{R}\lambda^{-1/2} +
O(\lambda^{-(1/2 + \delta)}).
\end{equation}
\end{proposition}

Once Proposition \ref{prop2.1} is proved, then we obtain Theorem \ref{thm2}
easily as follows.



\begin{proof}[Proof of Theorem \ref{thm2}]
 By Proposition \ref{prop2.1}, Theorem \ref{thm1} and Taylor expansion,
for $\lambda \gg 1$,
\begin{align*}
\Vert u_\lambda\Vert_q &=
\Big(\vert B_R \vert\Vert u_\lambda\Vert_\infty^q
-\frac{N\vert B_R\vert C_0}{R}\lambda^{-1/2} +
O(\lambda^{-(1/2 + \delta)})\Big)^{1/q} \\
&=\vert B_R\vert^{1/q} \Vert u_\lambda\Vert_\infty
\Big(1 -\frac{NC_0}{R\Vert u_\lambda\Vert_\infty^q}\lambda^{-1/2} +
O(\lambda^{-(1/2 + \delta)})\Big)^{1/q}\\
&=\vert B_R\vert^{1/q} \Big(\pi - g(\pi)\lambda^{-1} + o(\lambda^{-1})\Big)
 \Big(1 - \frac{NC_0}{qR\pi^q}\lambda^{-1/2} +
O(\lambda^{-(1/2 + \delta)})\Big)\\
&= \vert B_R\vert^{1/q}
\Big(\pi - \frac{NC_0}{\pi^{q-1}qR}\lambda^{-1/2}
+ O(\lambda^{-(1/2 + \delta)})\Big).
\end{align*}
Thus the proof is complete.
 \end{proof}

The basic idea to obtain Proposition \ref{prop2.1} is as follows.
In what follows, let an arbitrary $0 < \delta < 1/4$ be fixed.
Let $0 < R_{\lambda,\delta} < R$ satisfy
$u_\lambda(R_{\lambda,\delta})
= \Vert u_\lambda\Vert_\infty - \lambda^{-\delta}$.
By \eqref{e2.7}, we have
\begin{equation} \label{e2.9}
\begin{aligned}
&\vert B_R \vert\Vert u_\lambda\Vert_\infty^q
- \Vert u_\lambda\Vert_q^q \\
&= \vert S^{N-1}\vert\int_0^R r^{N-1}
(\Vert u_\lambda\Vert_\infty^q - u_\lambda(r)^q)
\frac{-u_\lambda'(r)}{\sqrt{2(I_\lambda(r)-II_\lambda(r))}}dr \\
&= \vert S^{N-1}\vert\int_0^{R_{\lambda,\delta}} r^{N-1}
(\Vert u_\lambda\Vert_\infty^q - u_\lambda(r)^q)
\frac{-u_\lambda'(r)}{\sqrt{2(I_\lambda(r)-II_\lambda(r))}}dr
\\
&\quad + \vert S^{N-1}\vert\int_{R_{\lambda,\delta}}^R r^{N-1}
(\Vert u_\lambda\Vert_\infty^q - u_\lambda(r)^q)
\frac{-u_\lambda'(r)}{\sqrt{2(I_\lambda(r)-II_\lambda(r))}}dr
\\
&:=A(\lambda) + B(\lambda).
\end{aligned}
\end{equation}
Therefore, to show Proposition \ref{prop2.1}, we have only to
estimate $A(\lambda)$ and $B(\lambda)$.

For $0 \le \theta \le \Vert u_\lambda\Vert_\infty$, we put
\begin{gather} \label{e2.10}
V_0 := 2\lambda(\cos\theta + 1),
\\
V_1 := 2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty), \label{e2.11}
\\
V_2 := 2(G(\theta) - G(\Vert u_\lambda\Vert_\infty))
- 2\int_0^{u_\lambda^{-1}(\theta)}\frac{N-1}{s}u_\lambda'(s)^2ds. \label{e2.12}
\end{gather}
By putting $\theta = u_\lambda(r)$, we have
\begin{gather} \label{e2.13}
A_\lambda = \vert S^{N-1}\vert
\int_{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
^{\Vert u_\lambda\Vert_\infty}
\frac{u_\lambda^{-1}(\theta)^{N-1}
(\Vert u_\lambda\Vert_\infty^q - \theta^q)}
{\sqrt{V_1 + V_2}}d\theta,
\\
B_\lambda = \vert S^{N-1}\vert
\int_0^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{u_\lambda^{-1}(\theta)^{N-1}
(\Vert u_\lambda\Vert_\infty^q - \theta^q)}
{\sqrt{V_1 + V_2}}d\theta. \label{e2.14}
\end{gather}
We estimate $A_\lambda$ first by using the following Lemma.

\begin{lemma}[\cite{s3}] \label{lem2.2}
 Assume that $0 < r_\lambda < R$
satisfies $u_\lambda(r_\lambda) \to \pi$ as $\lambda \to \infty$. Then
for $0 \le r \le r_\lambda$ and $\lambda \gg 1$
\begin{equation} \label{e2.15}
I_\lambda(r) = \xi_\lambda
(\Vert u_\lambda\Vert_\infty - u_\lambda(r))
+ \frac12(\lambda + g'(\pi))
(1 + o(1))(\Vert u_\lambda\Vert_\infty - u_\lambda(r))^2,
\end{equation}
\begin{equation} \label{e2.16}
\begin{aligned}
II_\lambda(r) &\le
\frac{N-1}{N}\xi_\lambda
(\Vert u_\lambda\Vert_\infty - u_\lambda(r))
\\
&\quad + \frac{N-1}{2(N+1)}(\lambda + g'(\pi))(1 + o(1))
(\Vert u_\lambda\Vert_\infty - u_\lambda(r))^2.
\end{aligned}
\end{equation}
Furthermore, $\xi_\lambda
= o\big(\lambda e^{-\sqrt{2\lambda(1 + o(1))/(N+1)}r_0}\big)$
as $\lambda \to \infty$.
\end{lemma}

\begin{lemma} \label{lem2.3}
$A(\lambda) = O(\lambda^{-(1/2 + \delta)})$ for $\lambda \gg 1$.
\end{lemma}

\begin{proof}
By Lemma \ref{lem2.2}, for $0 \le r \le R_{\lambda,\delta}$ and $\lambda \gg 1$
\begin{equation} \label{e2.17}
\begin{aligned}
\frac12u_\lambda'(r)^2 &= I_\lambda(r) - II_\lambda(r) \\
&\ge \frac{1}{N}\xi_\lambda
 (\Vert u_\lambda\Vert_\infty - u_\lambda(r))
 + \frac{1}{N + 1}(1 + o(1))(g'(\pi) + \lambda)
(\Vert u_\lambda\Vert_\infty - u_\lambda(r))^2 \\
&\ge C\lambda (\Vert u_\lambda\Vert_\infty - u_\lambda(r))^2.
\end{aligned}
\end{equation}
This implies that for
$\Vert u_\lambda\Vert_\infty
- \lambda^{-\delta} \le \theta \le \Vert u_\lambda\Vert_\infty$,
\begin{equation} \label{e2.18}
V_1 + V_2 \ge C\lambda (\Vert u_\lambda\Vert_\infty - \theta)^2.
\end{equation}
By this and \eqref{e2.13}, we obtain
\begin{align*}
A(\lambda) &\le \vert S^{N-1}\vert R^{N-1}
\int_{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
^{\Vert u_\lambda\Vert_\infty}
\frac{\Vert u_\lambda\Vert_\infty^q - \theta^q}
{\sqrt{\lambda(\Vert u_\lambda\Vert_\infty - \theta)^2}}d\theta
\\
&\le \frac{C}{\sqrt{\lambda}}
\int_{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
^{\Vert u_\lambda\Vert_\infty}
\frac{\Vert u_\lambda\Vert_\infty^q - \theta^q}
{\Vert u_\lambda\Vert_\infty - \theta}d\theta
\\
&= O\big(\lambda^{-(1/2 + \delta)}\big).
\end{align*}
Thus the proof is complete.
\end{proof}

We next estimate $B(\lambda)$. We put
\begin{equation} \label{e2.19}
K_1 := \vert S^{N-1}\vert
\int_0^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{u_\lambda^{-1}(\theta)^{N-1}
(\Vert u_\lambda\Vert_\infty^q - \theta^q)}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}d\theta
\end{equation}
and $K_2 := B(\lambda) - K_1$. Once we obtain the estimates of
$K_1$ and $K_2$, then the proof of Proposition \ref{prop2.1} is complete.

To calculate $K_2$, we need the following lemma.

\begin{lemma} \label{lem2.4}
 For $\lambda \gg 1$,
\begin{equation} \label{e2.20}
\int_0^R \frac{N-1}{s}u_\lambda'(s)^2ds
\le C\sqrt{\lambda}.
\end{equation}
\end{lemma}

\begin{proof}
Let an arbitrary $0 < \epsilon \ll 1$ be fixed. Then
\begin{equation} \label{e2.21}
\int_0^R \frac{N-1}{r}u_\lambda'(r)^2dr
= \int_0^{R-\epsilon} \frac{N-1}{r}u_\lambda'(r)^2dr
+ \int_{R-\epsilon}^R \frac{N-1}{r}u_\lambda'(r)^2dr.
\end{equation}
By Lemma \ref{lem2.2} and Theorem \ref{thm1},
\begin{equation} \label{e2.22}
\begin{aligned}
\int_0^{R-\epsilon} \frac{N-1}{r}u_\lambda'(r)^2dr
&= II_\lambda(R-\epsilon) \\
&\le  C\xi_\lambda (\Vert u_\lambda\Vert_\infty - u_\lambda(R-\epsilon))
+ C\lambda(\Vert u_\lambda\Vert_\infty - u_\lambda(R-\epsilon))^2
\\
&\le C\lambda^{-1}.
\end{aligned}
\end{equation}
By \eqref{e2.7} and putting $\theta = u_\lambda(r)$, for $\lambda \gg 1$,
we have
\begin{align*}
\int_0^R r^{N-1}u_\lambda'(r)^2dr
&\le  R^{N-1}
\int_0^R (-u_\lambda'(r))\sqrt{2\lambda(\cos u_\lambda(r)
- \cos\Vert u_\lambda\Vert_\infty)}dr
\\
&\le C\int_0^{\Vert u_\lambda\Vert_\infty} \sqrt{2\lambda(\cos\theta
- \cos\Vert u_\lambda\Vert_\infty)}d\theta
\le C\sqrt{\lambda}.
\end{align*}
By this, for $\lambda \gg 1$, we obtain
\begin{equation} \label{e2.24}
\begin{aligned}
\int_{R-\epsilon}^R \frac{N-1}{r}u_\lambda'(r)^2dr
&\le \frac{N-1}{(R-\epsilon)^N}\int_{R-\epsilon}^R
r^{N-1}u_\lambda'(r)^2dr \\
&\le C\int_0^R r^{N-1}u_\lambda'(r)^2dr \le C\sqrt{\lambda}.
\end{aligned}
\end{equation}
By this and \eqref{e2.22}, we obtain our conclusion.
\end{proof}

\begin{lemma} \label{lem2.5}
 $K_2 = O(\lambda^{-(1/2 + \delta)})$ for $\lambda \gg 1$.
\end{lemma}

\begin{proof}  Let an arbitrary $0 < \epsilon \ll 1$ be fixed.
Since $V_2 \le 0$,
by \eqref{e2.11},
\eqref{e2.12}, \eqref{e2.14} and \eqref{e2.19}, for $\lambda \gg 1$,
\begin{equation} \label{e2.25}
\begin{aligned}
K_2 &= \vert S^{N-1}\vert
\int_0^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
u_\lambda^{-1}(\theta)^{N-1}
(\Vert u_\lambda\Vert_\infty^q - \theta^q)
\big\{\frac{1}{\sqrt{V_1 + V_2}} - \frac{1}{\sqrt{V_1}}\big\}d\theta
\\
&\le CR^{N-1}\int_0^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
(\Vert u_\lambda\Vert_\infty^q - \theta^q)
\frac{\vert V_2\vert}{\sqrt{V_1}\sqrt{V_1+V_2}(\sqrt{V_1+V_2}+\sqrt{V_1})}
d\theta
\\
&\le CR^{N-1}\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
(\Vert u_\lambda\Vert_\infty^q - \theta^q)
\frac{\vert V_2\vert}{(V_1 + V_2)^{3/2}}
d\theta
\\
&\quad + CR^{N-1}\int_0^
{\Vert u_\lambda\Vert_\infty - \epsilon}
(\Vert u_\lambda\Vert_\infty^q - \theta^q)
\frac{\vert V_2\vert}{(V_1 + V_2)^{3/2}} d\theta
\\
&= K_{2,1} + K_{2,2}.
\end{aligned}
\end{equation}
We know from Lemma \ref{lem2.4} that $\vert V_2\vert \le C\sqrt{\lambda}$.
By this, Lemma \ref{lem2.2} and the same estimate as \eqref{e2.18}, we obtain
\begin{equation} \label{e2.26}
\begin{aligned}
K_{2,1} &\le CR^{N-1}\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
(\Vert u_\lambda\Vert_\infty^q - \theta^q)
\frac{\vert V_2\vert}{(V_1+V_2)^{3/2}}d\theta
\\
&\le C \int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
(\Vert u_\lambda\Vert_\infty^q - \theta^q)
\frac{\sqrt{\lambda}}
{(\sqrt{\lambda}(\Vert u_\lambda\Vert_\infty - \theta)^2)^{3/2}}d\theta
\\
&\le C\lambda^{-1}
\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{1}{(\Vert u_\lambda\Vert_\infty - \theta)^2}d\theta
\\
&\le O\left(\lambda^{-1+\delta}\right)
= o\big(\lambda^{-(1/2+\delta)}\big).
\end{aligned}
\end{equation}
We note that $0 < \delta < 1/4$. By \eqref{e2.11}, \eqref{e2.12}
and Lemma \ref{lem2.4},
for $0 \le \theta \le \Vert u_\lambda\Vert_\infty - \epsilon$,
\begin{equation} \label{e2.27}
V_1 + V_2 \ge C\lambda - C - C\sqrt{\lambda} \ge C\lambda.
\end{equation}
By this, Lemma \ref{lem2.4} and \eqref{e2.25}, we obtain
\begin{equation} \label{e2.28}
K_{2,2} \le CR^{N-1}\int_0^{\Vert u_\lambda\Vert_\infty - \epsilon}
\frac{\sqrt{\lambda}}{\lambda^{3/2}}d\theta \le C\lambda^{-1}.
\end{equation}
By \eqref{e2.25}, \eqref{e2.26} and \eqref{e2.28}, we obtain our conclusion.
Thus the proof is complete.
\end{proof}

We next calculate $K_1$. We put
$K_1 := L_1 + L_2$, where
\begin{equation} \label{e2.29}
L_1 := \vert S^{N-1}\vert R^{N-1}
\int_0^{\pi}\frac{\pi^q-\theta^q}
{\sqrt{2\lambda(\cos\theta + 1)}}d\theta
= \frac{N\vert B_R\vert C_0}{R}\lambda^{-1/2}
\end{equation}
and $L_2 := K_1 - L_1$. All we have to do is to calculate $L_2$.
To do this, we put $L_2 := D_1 + D_2 + D_3 + D_4$,
where
\begin{gather} \label{e2.30}
D_1 := -\vert S^{N-1}\vert
\int_0^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{(R^{N-1} - u_\lambda^{-1}(\theta)^{N-1})
(\Vert u_\lambda\Vert_\infty^q - \theta^q)}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}d\theta,
\\
D_2 := \vert S^{N-1}\vert\int_0^{\Vert u_\lambda\Vert_\infty
- \lambda^{-\delta}} \frac{R^{N-1}
(\Vert u_\lambda\Vert_\infty^q - \pi^q)}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}d\theta,
\label{e2.31} \\
\begin{aligned}
D_3 &:= \vert S^{N-1}\vert\int_0^{\Vert u_\lambda\Vert_\infty
- \lambda^{-\delta}} \frac{R^{N-1}
(\pi^q - \theta^q)}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}d\theta
\\
&\quad -\vert S^{N-1}\vert\int_0^{\Vert u_\lambda\Vert_\infty
- \lambda^{-\delta}} \frac{R^{N-1} (\pi^q - \theta^q)}
{\sqrt{2\lambda(\cos\theta + 1)}}d\theta,
\end{aligned} \label{e2.32}\\
D_4 :=-\vert S^{N-1}\vert
\int_{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}^{\pi}
\frac{R^{N-1}(\pi^q - \theta^q)}
{\sqrt{2\lambda(\cos\theta + 1)}}d\theta. \label{e2.33}
\end{gather}
The most essential term in $L_2$ is $D_1$. Therefore, we treat
$D_1$ after the estimates of $D_2, D_3$ and $D_4$.

\begin{lemma} \label{lem2.6}
 $\vert D_4\vert \le C\lambda^{-(1/2 + \delta)}$
for $\lambda \gg 1$.
\end{lemma}

\begin{proof}  Since $(\pi^q - \theta^q)/\sqrt{1+\cos\theta}$ is bounded
for $0 \le \theta \le \pi$, by Theorem \ref{thm1} and \eqref{e2.33},
\begin{align*}
\vert D_4\vert &\le C\lambda^{-1/2}
\int_{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}^{\pi}
\frac{\pi^q - \theta^q}{\sqrt{\cos\theta + 1}}d\theta
\\
&\le C\lambda^{-1/2}(\pi - \Vert u_\lambda\Vert_\infty + \lambda^{-\delta})
\le C\lambda^{-(1/2 + \delta)}.
\end{align*}
Thus the proof is complete.
\end{proof}


\begin{lemma} \label{lem2.7}
 $\vert D_2\vert \le C\lambda^{-3/2}\log\lambda$
for $\lambda \gg 1$.
\end{lemma}


\begin{proof}
Let an arbitrary $0 < \epsilon \ll 1$ be fixed.
By Taylor expansion, we see that for
$\Vert u_\lambda\Vert_\infty-\epsilon \le \theta \le
\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}$
\begin{equation} \label{e2.34}
\cos\theta - \cos\Vert u_\lambda\Vert_\infty \ge
C(\Vert u_\lambda\Vert_\infty - \theta)^2.
\end{equation}
By this and Theorem \ref{thm1},
\begin{equation} \label{e2.35}
\begin{aligned}
\vert D_2\vert &\le C\lambda^{-1}\int_
{\Vert u_\lambda\Vert_\infty-\epsilon}
^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{1}{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}
d\theta
\\
&\quad + C\lambda^{-1}\int_0^
{\Vert u_\lambda\Vert_\infty-\epsilon}
\frac{1}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}
d\theta
\\
&\le C\lambda^{-3/2}\int_{\Vert u_\lambda\Vert_\infty-\epsilon}
^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{1} {\Vert u_\lambda\Vert_\infty - \theta}
d\theta + C\lambda^{-3/2}
\\
&\le C\lambda^{-3/2}\log\lambda.
\end{aligned}
\end{equation}
Thus the proof is complete.
 \end{proof}

\begin{lemma} \label{lem2.8}
 $\vert D_3\vert \le C\lambda^{-7/2 + 2\delta}$
for $\lambda \gg 1$.
\end{lemma}


\begin{proof}
Let an arbitrary $0 < \epsilon \ll 1$ be fixed.
By \eqref{e2.10}, \eqref{e2.11}, \eqref{e2.32}, \eqref{e2.34}
and Theorem \ref{thm1},
\begin{equation} \label{e2.36}
\begin{aligned}
\vert D_3\vert &\le \int_0^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
(\pi^q - \theta^q)\left(\frac{1}{\sqrt{V_1}}
-\frac{1}{\sqrt{V_0}}\right)d\theta
\\
&\le C \int_0^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{1 + \cos\Vert u_\lambda\Vert_\infty}
{\sqrt{V_0}\sqrt{V_1}(\sqrt{V_0}+ \sqrt{V_1})}d\theta
\\
&\le C\lambda^{-2}
\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{1}{(2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty))
^{3/2}}d\theta
\\
&\quad +C\lambda^{-2}
\int_0^{\Vert u_\lambda\Vert_\infty - \epsilon}
\frac{1}{(2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty))
^{3/2}}d\theta
\\
&= C\lambda^{-7/2}\int_{\Vert u_\lambda\Vert_\infty - \epsilon}
^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{1}{(\Vert u_\lambda\Vert_\infty - \theta)^3}d\theta
+ C\lambda^{-7/2}
\\
&\le C\lambda^{-7/2 + 2\delta}.
\end{aligned}
\end{equation}
Thus the proof is complete.
\end{proof}

Now we calculate $D_1$. To do this, we need additional two lemmas.

\begin{lemma} \label{lem2.9}
For $\lambda \gg 1$
\begin{equation} \label{e2.37}
u_\lambda'(R)^2 = 4\lambda + O(\sqrt{\lambda}).
\end{equation}
\end{lemma}

\begin{proof}
By \eqref{e2.2}, Theorem \ref{thm1} and Lemma \ref{lem2.4}, we obtain
\begin{align*}
\frac12u_\lambda'(R)^2
&= \lambda(1 - \cos\Vert u_\lambda\Vert_\infty)
- \int_0^R \frac{N-1}{s}u_\lambda'(s)^2ds
- G(\Vert u_\lambda\Vert_\infty)
\\
&= 2\lambda + O(\sqrt{\lambda}).
\end{align*}
Thus the proof is complete.
\end{proof}

\begin{lemma} \label{lem2.10}
 For $\lambda \gg 1$
\begin{equation} \label{e2.38}
R - R_{\lambda,\delta} \le C\lambda^{-1/2 + \delta}.
\end{equation}
\end{lemma}

\begin{proof}
Since $u_\lambda(r)$ is decreasing for $0 \le r \le R$ by
\eqref{e2.1}, for $R_{\lambda,\delta} \le r \le R$
\begin{equation} \label{e2.39}
\cos u_\lambda(r) \ge
\cos(\Vert u_\lambda\Vert_\infty - \lambda^{-\delta})
= -1 + \frac12(1 + o(1))\lambda^{-2\delta}.
\end{equation}
By this, \eqref{e2.2} and Lemmas \ref{lem2.4} and \ref{lem2.9}, for
$R_{\lambda,\delta} \le r \le R$,
\begin{equation} \label{e2.40}
\begin{aligned}
\frac12u_\lambda'(r)^2 &= \frac12u_\lambda'(R)^2
+ \int_r^R \frac{N-1}{s}u_\lambda'(s)^2ds
+ \lambda(\cos u_\lambda(r) - 1) + G(u_\lambda(r))
\\
&\ge \frac12(4\lambda + O(\sqrt{\lambda}))
+ \lambda(-2 + \frac12(1 + o(1))\lambda^{-2\delta})
\\
&= \frac12\lambda^{1-2\delta}(1 + o(1)).
\end{aligned}
\end{equation}
Note that $0 < \delta < 1/4$. By this, for $\lambda \gg 1$, we obtain
\[
C\lambda^{(1-2\delta)/2}(R - R_{\lambda,\delta})
\le \int_{R_{\lambda,\delta}}^R -u_\lambda'(r)dr \le u(R_{\lambda,\delta})
< \pi.
\]
This implies our conclusion.
 \end{proof}

\begin{lemma} \label{lem2.11}
 $\vert D_1\vert \le C\lambda^{-(1/2 + \delta)}$
for $\lambda \gg 1$.
\end{lemma}

\begin{proof}
It is easy to see that for
$0 \le \theta \le \Vert u_\lambda\Vert_\infty - \lambda^{-\delta}$,
\begin{equation} \label{e2.41}
\frac{\Vert u_\lambda\Vert_\infty^q - \theta^q}
{\sqrt{\cos\theta - \cos\Vert u_\lambda\Vert_\infty}} \le C.
\end{equation}
By this and Lemma \ref{lem2.10},
\begin{align*}
\vert D_1 \vert
&=\vert S^{N-1}\vert
\int_0^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}
\frac{(R^{N-1} - u_\lambda^{-1}(\theta)^{N-1})
(\Vert u_\lambda\Vert_\infty^q - \theta^q)d\theta}
{\sqrt{2\lambda(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}}
\\
&\le C(R - R_{\lambda,\delta})\lambda^{-1/2}
\int_0^{\Vert u_\lambda\Vert_\infty - \lambda^{-\delta}}d\theta
\\
&= C\lambda^{-1+ \delta}
\le
C\lambda^{-(1/2 + \delta)}.
\end{align*}
We note here that $0 < \delta < 1/4$.
Thus the proof is complete.
\end{proof}

By Lemmas \ref{lem2.3}, \ref{lem2.5} \ref{lem2.6}, \ref{lem2.7},
\ref{lem2.8}, \ref{lem2.11},
we obtain Proposition \ref{prop2.1}. Thus the proof is complete.


\section{Proof of Theorem \ref{thm3}}

To prove Theorem \ref{thm3}, we have only to improve Lemma \ref{lem2.4}.

\begin{lemma} \label{lem3.1}
 For $\lambda \gg 1$
\begin{equation} \label{e3.1}
\int_0^R \frac{N-1}{r}u_\lambda'(r)^2dr =
\frac{4(N-1)}{R}\sqrt{\lambda} + o(\sqrt{\lambda}).
\end{equation}
\end{lemma}


\begin{proof} Let an arbitrary $0 < \epsilon \ll 1$ be fixed.
We consider \eqref{e2.21}.
Then by \eqref{e2.7} and Theorem \ref{thm1}, for $\lambda \gg 1$
\begin{equation} \label{e3.2}
\begin{aligned}
\int_{R-\epsilon}^R \frac{N-1}{r}u_\lambda'(r)^2dr
&\le \frac{N-1}{R-\epsilon}
\int_{R-\epsilon}^R u_\lambda'(r)^2dr \\
&\le \frac{N-1}{R-\epsilon}
\int_{R-\epsilon}^R
\sqrt{2\lambda(\cos u_\lambda(r) - \cos\Vert u_\lambda\Vert_\infty)}
(-u_\lambda'(r))dr
\\
&= \frac{N-1}{R-\epsilon}\sqrt{\lambda}
\int_0^{u_\lambda(R-\epsilon)}
\sqrt{2(\cos\theta - \cos\Vert u_\lambda\Vert_\infty)}d\theta
\\
&= \frac{N-1}{R-\epsilon}\sqrt{\lambda}(1 + o(1))
\int_0^{\pi}
\sqrt{2(\cos\theta + 1)}d\theta
\\
&= \frac{4(N-1)}{R-\epsilon}\sqrt{\lambda}(1 + o(1)).
\end{aligned}
\end{equation}
By the same argument as that just above, we obtain
\begin{equation} \label{e3.3}
\int_{R-\epsilon}^R \frac{N-1}{r}u_\lambda'(r)^2dr
\ge \frac{4(N-1)}{R}\sqrt{\lambda}(1 + o(1)).
\end{equation}
Since $0 < \epsilon \ll 1$ is arbitrary,
by \eqref{e2.21}, \eqref{e2.22}, \eqref{e3.2} and \eqref{e3.3},
we obtain \eqref{e3.1}.
Thus the proof is complete.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm3}]
 By Theorem \ref{thm1}, Lemma \ref{lem3.1} and \eqref{e2.2},
for $\lambda \gg 1$,
\begin{align*}
\frac12u_\lambda'(R)^2
&= \lambda(1 - \cos\Vert u_\lambda\Vert_\infty)
- \int_0^R \frac{N-1}{s}u_\lambda'(s)^2ds - G(\Vert u_\lambda\Vert_\infty)
\\
&=\lambda\Big(2 - \frac{1}{2}(1 + o(1))g(\pi)^2\lambda^{-2}\Big)
- \frac{4(N-1)}{R}\sqrt{\lambda} + o(\sqrt{\lambda})
\\
&= 2\lambda
- \frac{4(N-1)}{R}\sqrt{\lambda} + o(\sqrt{\lambda}).
\end{align*}
By this, we obtain Theorem \ref{thm3}. Thus the proof is complete.
\end{proof}

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\end{document}