\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 65, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/65\hfil Constant sign solutions]
{Constant sign solutions for second-order  $m$-point boundary-value problems}

\author[J. Yang \hfil EJDE-2013/65\hfilneg]
{Jingping Yang}  % in alphabetical order

\address{Jingping Yang \newline
Gansu Institute of Political Science and Law  \\
Lanzhou, 730070,  China}
\email{fuj09@lzu.cn}

\thanks{Submitted November 8, 2012. Published March 5, 2013.}
\subjclass[2000]{34B18, 34C25}
\keywords{Constant sign solutions; eigenvalue; bifurcation methods}

\begin{abstract}
 We will study the existence of constant sign solutions for the second-order
 $m$-point boundary-value problem
 \begin{gather*}
 u''(t)+f(t,u(t))=0,\quad t\in(0,1),\\
 u(0)=0, \quad  u(1)=\sum^{m-2}_{i=1}\alpha_i u(\eta_i),
 \end{gather*}
 where $m\geq3$, $\eta_i\in(0,1)$ and $\alpha_i>0$ for
 $i=1,\dots,m-2$, with   $\sum^{m-2}_{i=1}\alpha_i<1$, we obtain that
 there exist at least a positive and a negative
 solution for the above problem. Our approach is based on
 unilateral global bifurcation theorem.
\end{abstract}

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



\section{Introduction}

In recent years, there has been considerable interests in the
existence of nodal solutions of second-order $m$-point boundary
value problems (BVPs) of the form
\begin{equation}
\begin{gathered}
u''(t)+ f(u(t))=0,\quad t\in(0,1),\\
u(0)=0, \quad  u(1)=\sum^{m-2}_{i=1}\alpha_i u(\eta_i),
\end{gathered}\label{e1.1}
\end{equation}
see \cite{a1,a2,m1,r2,x1} and the references therein.

 Ma and  O'Regan \cite{m1} considered
 \eqref{e1.1} under the assumption $f\in C^1(\mathbb{R},\mathbb{R} )$ with
$sf(s)>0$ for $s\neq0$.  They obtained the existence of nodal solutions
for  $f_0, f_\infty\in (0,\infty)$, where
 $f_0=\lim_{u\to 0}\frac{f(u)}{u}$,
$f_\infty =\lim_{u\to \infty}\frac{f(u)}{u}$.

In 2011, An \cite{a2} considered the problem
\begin{equation}
\begin{gathered}
u''(t)+\lambda f(u(t))=0,\quad  t\in(0,1),\\
u(0)=0, \quad  u(1)=\sum^{m-2}_{i=1}\alpha_i u(\eta_i)
\end{gathered}\label{e1.2}
\end{equation}
under the assumption $f\in
C^1(\mathbb{R}\backslash\{0\}, \mathbb{R})\cap C(\mathbb{R},
\mathbb{R})$ with $sf(s)>0$ for $s\neq0$. She investigated the
global structure of nodal solutions of \eqref{e1.2} in the case
$f_0=\infty$, $f_{\infty}\in[0,\infty]$, by using  Rabinowit's global
bifurcation theorem.

From above results, we can see that the existence results are
largely based on the assumption that $f_0, f_\infty$ are constants
and nonlinearity term is autonomous. It is interesting to know what
will happen if $f_0, f_\infty$ are functions and the nonlinear term
is non-autonomous?

The  above results rely largely on the direct
computation of eigenvalues and eigenfunctions of the linear problem
associated with \eqref{e1.2}, hence, it can not be extended to the
more general problem.
In view of the fact that the principle eigenvalue can be easily
obtained by Krein-Rutman Theorem, in this paper, we obtain the
existence of constant sign solution for
\begin{equation}
\begin{gathered}
u''(t)+ f(t,u(t))=0,\quad  t\in(0,1),\\
u(0)=0, \quad  u(1)=\sum^{m-2}_{i=1}\alpha_i u(\eta_i)
\end{gathered}\label{e1.3}
\end{equation}
by relating it to the principle eigenvalue of the associated linear
problem. We make the following assumptions:
\begin{itemize}

\item[(H1)] $\lambda_1\leq a(t)\equiv\lim_{|s|\to+\infty}\frac{f(t,s)}{s}$
uniformly on $[0,1]$, and the
inequality is strict on some subset of positive measure in $(0,1)$;
where $\lambda_1$ denotes the principle eigenvalue of
\begin{equation}
\begin{gathered}
\psi''(t)+\lambda\psi(t)=0,\quad t\in(0,1),\\
\psi(0)=0, \quad  \psi(1)=\sum^{m-2}_{i=1}\alpha_i \psi(\eta_i);
\end{gathered}\label{e1.4}
\end{equation}

\item[(H2)]  $0\leq\lim_{|s|\to 0} \frac{f(t,s)}{s}\equiv c(t)\leq \lambda_1$
uniformly on $[0,1]$, and all the inequalities are strict on some
subset of positive measure in $(0,1)$;

\item[(H3)]  $f(t,s)s>0$ for all $t\in (0,1)$ and
$s\neq 0$.
\end{itemize}


By applying the bifurcation theorem of L\'{o}pez-G\'{o}mez
\cite[Theorem 6.4.3]{l1}, we will establish the following results.

\begin{theorem} \label{thm1.1}
Suppose that $f(t, u)$ satisfies
{\rm(H1)--(H3)}. Then \eqref{e1.3} possesses at
 least one positive and one negative solution.
\end{theorem}

 Similar result is obtained under the following assumptions.
\begin{itemize}
\item[(H1')] $\lambda_1\geq a(t)\equiv\lim_{| s|\to +\infty}
\frac{f(t,s)}{s}\geq0$ uniformly on $[0,1]$, and all the
inequalities are strict on some subset of positive measure in
$(0,1)$, where $\lambda_1$ denotes the principle eigenvalue of \eqref{e1.4};

\item[(H2')] $\lim_{| s|\to0} \frac{f(t,s)}{s}\equiv c(t)\geq \lambda_1$
 uniformly on $[0,1]$, and the inequality is strict on some subset of positive
measure in $(0,1)$.
\end{itemize}

\begin{theorem} \label{thm1.2}
Suppose that $f(t, u)$ satisfies {\rm (H1'), (H2'), (H3)}.
Then  \eqref{e1.3} possesses at
 least one positive and one negative solution.
\end{theorem}

 The existence of constant sign solutions of \eqref{e1.3} is related
to the  eigenvalue problem
\begin{equation}
\begin{gathered}
u''(t)+ \mu f(t,u(t))=0,\quad  t\in(0,1),\\
u(0)=0, \quad  u(1)=\sum^{m-2}_{i=1}\alpha_i u(\eta_i),
\end{gathered}\label{e1.5}
\end{equation}
where $\mu>0$ is a parameter. Therefore, we will study the
bifurcation phenomena for \eqref{e1.5} with crossing nonlinearity.
Moreover, the bifurcation point of \eqref{e1.5} is related to the principle
eigenvalues of the problem
\begin{equation}
\begin{gathered}
u''(t)+ \mu c(t)u(t)=0,\quad  t\in(0,1),\\
u(0)=0, \quad  u(1)=\sum^{m-2}_{i=1}\alpha_i u(\eta_i),
\end{gathered}\label{e1.6}
\end{equation}
it is well-known that there exists a principle eigenvalue
 $\mu_1(c(t))$ of \eqref{e1.6} (see \cite{z1}).

The rest of the paper is organized as follows:
in Section 2, we state some notations and preliminary results.
In Section 3, we  prove the main results.

\section{Notation and preliminary results}

To show the constant sign solutions of  \eqref{e1.5}, we
consider the operator equation
\begin{equation}
u=\mu Tu.\label{e2.1}
\end{equation}
This equations are usually called nonlinear eigenvalue problems.
L\'{o}pez-G\'{o}mez \cite{l1} studied a
nonlinear eigenvalue problem of the form
\begin{equation}
u=\mu Tu+H(\mu,u),\label{e2.2}
\end{equation}
where $H(\mu,u)=o(\|u\|)$ as $\|u\|\to 0$ uniformly for $\mu$ on
a bounded interval, and $T$ is a linear completely continuous operator
on a Banach space $X$.
A solution of \eqref{e2.2} is a pair $(\mu,u)\in \mathbb{R}\times X$,
which satisfies \eqref{e2.2}.
The closure of  the set nontrivial solutions of \eqref{e2.2} is denoted
by $\mathcal{C}$. Let $\Sigma(T)$ denote the
  set of eigenvalues of linear operator $T$.
L\'{o}pez-G\'{o}mez \cite{l1} established the following results.

\begin{lemma}[{\cite[Theorem 6.4.3]{l1}}] \label{lem2.1}
 Assume $\Sigma(T)$ is discrete. Let $\mu_0\in\Sigma(T)$ such
that $ind (I-\mu T,\theta)$ changes sign as $\mu$ crosses $\mu_0$,
then each of the components  $\mathcal{C}$ (denote the components of $S$
emanating of $(\mu, \theta)$ at $(\mu_0, \theta)$), satisfies
$(\mu_0,\theta)\in\mathcal{C}$, and either
\begin{itemize}
\item[(i)] $\mathcal{C}$ is unbounded in $\mathbb{R}\times X$;

\item[(ii)] there exist $\lambda_1\in\Sigma(T)\setminus\{\lambda_0\}$
such that $(\lambda_1,\theta)\in\mathcal{C}$; or

\item[(iii)] $\mathcal{C}$ contains a point
  $$
(\iota,y)\in\mathbb{R}\times (V\backslash\{\theta\}),
$$
where $V$ is the complement of
$\operatorname{span}\{\varphi_{\mu_0}\}$,
$\varphi_{\mu_0}$ denotes the eigenfunction corresponding to
  eigenvalue $\mu_0$.
\end{itemize}
\end{lemma}

\begin{lemma}[{\cite[Theorem 6.5.1]{l1}}] \label{lem2.2}
 Under the  assumptions:
\begin{itemize}
\item[(A)] $X$ is an ordered Banach space, whose positive cone,
denoted by $P$, is normal and has a nonempty interior;

\item[(B)] The family $\Upsilon(\mu)$ has the special form
 $$
\Upsilon(\mu)=I_X-\mu T,
$$
 where $T$ is a compact strongly positive operator, i.e.,
 $T(P\backslash\{\theta\})\subset $int P;

\item[(C)] The solutions of $u=\mu Tu+H(\mu,u)$ satisfy the strong
maximum principle.
\end{itemize}
 Then the following assertions are true:
\begin{itemize}
\item[(1)] $\operatorname{Spr}(T)$ is a simple eigenvalue of $T$, having
a positive  eigenfunction denoted  by $\psi_0>0$, i.e.,
$\psi_0\in\operatorname{int} P$, and there is no other
 eigenvalue of $T$ with a positive eigenfunction;

\item[(2)] For every $y\in $int P, the equation
 $$
u-\mu Tu=y
$$
has exactly one positive solution if
$\mu<\frac{1}{\operatorname{Spr} (T)}$, whereas it does not admit
a positive solution if $\mu\geq\frac{1}{\operatorname{Spr} (T)}$.
\end{itemize}
\end{lemma}

\begin{lemma}[{cite[Theorem 2.5]{b1}}] \label{lem2.3}
Assume $T: X\to X$ is a  linear completely continuous operator,
and 1 is not an eigenvalue of  $T$, then
 $$
\operatorname{ind}(I-T,\theta)=(-1)^\beta,
$$
where $\beta$ is the sum of the algebraic multiplicities of the eigenvalues of
 $T$ large than 1, and $\beta=0$ if $T$ has no eigenvalue of this
 kind.
\end{lemma}

Let $Y$ be the space $C[0,1]$ with the norm
$\|u\|_{\infty}=\max_{t\in[0,1]}|u(t)|$.
Let
$$
E=\{u\in C^1[0,1]: u(0)=0, \;
 u(1)=\sum^{m-2}_{i=1}\alpha_i u(\eta_i)\}
$$
with the norm
$$
\|u\|=\max_{t\in[0,1]}|u(t)|+\max_{t\in[0,1]}|u'(t)|.
$$
 Define $L:D(L)\to Y$ by setting
$$
Lu(t):=-u''(t),\quad t\in [0,1],\; u\in D(L),
$$
where
 $$
D(L)=\{u\in C^2[0,1]: u(0)=0, \;
 u(1)=\sum^{m-2}_{i=1}\alpha_i u(\eta_i) \}.
$$
Then $L^{-1}: Y\to E$ is compact.

Let $\mathbb{E}=\mathbb{R}\times E$ under the product topology. As
 in \cite{r1}, we add the point $\{(\mu,\infty)|\ \mu\in
 \mathbb{R}\}$ to our space $\mathbb{E}$. For any  $u\in C^1[0,1]$, if
 $u(x_0)=0$, then $x_0$ is a simple zero of $u$ if $u'(x_0)\neq 0$.
 For $\nu\in\{+,-\}$, define:
\begin{itemize}
\item $S_1^\nu$ is the set of functions such that
\begin{itemize}
\item[(i)] $u(0) = 0$, $\nu u'(0)>0;$

\item[(ii)] $u$ has constant sign in $(0,1)$.
\end{itemize}

\item $T_1^\nu$ is the set of functions such that
\begin{itemize}
\item[(i)] $u(0) = 0$, $\nu u'(0)>0$ and $u'(1)\neq0$;

\item[(ii)] $u'$ has exactly one simple zero point in $(0,1)$;

\item[(iii)] $u$ has a zero strictly between each two consecutive zeros of $u'$.
\end{itemize}
\end{itemize}
Obviously, if $u\in T_1^\nu$, then $u\in S_1^\nu$.
The sets  $T_1^\nu$ are disjoint and open in $E$, (see \cite[Remark 2.2]{r2}).
Finally, let
$\phi_1^\nu=\mathbb{R}\times T_1^\nu$.

Furthermore, let $\zeta\in C([0,1]\times\mathbb{R})$ be such that
$f(t,u)=c(t)u+\zeta(t,u)$
with
$$
\lim_{| u|\to0}\frac{\zeta(t,u)}{u}=0\quad
\text{uniformly on } [0,1].
$$
Let
$$
\bar{\zeta}(t,u)=\max_{0\leq | s|\leq u}|
\zeta(t,s)|\quad \text{for } t\in[0,1].
$$
Then $\bar{\zeta}$ is nondecreasing with respect to $u$ and
\begin{equation}
\lim_{ u\to 0^+}\frac{\bar{\zeta}(t,u)}{u}=0.\label{e2.3}
\end{equation}
From this equality, it follows that
$$
\frac{{\zeta}(t,u)}{\| u\|} \leq\frac{ \bar{\zeta}(t,|
u|)}{\| u\|} \leq \frac{ \bar{\zeta}(t,\|
u\|_\infty)}{\| u\|} \leq \frac{ \bar{\zeta}(t,\|
u\|)}{\| u\|}\to0,\,\,\ \  \text{as}\,\, \|
u\|\to 0
$$
uniformly for  $t\in[0,1]$.

Let us  study
\begin{equation}
Lu-\mu c(t) u=\mu \zeta(t,u)\label{e2.4}
\end{equation}
as a bifurcation problem from the trivial solution $u\equiv 0$.
Equation \eqref{e2.4} can be converted to the equivalent
equation
\begin{align*}
u(t)&= \int_0^1G(t,s)[\mu c(s) u(s)+\mu \zeta(s, u(s))]ds\\
&:=\mu L^{-1}[c(t) u(t)]+ \mu L^{-1}[\zeta(t, u(t))],
\end{align*}
where $G(t,s)$ denotes the Green's function of $Lu=0$.

We note that $\|L^{-1}[ \zeta(t, u(t))]\|=o(\|u\|)$ for $u$
near $0$ in $E$. Since
\begin{align*}
\|L^{-1}[\zeta(t, u(t))]\| 
&= \max_{t\in[0,1]}|\int_0^1G(t,s) \zeta(s, u(s))ds|
+\max_{t\in[0,1]}|\int_0^1G_t(t,s) \zeta(s,u(s))ds|\\
&\leq C \|\zeta(t, u(t))\|_{\infty}.
\end{align*}

\begin{lemma}[{\cite[Proposition 4.1]{r2}}] \label{lem2.4}
If $(\mu,u)\in\mathbb{E}$ is a
non-trivial solution of \eqref{e2.4}, then $u\in T_1^\nu$ for
$\nu\in\{+,-\}$.
\end{lemma}

\begin{lemma} \label{lem2.5}
 For $\nu\in\{+,-\}$, there exists a continuum
$\mathcal{C}_1^\nu\subset\mathbb{E}$ of solutions of \eqref{e2.4} with the
properties:
\begin{itemize}
\item[(i)] $(\mu_1(c(t)),\theta)\in \mathcal{C}_1^\nu;$

\item[(ii)] $\mathcal{C}_1^\nu\backslash\{(\mu_1(c(t)),\theta)\}\subset
\mathbb{R}\times T_1^\nu$;

\item[(iii)]  $\mathcal{C}_1^\nu$ is unbounded in $\mathbb{E}$, where
$\mu_1(c(t))$ denotes the principle eigenvalue of  \eqref{e1.6}.
\end{itemize}
\end{lemma}

\begin{proof} 
From above, we know that  problem \eqref{e2.4} is of the
form considered in \cite{l1}, and satisfies the general hypotheses
imposed in that paper.

 From \cite{z1}, we know that the principle eigenvalues of \eqref{e1.6} is simple. 
So for $\nu\in\{+,-\}$, combining Lemma 2.1  with Lemma 2.3, we
know that there exists a continuum,
$\mathcal{C}_1^\nu\subset\mathbb{E}$, of solutions of \eqref{e2.4} such
that:
\begin{itemize}
\item[(a)] $\mathcal{C}_1^\nu$ is unbounded and $(\mu_1(c(t)),\theta)\in
\mathcal{C}_1^\nu,
\mathcal{C}_1^\nu\backslash\{(\mu_1(c(t)),\theta)\}\subset\mathbb{E}$, or

\item[(b)]  $(\mu_j(c(t)),\theta)\in \mathcal{C}_1^\nu$, where
$j\in\mathbb{N},\mu_j(c(t))$ is another eigenvalue of  
\eqref{e1.6} if possible, or

\item[(c)] $\mathcal{C}_1^\nu$ contains a point
  $$
(\iota,y)\in\mathbb{R}\times (V\backslash\{\theta\}),
$$
  where $V$ is the complement of 
$\operatorname{span}\{\varphi_{1}\}$,
 $\varphi_{1}$ denotes the eigenfunction corresponding to
  principle eigenvalue $\mu_1(c(t))$.
\end{itemize}

We finally prove that the first choice  (a) is the only
possibility.
In fact, all functions belong to the continuum set
$\mathcal{C}_1^\nu$ are constant sign, this implies that it is
impossible to exist $(\mu_j(c(t)),\theta) \in \mathcal{C}_1^\nu,
j\in\mathbb{N},j\neq 1$, where $\mu_j(c(t))$ 
is another eigenvalue of \eqref{e1.6} if possible.
 If this happened, it will be contracted with the definition of $S_1^\nu$.


Next, we will prove (c) is impossible, suppose (c) occurs,  without
loss of generality, suppose there exists a point
$(\iota,y)\in\mathbb{R}\times
(V\backslash\{\theta\})\cap \mathcal{C}_1^+$.
 Define
$$
P=\{u\in C^1[0,1]: u(t)\geq 0, \; t\in [0,1]\},
$$
then $P$ is a normal cone and has a nonempty interior, and
$\mathcal{C}_1^+\backslash\{(\mu_1(c(t)),\theta)\}\subset
 \operatorname{int} P$.


Note that as the complement $V$ of Span$\{\varphi_1\}$ in $E$, we can take
$$
V:=R[c(t)I_E-\frac{1}{\mu_1(c(t))}L].
$$
Thus, for this choice of $V$, if the component $\mathcal{C}_1^+$
 contains a point 
$$
(\iota,y)\in\mathbb{R}\times
(V\backslash\{\theta\})\cap \mathcal{C}_1^+.
$$ 
Then  there exists $u\in E$ for which
$$
c(t)u-\frac{1}{\mu_1(c(t))}Lu=y>0,\quad \text{in } (0,1).
$$
Thus, for each sufficiently large $\eta>0$, we have that 
$c(t)u+\eta \varphi_1(t)>0$ in $(0,1)$ and 
$$
c(t)u+\eta c(t)\varphi_1(t)-\frac{1}{\mu_1(c(t))} L(u+\eta \varphi_1)=y>0\quad
\text{in } (0,1).
$$ 
Hence, by Lemma 2.2, we have
$$
\operatorname{Spr}(\frac{1}{\mu_1(c(t))} L)<1,
$$
which is impossible. since $Spr(L)=\mu_1(c(t))$.
\end{proof}



\section{Proof of  main results}

\begin{proof}[Proof of  Theorem 1.1]
  Theorem 1.2 is proved in similar manner.
It is clear that any solution of \eqref{e2.4} of the form $(1,u)$ yields a
solution $u$ of \eqref{e1.3}. We will show  $\mathcal{C}_1^\nu$ crosses the
hyperplane $\{1\}\times E$ in $\mathbb{R}\times E$.

By  $\mu_1(c(t))$ being strict decreasing  with respect to $c(t)$
(see \cite{l2}), where $\mu_1(c(t))$ is the principle eigenvalue of
\eqref{e1.6}, we have $\mu_1(c(t)) > \mu_1(\lambda_1) = 1$.

Let $(\mu_n, u_n) \in \mathcal{C}_{1}^\nu$ with $u_n\not\equiv 0$
satisfies
\[
\mu_n+\| u_n\|\to+\infty.
\]
We note that $\mu_n >0$ for all $n \in \mathbb{N}$, since
$(0,\theta)$ is the only solution of \eqref{e2.4} for $\mu = 0$ and
$\mathcal{C}_{1}^\nu\cap(\{0\}\times E)=\emptyset$.

\textbf{Step 1:} We show that if there exists a constant $M>0$, such that
$\mu_n\subset(0,M]$
for $n\in\mathbb{N}$ large enough, then $\mathcal{C}_1^\nu$ crosses
the hyperplane $\{1\}\times E$ in $\mathbb{R}\times E$.
In this case it follows that
$\|u_n\|\to\infty$.

Let $\xi\in C([0,1]\times\mathbb{R})$ be such that
\[
f(t,u)=a(t)u+\xi(t,u)
\]
with
\begin{equation}\label{e3.1}
\lim_{| u|\to+\infty}\frac{\xi(t,u)}{u}=0\quad
\text{uniformly on } [0,1].
\end{equation}
We divide the equation
\begin{equation}
Lu_n-\mu_n  a(t) u_n=\mu_n \xi(t,u_n)\label{e3.2}
\end{equation}
by $\|u_n\|$ and set $\bar{u}_n=\frac{u_n}{\|u_n\|}$. Since
$\bar{u}_n$ is bounded in $C^2[0,1]$, after taking a subsequence if
necessary, we have that $\bar{u}_n\to\bar{u}$ for some
$\bar{u}\in E$ with $\|\bar{u}\|=1$. By \eqref{e3.1}, using the similar
proof of \eqref{e2.3}, we have that
\[
\lim_{n\to+\infty}\frac{ \xi(t,u_n(t))}{\|
u_n\|}=0\quad \text{in} Y.
\]
Thus, we obtain
\[
-\bar{u}''-\overline{\mu}(a(t))a(t) \bar{u}=0,
\]
where $\overline{\mu}(a(t))=\underset{n\to+\infty}\lim \mu_n$.

It is clear that $\overline{u}\in
\overline{\mathcal{C}_{1}^\nu}\subseteq \mathcal{C}_{1}^\nu$,
since $\mathcal{C}_{1}^\nu$ is closed in $\mathbb{R}\times E$.
Therefore, $\overline{\mu}(a(t))$ is the principle eigenvalue of \eqref{e1.6}
corresponding to weight function $a(t)$.

 By the strict decreasing of $\overline{\mu}(a(t))$ with respect
to $a(t)$ (see \cite{l2}),  we have $\overline{\mu}(a(t)) <
\overline{\mu}(\lambda_1) = 1$. Therefore, $\mathcal{C}_{1}^\nu$
crosses the hyperplane $\{1\}\times E$ in $\mathbb{R}\times E$.

\textbf{Step 2:} We show that there exists a constant $M$ such that
$\mu_n \in(0,M]$ for $n\in \mathbb{N}$ large enough.
On the contrary, we suppose that
$\lim_{n\to +\infty}\mu_n=+\infty$.
On the other hand, we note that
\[
-u_n''=\mu_n\frac{f(t,u_n)}{u_n}u_n.
\]
We have $\mu_n\frac{f(t,u_n)}{u_n}>\lambda_1$ for
$n$ large enough and all $t\in (0,1]$. We get $u_n$ must change its sign in
$(0,1)$ for $n$ large enough, which contradicts the fact
that $u_n\in T_1^\nu$.
Therefore,
\[
\mu_n\leq M 
\]
for some constant positive  $M$ and $n\in \mathbb{N}$ sufficiently
large. 
\end{proof} 


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