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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 193, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/193\hfil Infinite semipositone problems]
{Infinite semipositone problems with a falling zero and
nonlinear boundary conditions}

\author[M. Mallick, L. Sankar, R. Shivaji, S. Sundar \hfil EJDE-2018/193\hfilneg]
{Mohan Mallick, Lakshmi Sankar, Ratnasingham Shivaji, Subbiah Sundar}

\address{Mohan Mallick \newline
Department of Mathematics,
IIT Madras, Chennai-600036, India}
\email{mohan.math09@gmail.com}

\address{Lakshmi Sankar \newline
Department of Mathematics,
IIT Palakkad, Kerala-678557, India}
\email{lakshmi@iitpkd.ac.in}

\address{Ratnasingham  Shivaji \newline
Department of Mathematics \& Statistics,
University of North Carolina at
Greensboro, NC 27412, USA}
\email{shivaji@uncg.edu}

\address{Subbiah Sundar \newline
Department of Mathematics,
IIT Madras, Chennai-600036, India}
\email{slnt@iitm.ac.in}

\dedicatory{Communicated by Pavel Drabek}

\thanks{Submitted October 15, 2018. Published November 27, 2018.}
\subjclass[2010]{35J25, 35J66. 35J75}
\keywords{Infinite semipostione; exterior domain; sub and super solutions; 
\hfill\break\indent nonlinear boundary conditions}

\begin{abstract}
We consider the problem 
 \begin{gather*}
 -u'' =h(t)\big(\frac{au-u^{2}-c}{u^\alpha}\big) , \quad t \in (0, 1),\\
 u(0) = 0, \quad u'(1)+g(u(1))=0,
 \end{gather*}
 where $a>0$, $c\geq 0$, $\alpha \in (0, 1)$, $h{:}(0, 1] \to (0, \infty)$
 is a continuous function which may be singular at $t=0$, but belongs to
 $L^1(0, 1)\cap C^1(0,1)$, and $g{:}[0, \infty) \to [0, \infty)$ is a continuous
 function. We discuss existence, uniqueness, and non existence results for
 positive solutions for certain values of $a$, $b$ and $c$.
\end{abstract}

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

\section{Introduction}

In this article, we consider the boundary-value problem
\begin{equation} \label{P}
\begin{gathered}
-u'' =h(t)\big(\frac{au-u^{2}-c}{u^\alpha}\big) , \quad t \in (0, 1),
\\u(0) = 0, u'(1)+g(u(1))=0,
\end{gathered}
\end{equation}
where $a>0$, $c\geq 0$, $\alpha \in (0, 1)$, and
$g{:}[0, \infty) \to [0, \infty)$ is a continuous function.
The function $h{:}(0, 1] \to (0, \infty)$ is a continuous function which satisfies:
\begin{itemize}
\item[(H1)] there exists $\epsilon_1>0, 0<\gamma<1-\alpha$, such that
 $h(s)\leq 1/s^\gamma$ for all $s \in (0, \epsilon_1)$,

\item [(H2)] $\inf_{s \in (0,1)}h(s) =\hat{h}> 0$.
\end{itemize}
Note that, for the nonlinear function $f(s)=(as-s^2-c)/s^\alpha$,
$\lim_{s\to 0^+} f(s)= -\infty$. This singularity together with the fact
that the solution needs to satisfy a Dirichlet boundary condition creates
a challenge in establishing the existence of positive solutions.
Such problems are referred in the literature as ``infinite semipositone" problems.
See \cite{ HSS, LSS, LSY, RSY, Z}, where infinite semipositone problems
have been studied when the nonlinearity $f$ only has a single zero beyond
which it is positive and increasing to infinity. The analysis is more
challenging when the reaction term $f$ has a second zero (falling zero)
beyond which it is negative. See \cite{GLSS,LSY2} where this study was
achieved in the case when Dirichlet boundary conditions persisted on the
entire boundary. In this paper, we extend this study to an even more challenging
situation, namely when a nonlinear boundary condition is involved on part of
the boundary.

Problems of the form \eqref{P} arise while studying radial solutions of
\begin{equation}\label{ext}
\begin{gathered}
 -\Delta u = K(|x |) \big(\frac{au-u^{2}-c}{u^{\alpha}}\big),\quad x \in \Omega, \\
\frac{\partial u}{\partial \eta}+g(u) = 0, \quad \text{if }|x | = r_0 ,\\
 u \to0, \quad \text{as }|x |\to\infty,
 \end{gathered} 
\end{equation}
where $\Omega=\{x\in \mathbb{R}^{n}:  |x |>r_0\}$ is an exterior domain, 
$n> 2$, $a, c, \alpha$ are as before, and $K:[r_0,\infty)\to(0,\infty)$ 
belongs to a class of continuous functions such that $\lim_{r\to \infty}K(r)=0$.
By using the transformation: $r=|x|$ and $s=(\frac{r}{r_0})^{(2-n)}$, we can 
reduce \eqref{ext} to \eqref{P}, where
 $h(s)= \frac{r_0^2}{(2-n)^2}s^{\frac{-2(n-1)}{n-2}}K(r_0s^{\frac{1}{2-n}})$ 
(see \cite{DKS}).
Note that if we assume $K \in C( [r_0,\infty), (0,\infty))$ and satisfies
$\frac{d_1}{r^{n+\sigma}}\leq K(r) \leq \frac{d_2}{r^{n+\sigma}}$ 
for some $d_1, d_2 >0$, and for
$\sigma \in ((n-2)\alpha, n-2)$, then $h$ satisfies our assumptions 
(H1) and (H2).

When the boundary condition at $|x|=r_0$ is replaced by a Dirichlet's condition, 
i.e.\ $u=0$, the same transformation reduces the problem to
\begin{equation}\label{Dir}
\begin{gathered}
-u'' =h(t)\big(\frac{au-u^{2}-c}{u^\alpha}\big) , \quad t \in (0, 1),\\
u(0) = 0,\quad  u(1)=0.
\end{gathered} 
\end{equation}
The existence of positive solutions of this Dirichlet problem was studied 
in \cite{GLSS}.
For given values of $a>0, \alpha \in (0, 1)$, the authors established the 
existence of positive solution for small values of $c$. In this paper, 
we extend this study to the case when a nonlinear boundary condition is 
satisfied at $|x|=r_0$.

In particular, we will show that \eqref{P} has a positive solution with $u(1)>0$,
which clearly shows that it is not a solution of \eqref{Dir}. 
Hence combining our result with the existence result obtained in \cite{GLSS}, 
we also see that the  problem
 \begin{gather*}
 -\Delta u = K(|x |) (\frac{au-u^{2}-c}{u^{\alpha}}), \quad  x \in \Omega,\\
u\big[\frac{\partial u}{\partial \eta}+g(u)\big] = 0, \quad \text{if }|x | = r_0 ,
 \\u\to0, \quad \text{as }|x |\to\infty,
 \end{gather*} 
has at least two positive radial solutions for certain values of $a$ and $c$.
 Existence of positive solutions to certain problems with such boundary 
conditions are discussed in \cite{GLS2,GLS}.

The study of such steady state reaction diffusion equations are of great 
importance in various applications. See in particular \cite{OSS} for a problem
arising  in ecology. See also \cite{BFC,Clark,GLS2,GLS}.
 Here we consider more challenging reaction diffusion models, namely, 
when nonlinear diffusion is involved (when the diffusion term is 
$u^\alpha \Delta u$ instead of $\Delta u$).

Below, we state our results for \eqref{P}. We first establish a non 
existence result for \eqref{P}. For this we assume 
\begin{itemize}
\item[(H3)] $h \in C^1\big((0, 1], (0, \infty)\big)$, and $h'(s)<0$ for $s >0$.
 \end{itemize}
Note that if the weight function $K$ in \eqref{ext} is such that $K$ is $C^1$ and
 $\frac{K(r^{-1})}{r^{2(n-1)}}$ is decreasing for $r>0$, then the corresponding 
$h$ satisfies (H3). A simple example of $K$ which satisfies our assumptions is
 $K(r)=\frac{d_1}{r^{n+\sigma}}$, where $d_1>0$, and 
$\sigma \in ((n-2)\alpha, n-2)$.

\begin{theorem}\label{thm2}
Assume $h$ satisfies {\rm (H1), (H3)}, and $g{:}[0,\infty)\to[0,\infty)$, 
is a continuous function. Then for given $a>0$ and $\alpha \in (0, 1)$, 
there exists $\hat{c}(a)=\frac{(3-\alpha)(1-\alpha)}{(2-\alpha)^2}\frac{a^2}{4}$ 
such that if $c>\hat{c}$, \eqref{P} has no nonnegative solution.
\end{theorem}

\begin{remark} \rm
Note that if $c>a^2/4$, then $f(s)=\frac{as-s^2-c}{s^{\alpha}}<0$ for all $s>0$ 
and this will immediately imply the non existence of nonnegative solution 
of \eqref{P}. This follows from the fact that, since $u(0)=0$ and 
$u'(1) \leq 0$, there exists a $\tilde{t}\in(0,1)$ such that 
$u''(\tilde{t})\leq 0$.
\end{remark}

\begin{remark} \rm
From the proof of Theorem\eqref{thm2}, we also see that, for a given $c>0$ and 
$\alpha \in (0, 1)$, there exists $\hat{a}(c)$ such that if $a<\hat{a}$, 
\eqref{P} has no nonnegative solution.
\end{remark}

Next, we state an existence result for \eqref{P} for the case when $c=0$.

\begin{theorem} \label{thm4}
Let $\alpha\in(0,1)$, $c=0$, and $g{:}[0,\infty)\to[0,\infty)$ is a continuous 
function. Assume $h{:}(0, 1] \to (0, \infty)$ is a continuous function which 
satisfies {\rm (H1)} and {\rm (H2)}. Then, there exists $\underline{a}>0$ 
such that if $a\geq\underline{a}$, \eqref{P} has a positive solution $u$ 
with $u(1)>0$.
\end{theorem}

\begin{remark} \rm
If $\hat{g}={\inf_{s\in[0,\infty)}g(s)}>0$, then integrating \eqref{P} from 
$0$ to $1$ with $c=0$, it is easy to see that for 
$a\leq[\frac{(2-\alpha)^{2-\alpha}}{(1-\alpha)^{1-\alpha}}\frac{\hat{g}}{\|h\|_1}
]^{\frac{1}{2-\alpha}}$, \eqref{P} has no positive solution.
\end{remark}

Under an additional assumption on $g$, we also establish the uniqueness
 of the positive solution obtained in Theorem \ref{thm4} for \eqref{P} when $c=0$. 
For this we assume
\begin{itemize}
\item[(H4)] $g(x)/x$ is nondecreasing for $x \in [0, \infty)$.
 \end{itemize}
Then we have the following uniqueness result.

 \begin{theorem}\label{thm3}
Let $a>0$, $c=0$, $\alpha \in (0, 1)$, and $h{:}(0, 1] \to (0, \infty)$ be a
continuous function which satisfies {\rm (H2)}. Assume also that 
$g{:}[0,\infty)\to[0,\infty)$ is a continuous function which satisfies 
{\rm (H4)}. Then \eqref{P} has at most one positive solution.
 \end{theorem}

 Finally, we state  our main existence result in this paper for \eqref{P}.

\begin{theorem}\label{thm1}
Let $\alpha\in(0,1)$ and $g{:}[0,\infty)\to[0,\infty)$ is a continuous function. 
Assume $h{:}(0, 1] \to (0, \infty)$ is a continuous function which satisfies 
{\rm (H1)} and {\rm (H2)}. Then, there exists $\bar{a}>0$, and for 
$a\geq\bar{a}$, $\bar{c}(a)>0$ such that for $c\leq\bar{c}$, \eqref{P} has 
a positive solution $u$ with $u(1)>0$. Further, this $\bar{c}$ is an 
increasing function of $a$ such that $\bar{c}(a)\to \infty$ as $a\to \infty$.
\end{theorem}

\begin{remark} \label{remA} \rm
From the proof of Theorem\eqref{thm1}, it is easy to see that, for any given 
$c\leq \bar{c}(\bar{a})$, there exists $a_*(c)$ such that for $a\geq a_*$, 
\eqref{P} has a positive solution.
\end{remark}

Figure \ref{demo1}  illustrates Theorem \ref{thm1} and 
Remark \ref{remA}. Here $\rho=\|u\|_{\infty}$.

 \begin{figure}[htb]
 \centering
\includegraphics[width=0.48\textwidth]{fig1a} % demo1.png
\includegraphics[width=0.48\textwidth]{fig1b} % demo2.png
 \caption{Bifurcation diagram of \eqref{P}:
left $a$ versus $\rho$, right $c$ versus $\rho$} \label{demo1} %\label{demo2}
\end{figure}

In the next section we recall a method of sub and super solutions established 
in \cite{LSSb}, which will be used to establish our existence results.
 We also provide some preliminary results about the existence of a 
positive eigenfunction for certain eigenvalue problems, which will 
be useful in the construction of our subsolution required in the proof 
of Theorem \ref{thm1}. The proofs of the theorems are
provided in the later sections. In the last section, we provide some 
exact bifurcation diagrams of positive solutions of \eqref{P} when $h(t) \equiv 1$.

\section{Preliminary results}

 We first discuss the method of sub and super solutions. By a subsolution of
 \eqref{P}, we mean a function $\psi\in C^2(0,1)\cap C^1[0,1]$ which satisfies
 \begin{equation}\label{sub}
\begin{gathered}
 -\psi''(t) \leq h(t)(\frac{a\psi(t)-\psi^{2}(t)-c}{\psi^{\alpha}(t)}),\quad 
 t \in (0,1),\\
 \psi(t)>0, \quad t\in(0,1] ,\\
 \psi'(1)+g(\psi(1))\leq0,\\
 \psi(0)=0,
 \end{gathered}
\end{equation}
and by a supersolution of \eqref{P}, we mean a function 
$\phi\in C^2(0,1)\cap C^1[0,1]$ which satisfies
 \begin{equation}\label{sup}
\begin{gathered}
 -\phi''(t) \geq h(t)(\frac{a\phi(t)-\phi^{2}(t)-c}{\phi^{\alpha}(t)}), \quad 
 t \in (0,1),\\
 \phi(t)>0, \quad t\in(0,1] ,\\
 \phi'(1)+g(\phi(1))\geq 0,\\
 \phi(0)=0.
 \end{gathered} 
\end{equation}

 \begin{lemma}[See \cite{LSSb}] \label{lem1}
If there exist a subsolution $\psi$ and a supersolution $\phi$ of \eqref{P} 
such that $\psi\leq\phi$, then \eqref{P} has at least one solution 
$u\in C^2(0,1)\cap C^1[0,1]$ satisfying $\psi\leq u\leq\phi$ in $[0,1]$.
 \end{lemma}

We note here that, in our case, the difficulty lies in the construction of a
positive subsolution, as the subsolution, $\psi$, needs to satisfy 
$\lim_{t \to 0^+}-\psi''(t)=-\infty$, and $-\psi''>0$ in a large part 
of the interior.

Next, we discuss the Sturm-Liouville problem
\begin{equation}
\begin{gathered}
y''(t)+\lambda y(t)=0,\quad t\in(0,1),\\
y(0)=0,\\
y'(1)+ly(1)=0,\label{egp}
\end{gathered}
\end{equation}
where $l>0$, and $\lambda$ is a real parameter.
 We first observe (see also \cite{mtbook}) that the following result holds.

\begin{lemma}\label{lem2}
 For a given $l>0$, the first eigenvalue of \eqref{egp}, 
$\lambda_1\in(\frac{\pi^2}{4},\pi^2)$, and the corresponding eigenfunction 
$\phi_1$ is positive, and is given by $\phi_1(t)=\sin\sqrt{\lambda_1}t$. 
Moreover, as $l\to 0$, $\lambda_1\to \frac{\pi^2}{4}$, and as
 $l\to \infty$, $\lambda_1\to \pi^2$.
\end{lemma}

\begin{proof}
The solution of the equation $y''+\lambda y=0$ is given by 
$\phi(x)=A\cos{\sqrt{\lambda}x}+B\sin{\sqrt{\lambda}x}$.
 Using the boundary conditions, we reduce that $\tan\eta=\frac{-1}{l}\eta$, 
where $\eta=\sqrt{\lambda}$. This equation does not possess an explicit solution. 
But the graphical solutions of this equation can be determined by plotting 
functions $y=\tan{\eta}$ and $y=-\frac{1}{l}\eta$ (see Figure \ref{f1}).

\begin{figure}[htb]
\begin{center}
 \includegraphics[width=0.6\textwidth]{fig2} % graph.png 
\end{center}
\caption{Graph of $\tan\eta$ vs $-1/(l\eta)$} \label{f1}
\end{figure}

From Figure \ref{f1}, it is clear that, there are infinitely many roots 
$\eta_n$ for $n=1,2,\dots$. To each root $\eta_n$, there corresponds an eigenvalue 
$\lambda_n=\eta_n^2$, $n=1,2,3,\dots$. Thus there exists a sequence of eigenvalues
$\lambda_1<\lambda_2<\lambda_3<\dots$ and the corresponding eigenfunctions
 are $\phi_n=\sin{\sqrt{\lambda_n}x}$. From the graph, we observe that the 
first eigenvalue $\lambda_1=\eta_1^2\in(\pi^2/4,\pi^2)$, and hence $\phi_1$ 
is positive. Also note that as $l\to \infty$, $\eta_1 \to \pi$ and, 
as $l\to 0$, $ \eta_1 \to \pi/2$.
\end{proof}

\section{Proof of Theorem \ref{thm2}}
We will first prove the following lemma.

\begin{lemma}\label{lem3}
Let $a>0$, $c\geq0$, $\alpha \in (0, 1)$, and $F(s)=\int_0^s f(t)\,\mathrm{d}t$,
where $f(s)=\frac{as-s^2-c}{s^{\alpha}}$. Let $h\in C((0,1),(0,\infty))$ 
satisfy {\rm (H1)} and {\rm (H3)}. If $F(s)< 0$ for all $s>0$, 
then \eqref{P} has no nonnegative solution.
\end{lemma}

\begin{proof} 
Let us assume that \eqref{P} has a nonnegative solution $u(t)$.
Since $u(0)=0$ and $u'(1) \leq 0$, there exists a $t_0>0$ such that 
$u'(t_0)=0$. Now define $E(t):=F(u(t))h(t)+\frac{[u'(t)]^2}{2}$. 
From (H1), there exists a $d>0$ such that $h(t) \leq \frac{d}{t^{\gamma}}$ 
for $t\in (0,1)$. Integrating \eqref{P} from $t$ to $t_0$ and using the fact 
for $s>0$, $f(s)\leq R$ for some $R>0$, we obtain
\begin{equation}\label{lema}
u'(t)=\int_t^{t_0} h(s)f(u(s))\,\mathrm{d}s
\leq\frac{dR}{1-\gamma}(t_0^{1-\gamma}-t^{1-\gamma})\leq\frac{dR}{1-\gamma}=R_0.
\end{equation}
Again integrating \eqref{lema} from $0$ to $t$, $t<t_0$, we have $u(t)<R_0t$, 
for $t\in(0,t_0)$. Since $f$ is integrable, there exist $k>0$ and $\epsilon>0$ 
such that $|F(u)|\leq ku$ for $u\in (0,\epsilon)$. Hence
$$
{\lim_{t\to0^+}}|F(u(t))|h(t)\leq {\lim_{t\to0^+}}kR_0dt^{1-\gamma}=0.
$$
This implies that ${\lim_{t\to0^+}E(t)}\geq0$.
Now note that $E'(t)=F(u)h'(t)$. From (H3), $h'(t)<0$ for $t\in (0,1)$,
 and $F(s)< 0$ for all $s>0$, $E'(t)>0$ for all $t>0$. 
Therefore $E(t)>0$ for all $t>0$. But $E(t_0)<0$, which is a contradiction.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
 We have
\[
F(s)=\int_0^s f(t)\,\mathrm{d}t=\int_0^s
 \frac{at-t^2-c}{t^{\alpha}}\,\mathrm{d}t
=s^{1-\alpha}\Big(\frac{a}{2-\alpha}s-\frac{1}{3-\alpha}s^2-\frac{c}{1-\alpha}\Big).
\]
The zeros of $F(s)$ are $s=0$ and 
\[
s=\frac{\frac{a}{2-\alpha}\pm\sqrt{\frac{a^2}{(2-\alpha)^2}
-\frac{4c}{(3-\alpha)(1-\alpha)}}}{\frac{2}{3-\alpha}}.
\]
 If $c>\hat{c}(a)$ then $\frac{a^2}{(2-\alpha)^2}-\frac{4c}{(3-\alpha)(1-\alpha)}<0$. 
This implies $F(s)$ has only one zero at $s=0$. Since 
${\lim_{s\to 0^{+}}}F'(s)=-\infty$ and $F(0)=0$, $F(s)<0$ for all $s>0$. 
Hence by Lemma \ref{lem3}, \eqref{P} has no nonnegative solution.
\end{proof}

\section{Proof of Theorem \ref{thm4}}

We first construct a subsolution for $\eqref{P}$ (when $c=0$).
 Let $\phi_1$ be the eigen function corresponding to the first eigenvalue 
$\lambda_1$ of the problem $-\phi''(t)=\lambda \phi(t), t\in (0, 1)$, 
$\phi(0)=\phi(1)=0$. Note that, $\phi_1(t)=\sin \pi t$, and $\lambda_1=\pi^2$.
Fix $k>0$ such that $k \geq \frac{-(g(1)+1)}{\phi_1'(1)}$. 
We now define our subsolution to be $\psi(t)=k \phi_1(t)+t$. 
Let $\underline{a}=\frac{\lambda_1(k+1)^{\alpha}}{\hat{h}}+(k+1)$. 
For $a\geq\underline{a}$, we will show that $\psi$ is a subsolution 
of \eqref{P}. To prove this, we need to show that 
$-\psi''=\lambda_1 k\phi_1 \leq h(t) (a \psi^{1-\alpha}- \psi^{2-\alpha})$, 
$\psi (0) \leq 0$ and $ \psi'(1)+g(\psi(1)) \leq 0$. We will first show that
\begin{equation} \label{v1}
\lambda_1(k\phi_1(t)+t) \leq\hat{h}(a(k\phi_1(t)+t)^{1-\alpha}
-(k\phi_1(t)+t)^{2-\alpha}),
\end{equation}
where $\hat{h}=\inf_{s \in (0,1)}h(s)$. This clearly implies 
$-\psi'' \leq h(t) (a \psi^{1-\alpha}-\psi^{2-\alpha})$ (since $\psi(t) \leq k+1$ 
for all $t$, $a \psi^{1-\alpha}-\psi^{2-\alpha}>0$).
 From the choice of $\underline{a}$,
 $$
\lambda_1(k+1)^{\alpha} \leq \hat{h} (a-(k+1)).
$$ 
From this, we obtain
 $$
\lambda_1(k \phi_1+t)^{\alpha} \leq \hat{h} (a-(k\phi_1+t)),
$$ 
and \eqref{v1} follows.
 Clearly $\psi(0)=0$. Also $\psi'(1)+g(\psi(1))=k\phi_1'(1)+1+g(1) \leq 0$, 
by the choice of $k$.
 Hence $\psi$ is a subsolution of \eqref{P}.
 Next we construct a supersolution of \eqref{P}. Let $e$ be the solution of
\begin{gather*}
-e''(t)=h(t),\quad t\in(0,1),\\
e(0)=e'(1)=0.
\end{gather*}
 Integrating the above equation from $t$ to $1$, we see that
 $e'(t)=\int_t^1 h(s)\,\mathrm{d}s>0$, and hence $e$ is an increasing
function for $t\in[0,1]$. Choose a constant $M>0$ such that 
$\frac{as-s^2}{s^{\alpha}}<M$, for all $s\geq0$. Then clearly $\phi=Me$ 
is a supersolution of \eqref{P}. Also since $e'(0)>0$ if we choose $M$ 
large enough then, $\psi(t)\leq\phi(t)$ for all $t\in[0,1]$. 
Hence, by Lemma \ref{lem1}, there exist a solution $u$ of \eqref{P} such 
that $\psi(t)\leq u(t)\leq\phi(t)$ for all $t\in[0,1]$. 
Clearly $u(1)>0$ since $\psi(1)>0$.

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

Let $u$ and $v$ be two positive solutions of \eqref{P} with $c=0$ such that 
$u\not\equiv v$. Without loss of generality let $t_1\in[0,1)$ be such that
 $v(t_1)-u(t_1)=0$, $v(t)-u(t)\geq0$ in $[t_1,1]$, and $v(t)-u(t)>0$ for some 
$(s_1, s_2)\subset [t_1,1]$. For $t\in(s_1, s_2)$, we have
\begin{align*}
-(uv''-vu'')
&= h(t)\Big(u\frac{av-v^2}{v^{\alpha}}-v\frac{au-u^2}{u^{\alpha}}\Big)\\
&=h(t)\frac{(av-v^2)(au-u^2)}{u^{\alpha} v^{\alpha}}
 \Big(\frac{u^{1+\alpha}}{au-u^2}-\frac{v^{1+\alpha}}{av-v^2}\Big).
\end{align*}
Since for any positive solution $u$, $\|u\|_{\infty}< a$, and 
$\tilde{f}(s)=\frac{s^{1+\alpha}}{as-s^2}$ is a strictly increasing function 
for $s\in(0,a)$, we see that $\int_{t_1}^1 -(uv''-vu'')\,\mathrm{d}t<0$.
Using $v(t_1)=u(t_1)$, $v'(t_1)\geq u'(t_1)$, and (H4), we obtain
\begin{align*}
&\int_{t_1}^1 -(uv''-vu'')(t)\,\mathrm{d}t\\
&=[-uv'+vu']_{t_1}^1\\
&=v(1)u'(1)-u(1)v'(1)-\left(v(t_1)u'(t_1)+u(t_1)v'(t_1)\right)\\
&=-v(1)g(u(1))+u(1)g(v(1))+u(t_1)v'(t_1)-u(t_1)u'(t_1)\\
&\geq-v(1)g(u(1))+u(1)g(v(1))\\
&\geq  u(1)v(1)\left( \frac{g(v(1))}{v(1)}- \frac{g(u(1))}{u(1)}\right)\geq 0,
\end{align*}
which a contradiction, and hence $u\equiv v$.

\section{Proof of Theorem \ref{thm1}}


\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig3} % graph1.png
\end{center}
\caption{Graph of $A_1(k)$ vs $A_2(k)$}\label{f2}
\end{figure}


We first construct a subsolution. For this, we fix a 
$\beta\in(1,\frac{2-\gamma}{1+\alpha})$. From (H1), it is clear that this 
interval is nonempty. Now, for $k\geq0$, we define
\begin{gather}
A_1(k):=2k+\frac{2\beta\pi^2k^{\alpha}}{\hat{h}},\label{T1.1}\\
A_2(k):=-\frac{3\pi}{4\sqrt{2}}+\frac{g(\frac{1}{k^{\beta-1}})}{\beta k^{2-\beta}}.
 \label{T1.2}
\end{gather}
 It is easy to see that $A_1(k)$ is an increasing function of $k$ and $A_2(k)$ 
is negative for $k$ large (see Figure \eqref{f2}). Let $r_{A_2}$ be the least 
nonnegative number such that $A_2(k)\leq 0$ for all $k\geq r_{A_2} $. 
Choose $\bar{k}=\max\{\sqrt{2},r_{A_2}\}$. Let $\bar{a}=A_2(\bar{k})$. 
Now, for given $a\geq\bar{a}$, there exists $\tilde{k}(a)\geq\bar{k}$ such 
that $a=A_1(\tilde{k})$.
From Lemma \ref{lem2}, note that there exist $\tilde{l}>0$ such that 
$\tilde{k}=\frac{1}{\phi_1(1)}$, where $\phi_1$ is the eigenfunction 
corresponding to the first eigenvalue $\lambda_1$ of
\begin{gather*}
y''(t)+\lambda y(t)=0,\quad t\in(0,1),\\
y(0)=0,\\
y'(1)+\tilde{l}y(1)=0.
\end{gather*}
We now define our subsolution $\psi$ to be $\psi:=\tilde{k}\phi_1^{\beta}$. 
Since $\phi_1(t)=\sin\sqrt{\lambda_1}t$, it is easy to see that $\phi_1$ has 
the following properties. There exist $\epsilon<\epsilon_1$ 
($\epsilon_1$ as in $H_1$) and $\mu>0$ such that $|\phi_1'|\geq\eta_1/2$ 
on $(0,\epsilon]$, where $\eta_1=\sqrt{\lambda_1}$, $\phi_1 \geq\mu$ on 
$(\epsilon,1)$, and $0\leq \phi_1(t)\leq \eta_1t$ for all $t\in(0,1)$. 
For $a\geq\bar{a}$, define
 \begin{align}
 \bar{c}(a)=\min\Big\{\tilde{k}^{1+\alpha}\beta(\beta-1)
\frac{\eta_1^{2-\gamma}}{4},\frac{1}{2}\tilde{k}\mu^{\beta}
\Big(a-\frac{\beta\lambda_1\tilde{k}^{\alpha}}{\hat{h}}\Big)\Big\}.\label{c}
 \end{align}
Note that $\bar{c}>0$ by the choice of $\tilde{k}$ and $\beta$. Next, we calculate
$$
-\psi''=\tilde{k}\lambda_1\beta\phi_1^{\beta}-\tilde{k}\beta(\beta-1)
\frac{\phi_1'^2}{\phi_1^{2-\beta}}.
$$
To prove $\psi$ is a subsolution, we need to establish
\begin{equation}
\tilde{k}\lambda_1\beta\phi_1^{\beta}-\tilde{k}\beta(\beta-1)
\frac{\phi_1'^2}{\phi_1^{2-\beta}}\leq h(t)
\Big(a\tilde{k}^{1-\alpha}\phi_1^{\beta(1-\alpha)}-\tilde{k}^{2-\alpha}
 \phi_1^{\beta(2-\alpha)}-\frac{c}{\tilde{k}^{\alpha}\phi_1^{\alpha\beta}}\Big)
 \label{T1.3}
\end{equation}
and $\psi'(1)+g(\psi(1))\leq 0$ (Clearly $\psi(0)=0$).

First we  show that \eqref{T1.3} satisfied. Note that
\begin{align*}
{\tilde{k}}\lambda_1\beta\phi_1^{\beta}
&= \frac{\hat{h}\tilde{k}\lambda_1\beta\phi_1^{\beta}}{\hat{h}}\\
&\leq h(t)\Big[a\tilde{k}^{1-\alpha}\phi_1^{\beta(1-\alpha)}
 -\frac{1}{2}\tilde{k}^{1-\alpha}\phi_1^{\beta(1-\alpha)}
 \Big(a-\frac{\tilde{k}^{\alpha}\lambda_1\beta\phi_1^{\alpha\beta}}{\hat{h}}\Big)\\
&\quad -\frac{1}{2}\tilde{k}^{1-\alpha}\phi_1^{\beta(1-\alpha)}
 \Big(a-\frac{\tilde{k}^{\alpha}\lambda_1\beta\phi_1^{\alpha\beta}}{\hat{h}}\Big)
 \Big].
\end{align*}
To prove \eqref{T1.3} holds in $(0,1)$, it is sufficient to show the following
 three inequalities hold:
\begin{gather}
-\frac{1}{2}\tilde{k}^{1-\alpha}\phi_1^{\beta(1-\alpha)}
 \Big(a-\frac{\tilde{k}^{\alpha}\lambda_1\beta\phi_1^{\alpha\beta}}{\hat{h}}\Big)
 \leq-\tilde{k}^{2-\alpha}\phi_1^{\beta(2-\alpha)}\quad\text{in }(0,1),\label{T1.5}\\
-\frac{1}{2}\tilde{k}^{1-\alpha}\phi_1^{\beta(1-\alpha)}
 \Big(a-\frac{\tilde{k}^{\alpha}\lambda_1\beta\phi_1^{\alpha\beta}}{\hat{h}}\Big)
 \leq-\frac{c}{\tilde{k}^{\alpha}\phi_1^{\alpha\beta}}\quad\text{in }(\epsilon,1),
 \label{T1.6} \\
-\tilde{k}\beta(\beta-1)\frac{\phi_1'^2}{\phi_1^{2-\beta}}
 \leq-h(t)\frac{c}{\tilde{k}^{\alpha}\phi_1^{\alpha\beta}}\quad\text{in } (0,\epsilon].
 \label{T1.7}
\end{gather}
From the definition of $a$, we have
$2\tilde{k}+\frac{\tilde{k}^{\alpha}\lambda_1\beta}{\hat{h}} < a$. Then 
\[
-\Big(a-\frac{\tilde{k}^{\alpha}\lambda_1\beta\phi_1^{\alpha\beta}}{\hat{h}}\Big)
 <-2\tilde{k}.
\]
 Hence
 \begin{equation}
\begin{aligned}
-\frac{1}{2}\tilde{k}^{1-\alpha}\phi_1^{\beta(1-\alpha)}
\Big(a-\frac{\tilde{k}^{\alpha}\lambda_1\beta\phi_1^{\alpha\beta}}{\hat{h}}\Big)
&<-\tilde{k}^{2-\alpha}\phi_1^{\beta(1-\alpha)} \\
&<-\tilde{k}^{2-\alpha}\phi_1^{\beta(2-\alpha)}\quad\text{in } (0,1). 
\end{aligned}\label{T1.8}
\end{equation}
Using $\phi_1\geq\mu$ in $(\epsilon,1)$, and 
$c\leq\frac{1}{2}\tilde{k}\mu^{\beta}
 (a-\frac{\beta\lambda_1\tilde{k}^{\alpha}}{\hat{h}})$,
\begin{equation}
\begin{aligned}
-\frac{1}{2}\tilde{k}^{1-\alpha}\phi_1^{\beta(1-\alpha)}
\Big(a-\frac{\tilde{k}^{\alpha}\lambda_1\beta\phi_1^{\alpha\beta}}{\hat{h}}\Big)
&\leq \frac{1}{\tilde{k}^{\alpha}\phi_1^{\alpha\beta}}
 \Big(\frac{-1}{2}\tilde{k}\phi_1^{\beta}
 \Big(a-\frac{\tilde{k}^{\alpha}\lambda_1\beta}{\hat{h}}\Big)\Big)\\
&\leq-\frac{c}{\tilde{k}^{\alpha}\phi_1^{\alpha\beta}}\quad\text{in }(\epsilon,1). 
\end{aligned} \label{T1.9}
\end{equation}
Next, we prove that \eqref{T1.7} holds in $(0,\epsilon]$. 
Since $|\phi_1'|\geq \eta_1/2$ and $2-\beta>\alpha\beta+\gamma$ we have
$$
-\tilde{k}\beta(\beta-1)\frac{\phi_1'^2}{\phi_1^{2-\beta}}
\leq -\frac{\tilde{k}^{1+\alpha}\beta(\beta-1)\eta_1^2}{4\tilde{k}^{\alpha}
\phi_1^{\alpha\beta}\phi_1^{\gamma}}
\leq-\frac{\tilde{k}^{1+\alpha}\beta(\beta-1)\eta_1^2}{4\tilde{k}^{\alpha}
\phi_1^{\alpha\beta}\eta_1^{\gamma}t^{\gamma}}.
$$
Since $h(t)\leq \frac{1}{t^{\gamma}}$ in $(0,\epsilon]$, and
 $c\leq\tilde{k}^{1+\alpha}\beta(\beta-1)\eta_1^{2-\gamma}/4$, it follows that
\begin{equation}
-\tilde{k}\beta(\beta-1)\frac{|\phi_1'|^2}{\phi_1^{2-\beta}}
\leq-h(t)\frac{c}{\tilde{k}^{\alpha}\phi_1^{\alpha\beta}}\quad\text{in }
(0,\epsilon]. \label{T1.10}
\end{equation}
Thus from \eqref{T1.8}, \eqref{T1.9} and \eqref{T1.10} we see that \eqref{T1.3}
 holds in $(0,1)$. 

Next we will show that $\psi'(1)+g(\psi(1))\leq 0$ and
$$
\psi'(1)+g(\psi(1))=\tilde{k}\beta\phi_1^{\beta-1}(1)\phi_1'(1)+g(\tilde{k}
\phi_1^{\beta}(1)).
$$
Since $\tilde{k}=\frac{1}{\phi_1(1)}$, it follows that
\[
\psi'(1)+g(\psi(1))
=\beta \tilde{k}^{2-\beta}\phi_1'(1)+g(\tilde{k}^{1-\beta})
=\beta \tilde{k}^{2-\beta}\Big(\phi'(1)+\frac{g(\tilde{k}^{1-\beta})}
{\beta \tilde{k}^{2-\beta}}\Big).
\]
Now note that, since $\tilde{k}>\sqrt{2}$, 
$\phi_1(1)=\sin\sqrt{\lambda_1}<\frac{1}{\sqrt{2}}$, which implies 
$\sqrt{\lambda_1}\in(\frac{3\pi}{4},\pi)$. 
Hence $\phi_1'(1)<-3\pi/(4\sqrt{2})$ and thus
\[
\psi'(1)+g(\psi(1))
\leq\beta \tilde{k}^{2-\beta}\Big(-\frac{3\pi}{4\sqrt{2}}
 +\frac{g(\frac{1}{\tilde{k}^{\beta-1})}}{\beta \tilde{k}^{2-\beta}}\Big)
\leq 0
\]
since $A_2(\tilde{k})\leq0$.
Therefore $\psi=\tilde{k}\phi_1^{\beta}$ is a subsolution of \eqref{P}. 
Next we will construct a supersolution of \eqref{P}. For this, we proceed
 as in the proof of Theorem \ref{thm4}. Let $e$ be the solution of
\begin{gather*}
-e''(t)=h(t),\quad t\in(0,1),\\
e(0)=e'(1)=0.
\end{gather*}
As discussed earlier, $e$ is an increasing function for $t\in[0,1]$.
 Choose a constant $M>0$ such that $\frac{as-s^2-c}{s^{\alpha}}<M$, 
for all $s\geq0$. Then clearly $\phi=Me$ is a supersolution of \eqref{P}. 
Also if we choose $M$ large enough then, $\psi(t)\leq\phi(t)$ for all 
$t\in[0,1]$. Hence, by Lemma \ref{lem1}, there exist a solution $u$ of \eqref{P} 
such that $\psi(t)\leq u(t)\leq\phi(t)$ for all $t\in[0,1]$. 
Clearly $u(1)>0$ since $\psi(1)>0$.

 We now show that $\bar{c}$, given by \eqref{c}, is an increasing function of $a$. 
By definition, $\tilde{k}$ increases as $a$ increases and hence
 $\tilde{k}^{1+\alpha}\beta(\beta-1)\frac{\eta_1^{2-\gamma}}{4}$ 
is an increasing function of $a$. Also,
\begin{align*}
\frac{d}{da}\Big(\frac{1}{2}\tilde{k}\mu^{\beta}
 (a-\frac{\beta\lambda_1\tilde{k}^{\alpha}}{\hat{h}})\Big)
&=\frac{1}{2}\frac{d\tilde{k}}{da}\mu^{\beta}
 \Big(a-\frac{\beta\lambda_1\tilde{k}^{\alpha}}{\hat{h}}\Big)
 +\frac{1}{2}\tilde{k}\mu^{\beta}\Big(1-\frac{\beta\lambda_1\alpha
 \tilde{k}^{\alpha-1}}{\hat{h}}\frac{d\tilde{k}}{da}\Big)\\
&=\frac{1}{2}\tilde{k}\mu^{\beta}+\frac{\mu^{\beta}}{2}\frac{d\tilde{k}}{da}
 \Big(a-\frac{(\alpha+1)\beta\lambda_1\tilde{k}^{\alpha}}{\hat{h}}\Big)\\
&>\frac{1}{2}\tilde{k}\mu^{\beta}+\frac{\mu^{\beta}}{2}\frac{d\tilde{k}}{da}
 \Big(a-\frac{2\beta\pi^2\tilde{k}^{\alpha}}{\hat{h}}\Big)>0.
\end{align*}
Hence $\bar{c}(a)$ is an increasing function of $a$ and $\bar{c}(a)\to\infty$ 
as $a\to \infty$.

\section{Numerical results}

In this section, we consider the boundary-value problem
\begin{equation}\label{P*}
\begin{gathered}
-u'' =\big(\frac{au-u^{2}-c}{u^\alpha}\big), \quad t \in (0, 1), \\
u(0)= 0, \quad u'(1)+g(u(1))=0,
\end{gathered} 
\end{equation}
where $a>0$, $c \geq 0$, $\alpha \in (0, 1)$, and 
$g{:}[0, \infty) \to [0, \infty)$, is a continuous function.
We plot the exact bifurcation diagram of positive solutions of \eqref{P*} 
($c$ versus $\|u\|_{\infty}$ and $a$ versus $\|u\|_{\infty}$) using Mathematica. 
For this, we adapt the quadrature method discussed in \cite{GLES, GMSS, LT}. 
Let $u(t)$ be a positive solution of \eqref{P*}. 
Let $F(z)=\int_0^zf(s)ds$, where $f(s)=\frac{as-s^2-c}{s^{\alpha}}$, 
$\rho:=\|u\|_\infty$, and $q=u(1)$. 
Following the arguments in \cite{GLES}, $u$ is a solution of \eqref{P*} 
if and only if $\rho, q$ satisfy:
 \begin{gather} \label{P15}
2\int_0^\rho\frac{ds}{\sqrt{F(\rho)-F(s)}}
 -\int_0^q\frac{ds}{\sqrt{F(\rho)-F(s)}}=\sqrt{2}, \\
\label{P8}
F(\rho)-F(q)= \frac{(g(q))^2}{2}.
\end{gather}
Let $\theta_1$ be the positive zero of $F$ (see figure \ref{figf})
 and $r_2$ be the falling zero of $f$ (see figure \ref{figf}).

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.4\textwidth]{fig4a} % 7.jpeg
\includegraphics[width=0.4\textwidth]{fig4b} % 8.jpeg
\end{center} 
\caption{Graph of $f(u)$ (left). Graph of $F(u)$ (right)} \label{figf} % on{ \label{figF}
\end{figure}

 We note that if $\rho\in (\theta_1,r_2)$ then the integrals in \eqref{P15} 
are well defined (see \cite{GLES} for details).
Now, using \eqref{P15} and \eqref{P8}, we are able to plot exact bifurcation 
diagram of positive solutions of \eqref{P*} by implementing a numerical
 root finding algorithm in Mathematica. Figures \ref{bif1}, \ref{bif3} 
are bifurcation diagrams $c$ versus $\rho$ for the cases $g(t) \equiv 1$ 
and $g(t)=t^2$ when $a=10$ and $a=15$. Figures \ref{bif5} has
bifurcation diagrams $a$ versus $\rho$ for the cases $g(t)=t^2$ when $c=0.1$ 
and $c=1$.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.49\textwidth]{fig5a} % 2.jpeg
\includegraphics[width=0.49\textwidth]{fig5b} % 1.jpeg
\end{center}
\caption{Bifurcation of \eqref{P*} when 
$g(t)\equiv 1$, $a=10$ (left); when 
$g(t)\equiv 1$, $a=15$ (right)}   \label{bif1} % \label{bif2}
\end{figure}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.49\textwidth]{fig6a} % 3.jpeg
\includegraphics[width=0.49\textwidth]{fig6b} % 4.jpeg
\end{center}
 \caption{Bifurcation of \eqref{P*} when $g(t)=t^2$, $a=10$ (left);
when $g(t)=t^2$, $a=15$ (right) } \label{bif3} % \label{bif4}
\end{figure}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.49\textwidth]{fig7a} % 5.jpeg
\includegraphics[width=0.49\textwidth]{fig7b} % 6.jpeg
\end{center}
\caption{Bifurcation of \eqref{P*} when $g(t)=t^2$, $c=0.1$ (left);
when $g(t)=t^2$, $c=1$ (right)} \label{bif5} % \label{bif6}
\end{figure}

Our bifurcation diagrams illustrate the existence result in Theorem \ref{thm1} 
for the case $h(t) \equiv 1$, $g(t) \equiv 1$ or $g(t)=t^2$, and $a=10$ or $15$.
 We see that for each $\alpha\in (0, 1)$, there exists a $\bar{c}>0$ such that 
for $c<\bar{c}$, \eqref{P*} has a positive solution. Also from the bifurcation 
diagrams (Figure \ref{bif5}) we can see that for given 
$c\leq\bar{c}(\bar{a})$, there exists $a_*(c)$ such that for $a>a_*$, 
\eqref{P*} has a positive solution. For $c=0$, the bifurcation diagrams show 
that the positive solution is unique which illustrates Theorem \ref{thm3}. 
The following observations can also be made from the bifurcation diagrams for 
the special cases considered. For $c \approx 0$, it appears that \eqref{P*} has 
unique positive solution and for a certain range of $c$, \eqref{P*} has multiple 
positive solutions. Also, for a fixed $c\leq\bar{c}(\bar{a})$ we observe that 
for large values of $a$, \eqref{P*} has unique positive solution and for a 
certain range of $a$, \eqref{P*} has multiple positive solutions. 
Proving these results for \eqref{P} (at least for certain cases of $g$) remains an
 open question.

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\end{document}