\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 102, pp. 1--6.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/102\hfil Existence of positive solutions]
{Existence of positive solutions for a singular
 $p$-Laplacian Dirichlet problem}

\author[W. Zhou \hfil EJDE-2008/102\hfilneg]
{Wenshu Zhou}

\address{Wenshu Zhou \newline
Department of Mathematics, Dalian Nationalities
University, 116600, China}
\email{pdezhou@126.com, wolfzws@163.com}

\thanks{Submitted December 17, 2007. Published July 30, 2008.}
\thanks{Supported by grants 20076209 from  Dalian Nationalities University,
 and 10626056 \hfill\break\indent from Tianyuan Fund}
\subjclass[2000]{34B18}
\keywords{$p$-Laplacian; singularity; positive solution;
 regularization technique}

\begin{abstract}
 By a argument based on regularization technique,
 upper and lower solutions method and Arzel\'a-Ascoli
 theorem, this paper shows  sufficient conditions of the
 existence of positive solutions of a Dirichlet problem
 for singular $p$-Laplacian.
\end{abstract}

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


\section{Introduction}

 This paper shows the existence of  positive solutions for the
 singular $p$-Laplacian equation
\begin{equation}
  \big(\phi_p(u')\big)'-\lambda\frac{|u'|^p}{u^m}+f(t, u')=0,\quad 0<t<1,
  \label{e1}
\end{equation}
subject to  Dirichlet boundary conditions
\begin{equation}
u(1)=u(0)=0, \label{e2}
\end{equation}
where $\phi_p(s)=|s|^{p-2}s$,  $p >1$, $\lambda$ and $m$  are
positive constants, and  $f$ is a continuous function.
 We call $u \in C^1[0,1]$ is  a solution
if $u>0$ in $(0, 1)$, $|u'|^{p-2}u' \in C^1(0,1)$, and it satisfies
\eqref{e1}--\eqref{e2}.

Such equation arises in the studies of some degenerate parabolic
equations and in Non-Newtonian fluids; see
 \cite{a2,b1,b2,b3,y2}. The interesting feature of \eqref{e1} is the lower term
both is singular at
$u=0$ and depends  on the first derivative.


Recently, the one-dimensional singular p-Laplacian differential
equations without dependence on the first derivative have been
studied extensively, see \cite{a1,j1,y1}  and references therein. When it
depends on the first derivative, however, it has not received
much attention, see \cite{j2,l1,o1,w1}.  Recently, the authors \cite{y3},
considered the equation
\begin{equation*}
 \big(\phi_p(u')\big)'-\lambda\frac{|u'|^p}{u}+g(t)=0,\quad  0<t<1,
\end{equation*}
subject to \eqref{e2}, and  proved, by the classical method of elliptic
regularization, that the problem admits one positive solution if
  $p \geq 2, \lambda>0$ and $g \in C[0, 1]$
with $g  >0$ on $[0, 1]$. In the present paper we extend the
  result and obtain the sufficient conditions of existence. Our argument is
based on  regularization technique, upper and lower solutions method
and Arzel\'a-Ascoli theorem.  In addition, an example is
also given to illustrate our main result.

\section{Main result}

The following hypotheses will be adopted in this section:
\begin{itemize}
\item[(H1)] $1 \leq m <p$.

\item[(H2)] $f(t, r)$ is a  positive,  continuous function in
 $[0,+\infty) \times \mathbb{R}$, and  there exist
constants $\alpha>0$, $\beta \in [0, 1)$ such that
$f(t, r) \leq \alpha +\beta|r|^{p-1}$, for all $(t, r)\in [0, 1]\times
\mathbb{R}$.

\item[(H3)] $ \lambda>\inf_{r\geq 1}{H}(r)$,
where $H(r):\mathbb{R}^+\to \mathbb{R}^+$ is defined by
\begin{equation*}
H(r)=\alpha r^{m-p}+\beta r^{m-1}.
\end{equation*}
\end{itemize}

\begin{remark} \label{rmk1} \rm
Let $m \in (1, p)$ and define
\begin{equation*}
 X_0=\Big(\frac{\alpha(p-m)}{\beta(m-1)}\Big)^{1/(p-1)};\quad
X_*=\begin{cases}
X_0,& X_0 \geq1\\
1.&X_0< 1
\end{cases}
\end{equation*}
Then   $\inf_{s\geq 1}H(s)=H(X_*)$.
     Indeed, since
$\lim_{s\to 0^+}H(s)=\lim_{s \to +\infty}H(s)=+\infty$, $H(s)$
must reach a minimum at some   $s \in (0,\infty)$ satisfying $H'(s)=0$.
Solving it gives $s=X_0$ and hence, $\inf_{s>0}H(s)=H(X_0)$.
Since $H'(s) \geq 0$ for all $s \geq X_0$, we see that
$\inf_{s \geq 1}H(s)=H(X_0)$ if $X_0 \geq 1$, and
$\inf_{s\geq 1}H(s)=H(1)$  if $X_0 < 1$.
\end{remark}

The main result of this paper is stated as follows.

\begin{theorem} \label{thm1}
Under Assumptions {\rm (H1)--(H3)},  problem \eqref{e1}--\eqref{e2}
has at least one solution.
\end{theorem}

\begin{remark} \label{rmk2} \rm
 If $m=1$ and $f\equiv 1$ (taking $\alpha=1,\beta=0$), then
 $\inf_{s\geq 1}H(s)=0$. Clearly,
 Theorem \ref{thm1} is an extension of the corresponding result of \cite{y3}.
\end{remark}

\subsection*{Proof of Theorem \ref{thm1}}
Let $\epsilon \in (0, 1)$, and define
$H_\epsilon(t, v,\xi): (0, 1)\times\mathbb{R}\times\mathbb{R}\to \mathbb{R}$
by
$$
H_\epsilon(t, v,\xi)= \lambda \frac{|\xi|^p}{[I_\epsilon(v)]^m}-f(t,\xi),
$$
where  $I_\epsilon(v) =v+\epsilon$ if $v \geq 0$,
$I_\epsilon(v) =\epsilon$ if $v < 0$. By (H2) and using the
inequality: $a^{p-1} \leq a^p+1$, for all $a \geq 0$,  we
have
\begin{equation*} %\label{0}
|H_\epsilon(t, v,\xi)| \leq
  \frac{\lambda}{\epsilon^{m}}|\xi|^p + \alpha+\beta|\xi|^{p-1}
\leq   \big( \frac{\lambda}{\epsilon^{m}}
 +\alpha +\beta \big)\mathcal{H}(|\xi|)
\end{equation*}
for all $(t, v, \xi) \in (0, 1)\times\mathbb{R}\times\mathbb{R}$,
where $\mathcal{H}(s)=1+s^p$ for $s \geq 0$. Denote
$\mathfrak{M}=\{u \in C^1(0, 1); |u'|^{p-2}u' \in C^1(0, 1)\}$, and
define $\mathscr{L}_\epsilon: \mathfrak{M}\to C(0, 1)$ by
\begin{equation*}
 ( \mathscr{L}_\epsilon
 u)(t)=-\big(\phi_p(u')\big)'+H_\epsilon(t,u,u'),\quad 0<t<1.
\end{equation*}
Consider the  problem:
 \begin{equation}
\begin{gathered}\label{01}
( \mathscr{L}_\epsilon  u)(t)=0,\quad 0<t<1,\\
u(1)=u(0)=0.
\end{gathered}
\end{equation}
We call $u \in \mathfrak{M}$ is an upper solution (lower solution)
of problem \eqref{01}  if $  \mathscr{L}_\epsilon  u  \geq 0$
($\leq 0$) in $(0, 1)$ and $u(t) \geq (\leq)0$ for
$t=0, 1$.


We will apply the upper and lower solutions method (see
\cite[Theorem 1 and Remark 2.4]{j2}) to show the existence of
solutions of problem \eqref{01}. Obviously,
$\int_0^{+\infty} \frac{s^{p-1}}{\mathcal{H}(s)} ds=+\infty$,
 thus the condition \cite[Eq. (2.3)]{j2} is satisfied. Then it
suffices to find a lower solution and an upper solution to obtain a
solution.

Let $ \inf_{s\geq1}H(s)\equiv\delta$. Then it follows
from the definition of infimum and $\lambda>\delta$ that for
$\delta_0=\frac{\lambda-\delta}{2}>0$, there exists $S^* \geq
1$ such that $H(S^*)<\delta+\delta_0<\lambda$.



\begin{lemma} \label{lem1}
 There exists a   constant $\epsilon_0\in(0, 1)$, such that for any
 $\epsilon \in (0, \epsilon_0)$, $U_\epsilon=S^*(t+\epsilon)$ is
an upper solution of \eqref{01}.
\end{lemma}

\begin{proof}  Noticing $U_\epsilon \geq \epsilon$ in $(0,1)$ and
$m \geq 1$, we have
\begin{align*}
\mathscr{L}_\epsilon U_\epsilon
&=-\big(|U_\epsilon'|^{p-2}U_\epsilon'\big)'
+\lambda\frac{|U_\epsilon'|^p}{(U_\epsilon+\epsilon)^m}- f(t, U_\epsilon')
\\
&=\frac{\lambda {S^*}^{p-m}}{(t+\epsilon+\epsilon/S^*)^m}-
f(t, S^*) \\
& \geq\frac{\lambda {S^*}^{p-m}}{(1+\epsilon+\epsilon/S^*)^m}-
\alpha-\beta {S^*}^{p-1}\\
&= {S^*}^{p-m}[\lambda-H(S^*)]
+r_\epsilon,\quad 0 <t < 1,
 \end{align*}
where $ r_\epsilon=\lambda
{S^*}^{p-m}[(1+\epsilon+\epsilon/S^*)^{-m}-1]$. Clearly,
$r_\epsilon\to 0$ ($\epsilon\to 0$). Since
$\lambda>H(S^*)$, there exists a constant $\epsilon_0 \in (0, 1)$,
such that for any $\epsilon \in (0, \epsilon_0)$ there holds
${S^*}^{p-m}  [\lambda-H(S^*) ] +r_\epsilon \geq 0$. So that we
obtain $ {\mathscr{L}_\epsilon}U_\epsilon  \geq 0$ in $(0, 1)$
for all $\epsilon \in (0, \epsilon_0)$.
The lemma follows.
\end{proof}


\begin{lemma} \label{lem2}
Let $W= C \Phi^{\alpha}$, where
$\alpha=\frac{p}{p-m}$, $\Phi(t) $ is defined by
$$
 \Phi(t)=
\frac{p-1}{p}\big[\big(\frac{1}{2}\big)^{p/(p-1)}-\big|
\frac{1}{2}-t\big|^{p/(p-1)}\big],
\quad  0 \leq t\leq 1,
$$
and $ C \in (0,1)$ such that $C\alpha <1$ and
$(C\alpha)^{p-1}   +\lambda C^{p-m}\alpha^p \leq \min_{[0, 1]
\times [-1,1]}f(s, r)$.
 Then $W$ is a lower solution of problem \eqref{01}.
\end{lemma}

\begin{proof}
It is easy to check that $\Phi$ has the
 following properties:
\begin{itemize}
\item[(a)] $\Phi>0$ in $(0, 1)$, $\Phi \in C^1[0,1]$.

\item[(b)] $(|\Phi'|^{p-2} \Phi')'=-1$ in $(0, 1)$,
$\Phi(1)=\Phi(0)=0$.

\item[(c)] $\Phi(t)\leq t$, $|\Phi'(t)|\leq 1$,  for all $t\in [0,1]$.

\end{itemize}
Using these properties of $\Phi$, by some calculations, we have
 \begin{align*}
\mathscr{L}_\epsilon W
&=-\big(|W'|^{p-2}W\big)' +\lambda
 \frac{|W'|^p}{(W+\epsilon)^m}- f(t, W')\\
&\leq-\big(|W'|^{p-2}W\big)' +\lambda
 \frac{|W'|^p}{W^m}- f(t, W')\\
&=-(C\alpha)^{p-1}\Phi^{(\alpha-1)(p-1)}\big(|\Phi'|^{p-2}\Phi'\big)'\\
&\quad -(C\alpha)^{p-1}(\alpha-1)(p-1)\Phi^{(\alpha-1)(p-1)-1}|\Phi'|^p\\
&\quad +\lambda C^{p-m}\alpha^p|\Phi'|^p-  f(t, C\alpha \Phi^{\alpha-1}\Phi')\\
&\leq (C\alpha)^{p-1} \Phi^{(\alpha-1)(p-1)}
+\lambda C^{p-m}\alpha^p|\Phi'|^p- \min_{[0, 1]\times [-1, 1]}f(s, r)\\
&\leq(C\alpha)^{p-1} +  \lambda C^{p-m}\alpha^p - \min_{[0,
1]\times [-1, 1]}f(s, r) \leq 0,\quad 0<t<1.
\end{align*}
Thus the  lemma follows.
\end{proof}


According to \cite[Theorem 1 and Remark 2.4]{j2},  for fixed
$\epsilon \in (0, \epsilon_0)$  problem \eqref{01} has a
solution $u_\epsilon \in C^1[0, 1]$ satisfying
$|u_\epsilon'|^{p-2}u_\epsilon' \in C^1(0,1)$ and
\begin{equation}\label{05}
U_\epsilon \geq u_\epsilon \geq W>0,\quad  t\in(0,1).
\end{equation}
Hence $u_\epsilon$ satisfies
\begin{equation} \label{07}
 -\big(|u_\epsilon'|^{p-2}u_\epsilon'\big)'
+\lambda\frac{|u_\epsilon'|^p}{(u_\epsilon+\epsilon)^m}
- f(t, u_\epsilon')= 0,\quad t \in (0, 1).
\end{equation}


\begin{lemma} \label{lem3}
 For all $\epsilon \in (0, \epsilon_0)$, we have
\begin{equation} \label{31}
|u_{\epsilon}'(t) | \leq [\alpha (1-\beta)^{-1}]^{1/(p-1)},\quad
 \forall t \in [0,1].
\end{equation}
\end{lemma}

\begin{proof}
 Noticing that $u_{\epsilon}(1)=u_{\epsilon}(0)=0 $ and
$u_{\epsilon} \geq0$ on $[0, 1]$, we have
\begin{equation} \label{12}
u_{\epsilon}'(0)\geq0 \geq u_{\epsilon}'(1).
\end{equation}
 From \eqref{07}, we obtain
\begin{equation}\label{08}
  \big(|u_\epsilon'|^{p-2}u_\epsilon'\big)'+ \alpha
 +\beta|u_\epsilon'|^{p-1} \geq  0,\quad t \in (0, 1).
\end{equation}
Let $\chi=\phi_p(u_\epsilon')$. Then  we obtain from \eqref{08},
$ \chi'+ \alpha+\beta|\chi| \geq   0$, $t \in (0, 1)$;
i.e.,
$ \big(\int_0^{\chi(t)} \frac{1}{\alpha+\beta|s| }ds+t\big)' \geq  0$,
$t \in (0, 1)$. This and \eqref{12} give
$1\geq \int_0^{\chi(t)} \frac{1}{\alpha+\beta|s| }ds+t  \geq  0$,
$ t \in [0, 1]$,
 hence $ \big|\int_0^{\chi(t)} \frac{1}{\alpha+\beta|s| }ds\big|\leq 1$,
$ t \in [0, 1]$. Using the inequality:
$ |\int_0^{y} \frac{1}{\alpha+\beta|s| }ds| \geq \frac{|y|}{\alpha+\beta|y|}$
($y \in \mathbb{R}$), we deduce that
$|\chi| \leq   \alpha+\beta|\chi|$, $t \in [0, 1]$;
that is, $ |\chi| \leq \alpha(1- \beta)^{-1}$ on $[0, 1]$.
The lemma is proved.
\end{proof}

\begin{lemma} \label{lem4}
 For each $\delta \in (0,1/2)$, there exists a
positive constant $C_\delta$ independent of $\epsilon$, such that
for all $\epsilon \in (0, \epsilon_0)$
\begin{equation}\label{22}
|u_\epsilon'(t_2)-u_\epsilon'(t_1)| \leq
C_\delta|t_2-t_1|^{\gamma},\quad \forall t_2,t_1 \in [\delta,
1-\delta],
\end{equation}
where $\gamma=1/(p-1)$ if $p \geq 2$; $\gamma=1$ if $1<p<2$.
\end{lemma}

\begin{proof}
By \eqref{05} and \eqref{31}, it is easy to derive
from \eqref{07} that for any $ \delta\in (0,1/2)$ there exists a
constant $C_\delta>0$ independent of $\epsilon$, such that for all
$\epsilon \in (0, \epsilon_0)$
\begin{equation} \label{30}
\big|(|u_\epsilon'|^{p-2} u_\epsilon')'\big| \leq
C_\delta,\quad \delta \leq t \leq 1-\delta.
\end{equation}
Recalling the inequality (see \cite{d1})
 \begin{equation*}
 (|\eta|^{p-2}\eta-|\eta'|^{p-2}\eta') \cdot (\eta-\eta')
 \geq \begin{cases}
 C_1 |\eta - \eta '|^p,& p \geq 2\\
 C_2 (|\eta|+|\eta'|)^{p-2}|\eta - \eta '|^2,&1<p<2
\end{cases}
\end{equation*}
for each $\eta \in \mathbb{R}$, where $C_i(i=1,2)$ are  positive
constants depending only on $p$, we derive, by \eqref{30}, that if
$p\geq 2$, then
\begin{align*}
|u_{\epsilon}'(t_2)-u_{\epsilon}'(t_1)|^p
&\leq C_2^{-1}[u_{\epsilon}'(t_2)-u_{\epsilon}'(t_1)]\cdot
[|u_{\epsilon}'(t_2)|^{p-2}u_{\epsilon}'(t_2)-|u_{\epsilon}'(t_1)|^{p-2}
u_{\epsilon}'(t_1)]\\
&\leq C_\delta|u_{\epsilon}'(t_2)-u_{\epsilon}'(t_1)||t_2-t_1|,\quad
\forall t_2,t_1\in [\delta, 1-\delta],
 \end{align*}
hence
\begin{equation*}
 |u_{\epsilon}'(t_2)-u_{\epsilon}'(t_1)| \leq
C_\delta |t_2-t_1|^{1/(p-1)},\quad \forall t_2,t_1\in [\delta,
1-\delta],
\end{equation*}
and if  $p \in (1, 2)$,  then
\begin{align*}
&|u_{\epsilon}'(t_2)-u_{\epsilon}'(t_1)|^2[|u_{\epsilon}'(t_2)|
+|u_{\epsilon}'(t_1)|]^{p-2}\\
&\leq C_2^{-1}[u_{\epsilon}'(t_2)-u_{\epsilon}'(t_1)]\cdot
[|u_{\epsilon}'(t_2)|^{p-2}u_{\epsilon}'(t_2)
 -|u_{\epsilon}'(t_1)|^{p-2}u_{\epsilon}'(t_1)]\\
&\leq C_\delta|u_{\epsilon}'(t_2)-u_{\epsilon}'(t_1)||t_2-t_1|,\quad
\forall t_2,t_1\in [\delta, 1-\delta].
 \end{align*}
Then, \eqref{31} yields
\begin{equation*}
 |u_{\epsilon}'(t_2)-u_{\epsilon}'(t_1)|    \leq
C_\delta|t_2-t_1|[|u_{\epsilon}'(t_2)|+|u_{\epsilon}'(t_1)|]^{2-p}
\leq   C_\delta|t_2-t_1|
\end{equation*}
for all $t_2, t_1\in [\delta, 1-\delta]$. This  completes the proof.
\end{proof}

By \eqref{31} and \eqref{22} and using Arzel\'a-Ascoli
theorem,   there exist  a subsequence of $\{u_{\epsilon}\}$, still
denoted by $\{u_\epsilon\}$, and a function $u \in C^1(0,1)\cap
C[0,1]$ such that,  as $\epsilon \to 0$,
\begin{equation}
\begin{gathered}\label{21}
 u_\epsilon \to u,\quad \text{uniformly in } C[0,1],\\
u_\epsilon \to u,\quad \text{uniformly in } C^1[\delta,1-\delta],
\end{gathered}
\end{equation}
where $\delta \in (0,1/2)$, and hence from
$u_\epsilon(1)=u_\epsilon(0)=\epsilon $ and \eqref{05} we derive
that $u(1)=u(0)=0$, and $ u(t) \geq C\Phi^{p/(p-m)}$,
$t\in[0, 1]$;  therefore $u>0$ in $(0, 1)$.

We now show that $u$ satisfies \eqref{e1}. Integrating \eqref{07} over
$(t_0,
 t)$ gives
\begin{equation*}
   |u_\epsilon'(t)|^{p-2}u_\epsilon'(t)
=\int_{t_0}^t\Big(
   \lambda\frac{|u_\epsilon'|^p}{(u_\epsilon+\epsilon)^m}
   -   f(s, u_\epsilon')\Big)ds+|u_\epsilon'(t_0)|^{p-2}u_\epsilon'(t_0),
\end{equation*}
and  letting $\epsilon \to 0$ and using Lebesgue's dominated
convergence theorem yield
\begin{equation}\label{23}
|u'(t)|^{p-2}u'(t) =\int_{t_0}^t\Big( \lambda\frac{|u'|^p}{u^m}-
    f(s, u')\Big)ds+|u'(t_0)|^{p-2}u'(t_0).
\end{equation}
This shows that $|u'(t)|^{p-2}u'(t) \in C^1(0, 1)$ and \eqref{e1} is
satisfied.

 It remains to show that $u \in C^1[0, 1]$.
Integrating \eqref{07} over $(0, 1)$
 and   using   \eqref{31} and \eqref{12}, we derive that
\begin{equation*}
\int_0^1 \frac{|u_\epsilon'|^p}{(u_\epsilon+\epsilon)^m}dt \leq
\frac{1}{\lambda}\min_{[0, 1]\times[-Y, Y]}f(t, r),\quad
Y:=\big(\frac{\alpha}{1-\beta}\big)^{1/(p-1)},
\end{equation*}
 and letting $\epsilon\to 0$
 and using Fatou's lemma and \eqref{21}, we obtain
\begin{equation*}
\int_0^1 \frac{|u'|^p}{u^m}dt \leq \frac{1}{\lambda}
\min_{[0, 1]\times[-Y, Y]}f(t, r).
\end{equation*}
So, $\frac{|u'|^p}{u^m}  \in L^1[0,1]$. By \eqref{23}, the function
 $\omega(t)=|u'(t)|^{p-2}u'(t) = \phi_{p}(u'(t))$
 is absolutely continuous on $[0,1]$.
Since $u'(t)=\phi_q(\omega(t))$($\frac1p + \frac1q =1$),
  $u' \in C[0,1]$.  The proof of Theorem \ref{thm1} is complete.

\subsection*{Example}
Let $\lambda>4/27$. Consider the  problem
\begin{equation}
 \begin{gathered}\label{13}
 (|u'|^3 u')'-
  \lambda\frac{|u'|^5}{u^{2}}+\frac{(\frac{1}{2}+\frac{\sqrt{6}}{18}|u'|^2)^{2}}{2t+2
\cos t-1}+\frac{\sin(\pi t)}{\sqrt{4+|u'|^3}}=0,\quad 0<t<1,\\
 u(1)=u(0)=0.
\end{gathered}
\end{equation}
Let $ p=5$,  $m=2$,
$$
f(t, r)=\frac{(\frac{1}{2}+\frac{\sqrt{6}}{18}r^2)^{2}}{2t+2 \cos
t-1}+\frac{\sin(\pi t)}{\sqrt{4+|r|^3}}.
$$
Since $(2t+2\cos t-1)'=2(1-
 \sin t)\geq 0$, $1+2\cos 1 \geq 2t+2\cos t-1 \geq
1$ for all $ t \in [0, 1]$ and hence,  noticing $0 \leq
\frac{\sin(\pi t)}{\sqrt{4+|r|^3}} \leq \frac12$ for $(t, r)
\in [0, 1]\times \mathbb{R}$, we obtain
\begin{equation*}
\frac{1}{1+2\cos1}\big(\frac{1}{2}+\frac{\sqrt{6}}{18}r^2\big)^{2}
\leq f(t, r)\leq \big(\frac{1}{2}+\frac{\sqrt{6}}{18}r^2\big)^{2}
+\frac{1}{2},
\end{equation*}
for $ (t, r) \in [0, 1] \times \mathbb{R}$.
By the inequality $(a+b)^2 \leq 2(a^2+b^2)$,
we have
\begin{equation*}
   f(t, r)\leq 1+\frac{1}{27}r^4,\quad (t, r) \in [0, 1] \times \mathbb{R}.
\end{equation*}
Let $\alpha=1$, $\beta=\frac{1}{27}$. Then $X_0=3$, and therefore
$X_*=3$ and  $\inf_{s \geq 1}H(s)=H(X_*)=H(X_0)=\frac{4}{27}$
(see Remark \ref{rmk1}).   Thus all assumptions
of Theorem \ref{thm1} are satisfied for any $\lambda>\frac{4}{27}$, so
problem \eqref{13} has at least one solution.

\subsection*{Acknowledgments}
The author  wants to thank the anonymous referee for
pointing out some errors on the original manuscript.


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