\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 04, 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 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/04\hfil Positive solutions for elliptic equations] {Positive solutions for elliptic equations with singular nonlinearity} \author[J. Shi, M. Yao\hfil EJDE-2005/04\hfilneg] {Junping Shi, Miaoxin Yao} % in alphabetical order \address{Junping Shi \hfill\break Department of Mathematics, College of William and Mary \\ Williamsburg, VA 23187, USA \hfill\break Department of Mathematics, Harbin Normal University, Harbin, Heilongjiang, China} \email{shij@math.wm.edu} \address{Miaoxin Yao\hfill\break Department of Mathematics, Tianjin University\hfill\break and Liu Hui Center for Applied Mathematics, Nankai University \& Tianjin University\\ Tianjin, 300072, China} \email{miaoxin@hotmail.com} \date{} \thanks{Submitted August 15, 2004. Published January 2, 2005.} \thanks{J. Shi was supported by NSF grant DMS-0314736, and by a grant from Science Council \hfill\break\indent of Heilongjiang Province, China } \subjclass[2000]{35J25, 35J60} \keywords{Singular nonlineararity; elliptic equation; positive solution; \hfill\break\indent monotonic iteration} \begin{abstract} We study an elliptic boundary-value problem with singular nonlinearity via the method of monotone iteration scheme: \begin{gather*} -\Delta u(x)=f(x,u(x)),\quad x \in \Omega,\\ u(x)=\phi(x),\quad x \in \partial \Omega , \end{gather*} where $\Delta$ is the Laplacian operator, $\Omega$ is a bounded domain in $\mathbb{R}^{N}$, $N \geq 2$, $\phi \geq 0$ may take the value $0$ on $\partial\Omega$, and $f(x,s)$ is possibly singular near $s=0$. We prove the existence and the uniqueness of positive solutions under a set of hypotheses that do not make neither monotonicity nor strict positivity assumption on $f(x,s)$, which improvements of some previous results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{cor}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} Let $\Omega$ be a bounded smooth domain in $\mathbb{R}^{N}$, $N \geq 2$. We assume that the boundary $\partial \Omega$ of $\Omega$ is of $C^{2,\theta}$ for some $\theta \in (0,1)$. Let $\phi (x)$ be a nonnegative function belonging to $C^{2,\theta}(\partial \Omega)$ and $f(x,s)$ be a function defined on $\overline{\Omega} \times (0,+\infty )$ which is locally H\"older continuous with exponent $\theta$. We consider the existence and the uniqueness of positive solutions for the nonlinear boundary-value problem \begin{gather} -\Delta u(x)=f(x,u(x)),\quad x \in \Omega , \label{eq:pp1}\\ u(x)=\phi(x),\quad x \in \partial \Omega , \label{eq:pp2} \end{gather} where $\Delta$ is the Laplacian operator. A positive solution of problem \eqref{eq:pp1}-\eqref{eq:pp2} is a function $u(x) \in C^{0}(\overline{\Omega}) \cap C^{2}(\Omega)$ satisfying \eqref{eq:pp1}-\eqref{eq:pp2} and $u(x) > 0$ for $x \in \Omega$. Many articles treat the problem of the existence and/or the uniqueness of positive solutions for \eqref{eq:pp1}-\eqref{eq:pp2} under a variety of hypotheses on function $f(x,s)$. When $f(x,s)$ is locally Lipschiz in $\Omega \times [0, +\infty )$, the existence and uniqueness of positive solutions (for some cases) are well understood. However, if there is a sequence $\{(x_{i},s_{i})\}$ in $\Omega \times (0,+\infty)$, for which $x_{i}$ converges to some point in the set $\{x\in\partial\Omega | \phi(x)=0\}$ and $s_{i}$ tends to $0$ as $i\to +\infty$, such that $f(x_{i},s_{i})\to\infty$, then problem \eqref{eq:pp1}-\eqref{eq:pp2} is singular, it does not have a solution in $C^2(\overline{\Omega})$, and the existence or uniqueness results do not follow from the results obtained for nonsingular equations in the literature. It is well-known that such singular elliptic problems arise in the contexts of chemical heterogeneous catalysts, non-Newtonian fluids and also the theory of heat conduction in electrically conducting materials, see \cite{CK, CN, DMO, FM} for a detailed discussion. In \cite{FM}, the existence of a positive solution of such a singular problem is established under a set of assumptions in which $f(x,s)$ is assumed to be non-increasing in $s$. Thus if $f(x,s)$ is defined, say, by $$f(x,s)=g(x)\ln^{2}s, \label{eq:ln}$$ or by $$f(x,s)=g(x)s^{-\alpha}+h(x)s^{\beta}-k(x)s^{\rho}, \label{eq:ap}$$ where $\alpha>0, \beta\in (0,1), \rho\geq 1$ , $g$ and $k$ are nonnegative H\"older continuous functions, then the existence of positive solutions does not follow from the results in \cite{FM}. the authors in \cite{St} and \cite{CRT} treat the singular problem with no monotonicity assumption on $f(x,s)$, and the results there may imply the existence of positive solutions even when $f(x,s)$ is given by (\ref{eq:ln}), (\ref{eq:ap}), in which $k(x)=0, g(x), h(x)>0$ for $x\in\overline{\Omega}$, or, by (see \cite{St}) $$f(x,s)=1+\{1+\cos\frac{1}{s}\}s^{1/2}e^{1/s}. \label{eq:cos1}$$ Some uniqueness results are also given in \cite{St} and \cite{CRT}. However, the method of proof in \cite{St} and \cite{CRT} requires that $f(x,s)$ be strictly positive near $s=0$, i.e., $f(x,s)$ is bounded away from $0$ as $s\to0^{+}$, for $x\in\overline{\Omega}$ (See $(H_{2}), (H_{2}')$ in \cite{St} and $(g_{1})$ in \cite{CRT}). Therefore, if $f(x,s)$ is given, say, by (\ref{eq:ln}), (\ref{eq:ap}), with $g(x)$ and $h(x)$ vanishing on some non-empty subset of $\Omega$, or given by $$f(x,s)=s^{1/2}e^{\frac{1}{s}(1+\cos\frac{1}{s})}, \label{eq:cos0}$$ then no conclusion regarding the existence of positive solutions can be derived from the results in \cite{St} and \cite{CRT}. For the special case where $f(x,s)=g(x)s^{-\alpha}$ in which $g$ is a sufficiently regular function and is positive in $\Omega$, and $\alpha>0$, \cite{LM} gives some results when $g(x)$ is vanishing or tending to $\infty$ near $\partial\Omega$ with a suitable rate, and the positivity of $f(x,s)$ for $x\in\Omega$ is still assumed. Recently the case where $f(x,s)=g(x)s^{-\alpha}+h(x)s^p$ is studied with $p\in(0,1)$ and the restriction that $\alpha\in (0,\frac{1}{N})$, also assumed the positivity hypotheses on functions $g(x)$ and $h(x)$ on whole $\Omega$. In the present article, neither monotonicity nor positivity on whole $\Omega$ is assumed for $f(x,s)$, and the results are more general, implying the existence of positive solutions for \eqref{eq:pp1}-\eqref{eq:pp2} even with $f(x,s)$ given by any of (\ref{eq:ln})--(\ref{eq:cos0}), where $g(x)$ and $h(x)$ may be $0$, and even $h(x)$ may be negative, in some subset of $\Omega$ . Also a uniqueness result is obtained. If we assume that for each $x\in \Omega$ either $s^{-1}f(x,s)$ is strictly decreasing in $s$ for $s>0$, or $f(x, s)$ and $s^{-1}f(x,s)$ are both nonincreasing in $s$, and that function $f$ satisfies some certain conditions in addition to the conditions for existence results, then we can further prove that the solution is unique. When $f(x,s)$ is locally Lipschiz in $\Omega \times [0, +\infty )$ and hence not singular, and $s^{-1}f(x,s)$ is strictly decreasing in $s$ for $s>0$ at every $x$ in $\Omega$, this kind of uniqueness result is well-known (see for example, \cite{OS}), however, our result extends it to include singular nonlinearity cases, which covers the special case where $f$ is given by (\ref{eq:ap}), and also applies to the case where $s^{-1}f(x,s)$ need'nt be strictly decreasing in $s$ for all $x$ in $\Omega$. The precise hypotheses and main results are stated in Section \ref{sec2}, and the proof for the results is given in Section \ref{sec3}. The proof for the existence results is based on a monotone convergence argument with solutions of (\ref{eq:pp1}) corresponding to the boundary data $\phi(x)+\frac {1}{k}$, which are obtained by using a monotone iteration scheme started with certain supersolutions and subsolutions particularly chosen; the proof for the uniqueness result makes use of a comparison lemma, which stems from some idea of a lemma in \cite{ABC}. \section{Hypotheses and Main Results}\label{sec2} We assume that the function $f$ that defines the nonlinear term in (\ref{eq:pp1}) satisfies the following conditions: \begin{itemize} \item [(F1)] $f$: $\overline{\Omega}\times(0,+\infty)\to\mathbb{R}$ is H\"older continuous with exponent $\theta \in (0,1)$ on each compact subset of $\overline{\Omega}\times(0,+\infty)$. \item [(F2)] $$\limsup_{s\to+\infty}\Big(s^{-1} \max_{x\in\overline{\Omega}}f(x,s)\Big) <\lambda_{1},$$ where $\lambda_{1}$ is the first eigenvalue of $-\Delta$ on $\Omega$ with Dirichlet boundary value. \item [(F3)] For each $t>0$, there exists a constant $D(t)>0$ such that $$f(x,r)-f(x,s)\geq -D(t)(r-s)$$ for $x\in\overline{\Omega}$ and $r\geq s\geq t$. (Without loss of generality we assume that $D(s)\leq D(t)$ for $s\geq t>0$.) \end{itemize} For the case in which $\phi(x)\not\equiv 0$ on $\partial\Omega$, we have the following result. \begin{theorem} \label{thm1} Suppose that $f$ satisfies (F1)--(F3) and $\phi\in C^{2,\theta} (\partial\Omega)$. If $\phi(x)\geq 0$ and $\phi(x)\not\equiv 0$ on $\partial\Omega$, and if there exist $\gamma, \delta >0$ such that $$f(x,s)\geq -\gamma s, \quad\mbox{for}\quad x\in \overline{\Omega} \; s\in (0,\delta) , \label{eq:t1}$$ then there exists at least one positive solution $u(x)$ of problem~{\rm (\ref{eq:pp1})~(\ref{eq:pp2})} such that for any compact subset $G$ of $\Omega\cup\{x\in\partial\Omega | \phi(x)>0\}$, $u(x)\in C^{2,\theta}(G)$. \end{theorem} For the general case where $\phi(x)$ may be $0$ for all $x\in\partial\Omega$, we have the following theorems. \begin{theorem}\label{thm2} Suppose that $f$ satisfies (F1)--(F3) and $\phi\in C^{2,\theta}(\partial\Omega)$. If $\phi(x)\geq 0$ on $\partial\Omega$ and if there exist positive numbers $\delta, \gamma$ and a nonempty open subset $\Omega_{0}$ of $\Omega$ such that \begin{gather} f(x,s)\geq -\gamma s, \quad\mbox{for } x\in \overline{\Omega} \; s\in (0,\delta), \label{t21} \\ s^{-1}f(x,s)\to +\infty \quad\mbox{as } s\to 0^{+} \mbox{ uniformly for } x \in\Omega_{0}, \label{t22} \end{gather} then the conclusion of Theorem~{\rm \ref{thm1}} holds. \end{theorem} \begin{theorem}\label{thm3} Suppose that $f$ satisfies (F1)--(F3) and $\phi\in C^{2,\theta}(\partial\Omega)$. If $\phi(x)\geq 0$ on $\partial\Omega$ and if there exists $\delta >0$ such that $$f(x,s)\geq \lambda_{1}s \quad\mbox{for } x\in\overline{\Omega} \; s\in(0,\delta) , \label{t3}$$ then the conclusion of Theorem~{\rm \ref{thm1}} holds. \end{theorem} The following theorem concerns to the uniqueness of positive solutions for problem \eqref{eq:pp1}-\eqref{eq:pp2}. We use the hypotheses \begin{itemize} \item [(F4)] Either $f(x,s)$ is nonincreasing in $s$ for each $x$ in $\Omega$, or, $s^{-1}f(x, s)$ is strictly decreasing in $s$ for each $x$ in an open subset $\Omega_0$ of $\Omega$ and both $f(x, s)$ and $s^{-1}f(x, s)$ are nonincreasing in $s$ for all $x$ in the remainder part $\Omega -\Omega_0$, \item [(F5)] The function $$F(s, t)=\max_{d(x)=s}|f(x, t)|, \quad\mbox{with } d(x)=\mathop{\rm dist}(x,\partial\Omega),$$ either is bounded on $(0, \delta)\times(0, \delta)$, or is a sum of such a bounded function and some function that is decreasing in $t$ on $(0, \delta)$ for any $s\in (0, \delta)$, and $$\int_{0}^{\delta}F(s, c_0s)ds < +\infty, \quad \mbox{for all } c_0 \in (0, 1).$$ \end{itemize} \begin{theorem}\label{thm4} Under the assumption of any of Theorems \ref{thm1}--\ref{thm3}, if in addition the function $f(x, s)$ satisfies (F4) and (F5), then problem \eqref{eq:pp1}-\eqref{eq:pp2} has one and only one positive solution in $C^{0}(\overline{\Omega})\cap C^{2,\theta}(\Omega)$. \end{theorem} \noindent{\bf Remarks.} \begin{itemize} \item[(1)] Examples of $f(x, s)$ , at a point $x$, satisfying the condition in (F4) that both $f(x, s)$ and $s^{-1}f(x, s)$ are non-increasing in $s$, are $f(x, s)= f_1(x)s^{\rho_1}$ for $s>0$ with $\rho_1 \leq 0$ and $f_1(x)\geq 0$ , $f(x,s)=f_2(x)s^{\rho_2}$ for $s>0$ with $\rho_2 \geq 1$ and $f_2(x)\leq 0$, and so on. \item[(2)] If $f(x, s)$ is a sum of a function $f_1(x, s)$ that is bounded on $\Omega \times (0, \delta)$ and some function $f_2(x, s)$ that is decreasing in $s$ on $(0, \delta)$ for any $x\in\Omega$, and if for any $c_0 \in (0, 1)$, there exists $\alpha_0 <1$ such that $$|f_2\left(x, c_0d(x)\right)|=O\left((d(x))^{-\alpha_0}\right), \quad \mathop{\rm as } d(x)\to 0,$$ then (F5) is obviously satisfied. \end{itemize} By the above remarks, we can easily derive from Theorems \ref{thm2} and \ref{thm4} the following corollary, in which $h^+$ and $h^-$ stand respectively for the positive part and the negative part of $h$, i.e., $h^+(x)=\max\{h(x), 0\}$, $h^-(x)=\max\{-h(x), 0\}$. \begin{cor} \label{cor1} The singular nonlinear elliptic problem \begin{gather*} \Delta u+g(x)u^{-\alpha}+h(x)u^{\beta}-k(x)u^{\rho}=0, \quad x\in\Omega,\\ u(x)=0, \quad x\in\partial\Omega, \end{gather*} with $\beta\in (0, 1)$, $\rho\geq 1$, and $\alpha >0$, possesses a positive solution $u$ in $C^{0}(\overline{\Omega})\cap C^{2,\theta}(\Omega)$, provided that functions $g, h$ and $k$ are $\theta-$H\"older continuous on $\overline{\Omega}$ , $g, k$ are nonnegative, $g+h^+$ is not identically zero, and $h^-(x)\leq\sigma_0g(x),\forall x\in \Omega$, for some constant $\sigma_0>0$. If in addition the function $h$ is non-negative or non-positive on whole $\Omega$, and for some $\alpha_0 <1$, $$g(x)=O\left((d(x))^{\alpha-\alpha_0}\right), \quad \mbox{as } d(x)\to 0, x\in \Omega,$$ then the solution $u$ is unique. \end{cor} This is an example in which the behavior of a coefficient function near the boundary affects the existence and uniqueness of solutions. Moreover, the result here makes improvement to some results in the literature \cite{CRT} \cite{FM} \cite{St} and \cite{SW}. \section{Proof of Results}\label{sec3} Let $(P_{k})$ denote the boundary-value problem: $$\label{eq:pk1-2} \begin{gathered} -\Delta u(x)=f(x,u(x)),\quad x\in\Omega , \\ % \label{eq:pk1} u(x)=\phi(x)+\frac{1}{k},\quad x\in\partial\Omega , %\label{eq:pk2} \end{gathered}$$ where $k$ is a positive integer. We say that a function $u$ is a supersolution, or a subsolution, of \eqref{eq:pk1-2} if $u$ belongs to $C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$ and satisfies \eqref{eq:pk1-2} with sign $=$ replaced by signs $\geq$, or $\leq$, respectively. In this Section we first prove Theorem \ref{thm1} in detail, then we outline the proofs for Theorems \ref{thm2} and \ref{thm3}. After we state and prove a lemma we finally prove Theorem \ref{thm4}. \begin{proof}[Proof of Theorem \ref{thm1}] {\it Step 1.} Let $m,k$ be positive integers and denote by $\psi_{m,k}(x)$ (resp. $\psi_{m,\infty}(x)$) the unique solution in $C^{2}(\overline{\Omega})$ of problem \begin{gather*} -\Delta\psi(x)+\gamma\psi(x)=0,\quad x\in\Omega , \\ \psi(x)=\frac{1}{m}\phi(x)+\frac{1}{k},\quad x\in\partial\Omega , \\ (\mbox{\rm resp.}\quad \psi(x)=\frac{1}{m}\phi(x),\quad x\in\partial\Omega .) \end{gather*} Then it follows from the estimates of Schauder type \cite{GT} and the maximum principle for $-\Delta +\gamma$ that there exists a positive integer $m_{0}$ such that \begin{gather*} 0<\psi_{m_{0},\infty}(x)<\psi_{m_{0},k}(x),\quad x\in\Omega,\; k\geq m_{0}, \\ 0<\psi_{m_{0},k+1}(x)<\psi_{m_{0},k}(x)<\delta,\quad x\in\overline{\Omega}, \; k\geq m_{0}. \end{gather*} Hence, by (\ref{eq:t1}), $\psi_{m_{0},k}(x)$ is a subsolution of \eqref{eq:pk1-2} for every $k\geq m_{0}$. Let $$\delta_{k}=\min_{x\in\overline{\Omega}}\psi_{m_{0},k}(x),$$ we have $$0<\delta_{k+1}<\delta_{k},\;\; k\geq m_{0}.$$ By (F2) we may take $\lambda_{0}>0$ such that $$\limsup_{s\to+\infty}\big( s^{-1}\max_{x\in\overline{\Omega}}f(x,s)\big) <\lambda_{0}<\lambda_{1},$$ and then consider the problem \begin{gather*} -\Delta\xi(x)\geq \lambda_{0}\xi(x),\quad x\in\Omega , \\ \xi(x)>0,\quad x\in\overline{\Omega}. \end{gather*} The existence of solutions to this problem is established in \cite{Se}. Let $\xi(x)$ be such a function and $k_{0}$ be a positive integer sufficiently large. Then it's easy to verify that $k_{0}\xi(x)$ is a supersolution of \eqref{eq:pk1-2} for every $k\geq m_{0}$, and we may have $$k_{0}\xi(x)\geq \psi_{m_{0},k}(x)+\max_{x\in\overline{\Omega}}\phi(x), \quad x\in\overline{\Omega},\;k\geq m_{0}.$$ \noindent {\it Step 2.} We define the iteration scheme below, as in the standard supersolution and subsolution argument, \begin{gather*} -\Delta w_{n}(x)+D(\delta_{m_{0}})w_{n}(x) =f(x,w_{n-1}(x))+ D(\delta_{m_{0}})w_{n-1}(x),\quad x\in\Omega , \\ w_{n}(x)=\phi(x)+\frac{1}{m_{0}},\quad x\in\partial \Omega , \end{gather*} noting that (F3) implies that for each $x\in\Omega, s\mapsto f(x,s)+D(\delta_{m_{0}})s$ is an increasing function on $[\delta_{m_{0}},+\infty)$. Thus, as in the proof of Theorem 1 in \cite{Aman}, by setting $w_{0}(x)=\psi_{m_{0},m_{0}}(x)$ (or $k_{0}\xi(x)$) for $x\in\Omega$, we obtain a monotonic sequence that converges to a solution $u_{m_{0}}(x)\in C^{2}(\overline{\Omega})$ of $(P_{m_{0}})$ such that $$\psi_{m_{0},m_{0}}(x)\leq u_{m_{0}}(x)\leq k_{0}\xi(x),\quad x\in\overline{\Omega}.$$ Using the same iteration scheme with $m_{0}$ replaced by $m_{0}+1$, and setting $w_{0}(x)=\psi_{m_{0},m_{0}+1}(x)$ (or $u_{m_{0}}(x)$), we can obtain, as in above, a positive solution $u_{m_{0}+1}(x)\in C^{2}(\overline{\Omega})$ of $(P_{m_{0}+1})$. Furthermore, by the maximum principle for $-\Delta +D(\delta_{m_{0}+1})$, we have $$\psi_{m_{0},m_{0}+1}(x)\leq u_{m_{0}+1}(x)\leq u_{m_{0}}(x), \quad x\in\overline{\Omega}.$$ Hence, by repeating the above process, we obtain the sequence $\{u_{k}(x)\}_{k\geq m_{0}}$ satisfying $$\psi_{m_{0},\infty }(x)\leq u_{k+1}(x)\leq u_{k}(x)\leq k_{0}\xi(x),\quad x\in\overline{\Omega}, \;k\geq m_{0}. \label{uk}$$ and $u_{k}(x)$ solves \eqref{eq:pk1-2} for any $k\geq m_{0}$. \noindent {\it Step 3.} We can define function $u$ by $$u(x)=\lim_{k\to +\infty}u_{k}(x),\quad x\in\overline{\Omega},$$ because $\{u_{k}(x)\}_{k\geq m_{0}}$ is a decreasing sequence uniformly bounded from below by $\psi_{m_{0},\infty}(x)$ on $\overline{\Omega}$. Now, we have from (\ref{uk}) that $$\psi_{m_{0},\infty}(x)\leq u(x)\leq k_{0}\xi(x),\quad x\in\overline\Omega.$$ Thus, if $G$ is a compact subset of $\Omega\cup \{x\in\partial\Omega | \phi(x)>0\}$, then there exist two positive constants $E_{1}(G)$ and $E_{2}(G)$ such that $$E_{1}(G)\leq u(x)\leq E_{2}(G),\quad x\in G,\;k\geq m_{1}.$$ Therefore, using the same reasoning as that in \cite{St} and \cite{LM} and the Schauder theory as stated in \cite{GT}, we conclude that $u(x)$ satisfies (\ref{eq:pp1}) and belong to $C^{2,\theta}(G)$. On the other hand, by the hypotheses about function $f$, the number $$H:=\inf_{k\geq m_{0}}\{\min_{x\in\overline{\Omega}}f(x,u_{k}(x))\}$$ exists, hence by the maximum principle we have $$Q(x)\leq u_{k}(x), \quad x\in\overline{\Omega}, \;k\geq m_{0},$$ and hence $$Q(x)\leq u(x),\quad x\in\overline{\Omega},$$ where $Q(x)$ is the solution of problem \begin{gather*} -\Delta Q(x)=H,\quad x\in\Omega , \\ Q(x)=\phi(x) , \quad x\in\partial\Omega . \end{gather*} Furthermore, it is easy to see that if $x_{0}\in\partial\Omega$, then for any $\varepsilon>0$ there exist $r_0>0$ and an integer $m_{1}\geq m_{0}$ such that $$Q(x)\leq u_{k}(x)\leq \phi(x_{0})+\varepsilon,$$ for all $k\geq m_{1}$ and $x\in\Omega$ for which $|x-x_{0}| < r_0$. Therefore, $u(x)$ is continuous on $\overline{\Omega}$ satisfying (\ref{eq:pp2}). This completes the proof. \end{proof} Functions $\psi_{m_{0},\infty }(x)$ and $\psi_{m_{0},k}(x)$ play an important role in the proof above. For the proof of Theorem \ref{thm2} or \ref{thm3}, we only show the way for obtaining these two functions, the remainder of the proof is almost the same as that of Theorem \ref{thm1} and is omitted. \begin{proof}[Proof of Theorem \ref{thm2}] Choose an $\eta(x)\in C_{0}^{\infty}(\Omega)$ such that $0\leq \eta(x)\leq 1$ for $x\in\overline{\Omega}$, $\eta (x)\not\equiv 0$, and $\mathop{\rm supp}\eta \subset\Omega_{0}$. By (F1) and (\ref{t22}), there exist $c_{1}, c_{2}>0$ such that $c_1<\delta$, hence $f_{\gamma}(x, c_1)\geq 0, x \in \overline{\Omega}$, here $f_{\gamma}(x, s)\equiv f(x, s)+\gamma s$, and $$c_{1}\leq f_{\gamma}(x,c_{1})\leq c_{2},\;x\in\Omega_{0}, \label{eq:t17}$$ then we denote by $\psi_{m,k}(x)$ ( resp. $\psi_{m,\infty}(x)$) the unique solution in $C^{2}(\overline{\Omega})$ of the problem \begin{gather*} -\Delta\psi(x)+\gamma \psi(x)=\frac{1}{m}\eta(x)f_{\gamma}(x,c_{1}),\quad x\in\Omega ,\\ \psi(x)=\frac{1}{k},\quad x\in\partial\Omega.\\ \mbox{(resp.}\quad \psi(x)= 0 , \quad x\in\partial\Omega.) \end{gather*} We have for all $m,k\geq 1$ that \begin{gather*} \psi_{m,\infty}(x)=\frac{1}{m}\psi_{1,\infty}(x),\quad x\in\overline{\Omega} \\ \psi_{m,k}(x)\geq \psi_{m,\infty}(x) > 0,\quad x\in\Omega \\ \psi_{m,k}(x)\geq \psi_{m^{*},k^{*}}(x) > 0,\quad x\in \overline{\Omega}, \mbox{if $m^{*}\geq m$ and $k^{*}\geq k$}. \end{gather*} Clearly there exist $d_{1}, d_{2}>0$ such that $$d_{1}\leq \psi_{1,\infty}(x)\leq d_{2} \quad\mbox{for } x\in \mathop{\rm supp}\eta. \label{eq:t18}$$ By the Schauder estimates \cite{GT}, we can make $\psi_{m,k}(x)$, uniformly for $x\in\overline{\Omega}$, as small as we want by taking $m$ and $k$ both large enough. Hence there exists integer $m_{0}$ such that $$\frac{f_{\gamma}( x, \psi_{m,k}(x))}{\psi_{m,k}(x)}\geq \frac{c_{2}}{d_{1}}, \quad m, k\geq m_{0}, \; x\in \mathop{\rm supp}\eta ,$$ by (\ref{t22}), now by \eqref{t21}, $$f_{\gamma}(x,\psi_{m,k}(x))\geq 0,\quad x\in\Omega .$$ Therefore, if $x\in \mathop{\rm supp}\eta$ and $m,k\geq m_{0}$, \begin{align*} -\Delta\psi_{m,k}(x)-f( x,\psi_{m,k}(x)) &= \frac{1}{m}f_{\gamma}(x,c_{1})\big[\eta(x)-\frac{f_{\gamma}(x,\psi_{m,k}(x))} {\psi_{m,k}(x)}\frac{\psi_{m,k}(x)}{\frac{1}{m}f_{\gamma}(x,c_{1})}\big]\\ &\leq \frac{1}{m}f_{\gamma}(x,c_{1}) \big[\eta(x)-\frac{c_{2}}{d_{1}} \frac{\psi_{1,\infty}(x)}{f_{\gamma}(x,c_{1})}\big]\\ &\leq 0; \end{align*} (by \eqref{eq:t17} and \eqref{eq:t18}). If $x\in\Omega\backslash\mathop{\rm supp}\eta$, $$-\Delta\psi_{m,k}(x)-f\big((x,\psi_{m,k}(x)\big) =-f_{\gamma}\big(x,\psi_{m,k}(x)\big) \leq 0.$$ Thus, if $m,k\geq m_{0}$, then $\psi_{m,k}(x)$ is a subsolution of \eqref{eq:pk1-2}. Therefore, the functions $\psi_{m_{0},k}(x)$ and $\psi_{m_{0},\infty}(x)$ meet the needs. \end{proof} \begin{proof}{Proof of Theorem \ref{thm3}} We point our that it suffices to let $\psi_{m,\infty}(x)$ be the function $m^{-1}\Phi_{1}(x)$ and $\psi_{m,k}(x)$ be the unique solution of the problem \begin{gather*} -\Delta\psi(x)=\frac{\lambda_{1}}{m}\Phi_{1}(x),\quad x\in\Omega , \\ \psi(x)=\frac{1}{k},\quad x\in\partial\Omega\, \end{gather*} where $\Phi_1$ is the first eigenfunction of $-\Delta$ with zero boundary value which satisfies $\max_{x\in\Omega} \Phi_1(x)=1$. \end{proof} To prove Theorem \ref{thm4}, we need the following lemma, which is an extension of a lemma in \cite{ABC}. \begin{lemma}\label{lm1} Let $\Omega$ be a domain with a $C^2$ boundary $\partial \Omega$ or no boundary in $\mathbb{R}^{N}$, $N \geq 2$. Suppose that $f:\Omega \times (0, +\infty )\to \mathbb{R}$ is a continuous function such that the assumption (F4) is satisfied, and let $w,v \in C^2(\Omega)$ satisfy: \begin{itemize} \item[(a)] $\Delta w + f(x, w)\leq 0 \leq \Delta v + f(x, v)$ in $\Omega$ \item[(b)] $w, v > 0$ in $\Omega$, $\liminf_{|x|\to +\infty}\big(w(x)-v(x)\big)$, and $\liminf_{x\to \partial\Omega}\big(w(x)-v(x)\big)\geq 0$ \item[(c)] $\Delta v \in L^1(\Omega)$. \end{itemize} Then $w(x)\geq v(x)$ for all $x \in \Omega$. \end{lemma} \begin{proof} The proof for the case where $f(x,s)$ is non-increasing in $s$ at each $x$ in $\Omega$ is trivial, so we only prove for the second case in assumption (F4). Without loss of generality, we assume that $\Omega = \Omega_1 \cup \Omega_2$ in which $$\Omega_1 = \{x \in \Omega : f(x, s)\;{\rm and }\;s^{-1}f(x, s) \mbox{ are nonincreasing in } s\},$$ $\Omega_1\neq \Omega$ and $$\Omega_2 = \Omega_0-\Omega_1$$ which is an anon-empty and open subset of $\Omega$, since $\Omega_1$ is a relative closed subset of $\Omega$. To prove the lemma by contradiction, we let $S_\delta$ be the set $\{x \in \Omega \mid w(x)0$ such that $S_{\sigma}\not=\emptyset$ and $\overline{S_{\sigma}}\subset \Omega$ . If $S_\sigma \cap\Omega_2=\emptyset$, then $\overline{S_{\sigma}}\subset \Omega_1$. Noting that, at the boundary of $S_{\sigma}, w(x)=v(x)-\sigma$, and that, for $x\in s_\sigma$, $$\Delta ( w(x)-(v(x)-\sigma))\leq f(x, v(x))-f(x, w(x)\leq 0$$ by the assumption on $f(x,s)$ for $x\in\Omega_1$ and the condition $(a)$, one could have $w(x)\geq v(x)-\sigma$ for all $x\in s_\sigma$ by the aid of the maximum principle applied on $S_{\sigma}$. But this is a contradiction to the definition of $S_\sigma$. If $S_{\sigma} \cap \Omega_2 \not=\emptyset$, then it is easily seen from the assumption on $f(x,s)$ for $x\in \Omega_0$ that there exist $\varepsilon_0 > 0$ and a closed ball $\overline{B} \subset (S_{\sigma} \cap \Omega_2)$ such that $$v(x)-w(x)\geq \varepsilon_0,\; x \in {\rm B},\label{eqn1.1}$$ and $$\delta_0:=\int_{B}vw\big(\frac{f(x, w)}{w}-\frac{f(x, v)}{v}\big)dx >0.\label{eqn1.2}$$ Let $$M = \max \{1, \|\Delta v \|_{L^1(\Omega)}\}, \quad \varepsilon = \min \{ 1, \varepsilon_0, \frac{\delta_0}{4M}\}.$$ Let $\theta$ be a smooth function on $\mathbb{R}$ such that $\theta(t)=0$ if $t \leq 1/2$, $\theta (t)=0$ if $t \geq 1, \theta (t) \in (0, 1)$ if $t \in (\frac{1}{2}, 1)$, and $\theta'(t) \geq 0$ for $t \in \mathbb{R}$. Then, for $\varepsilon > 0$, define the function $\theta_{\varepsilon}(t)$ by $$\theta_{\varepsilon}(t)=\theta \big(\frac{t}{\varepsilon}\big), \;t \in \mathbb{R}.$$ It then follows from condition (a) and the fact that $\theta_{\varepsilon}(t) \geq 0$ for $t \in \mathbb{R}$ that $$(w\Delta v - v\Delta w)\theta_{\varepsilon}(v-w) \geq vw\big(\frac{f(x, w)}{w}-\frac{f(x, v)}{v}\big)\theta_{\varepsilon}(v-w), \quad x \in \Omega.$$ On the other hand, by the continuity of $w, v$ and $\theta_{\varepsilon}$, and condition (b), we can take an open set $D$ with a smooth boundary such that ${\overline{\rm B}}\subset D \subset S_{\delta}$, here $\delta = \min\{\sigma, \frac{\varepsilon}{4}\}$, and $v(x)-w(x) \leq \frac{\varepsilon}{2}$, for all $x \in S_0-D$. Then we have $$\int_{D}(w\Delta v - v\Delta w)\theta_{\varepsilon}(v-w)dx \geq \int_{D}vw\big(\frac{f(x, w)}{w}-\frac{f(x, v)}{v}\big) \theta_{\varepsilon}(v-w)dx.$$ Denote $$\Theta_{\varepsilon}(t)=\int_{0}^{t}s\theta'(s)ds,\;t \in \mathbb{R},$$ then it is easy to verify that $$0\leq \Theta_{\varepsilon}(t)\leq 2\varepsilon ,\quad t \in \mathbb{R},\quad \mbox{and}\quad \Theta_{\varepsilon}(t)=0, \;\mbox{if } t<\frac{\varepsilon}{2}.\label{eqn1.3}$$ Therefore, \begin{align*} &\int_{D}(w\Delta v - v\Delta w)\theta_{\varepsilon}(v-w)dx\\ & = \int_{\partial D} w\theta_{\varepsilon}(v-w) \frac{\partial v}{\partial n}ds -\int_{D}(\nabla v \cdot \nabla w)\theta_{\varepsilon}(v-w)dx\\ &\quad -\int_{D}w\theta_{\varepsilon}'(v-w)\nabla v \cdot (\nabla v-\nabla w)dx -\int_{\partial D}v\theta_{\varepsilon}(v-w)\frac{\partial w}{\partial n}ds\\ &\quad+\int_{D}(\nabla w \cdot \nabla v)\theta_{\varepsilon}(v-w)dx +\int_{D}v\theta_{\varepsilon}'(v-w)\nabla w \cdot (\nabla v-\nabla w)dx\\ & = \int_{D}v\theta_{\varepsilon}'(v-w)(\nabla w - \nabla v) \cdot (\nabla v-\nabla w)dx\\ &\quad +\int_{D}(v-w)\theta_{\varepsilon}'(v-w)\nabla v \cdot (\nabla v-\nabla w)dx\\ & \leq \int_{D}\nabla v \cdot \nabla\left(\Theta_{\varepsilon}(v-w)\right)dx\\ & = \int_{\partial D}\Theta_{\varepsilon}(v-w)\frac{\partial v}{\partial n}ds -\int_{D}\Theta_{\varepsilon}(v-w)\Delta vdx\\ & \leq 2\varepsilon\int_{D}|\Delta v|dx\quad \mbox{( by \eqref{eqn1.3})} \\ & \leq 2\varepsilon M < \frac{\delta_0}{2}\,. \end{align*} However, \begin{align*} \int_{D}vw\big(\frac{f(x, w)}{w}-\frac{f(x, v)}{v}\big)\theta_{\varepsilon}(v-w)dx & \geq \int_{{\rm B}}vw\big(\frac{f(x, w)}{w}-\frac{f(x, v)}{v}\big) \theta_{\varepsilon}(v-w)dx\\ & = \int_{{\rm B}}vw\big(\frac{f(x, w)}{w}-\frac{f(x, v)}{v}\big)dx \quad \mbox{(by \eqref{eqn1.1})}\\ & \geq \delta_0 \quad \mbox{ by \eqref{eqn1.2})}, \end{align*} which is a contradiction. Thus $S_0$ must be empty, and the lemma is proved. \end{proof} \begin{proof}[Proof of Theorem \ref{thm4}] Let $u_{1},u_{2}\in C^{0}(\overline{\Omega})\cap C^{2}(\Omega)$ be two positive solutions of problem~\eqref{eq:pp1}-\eqref{eq:pp2}. We prove that $u_{1}(x)=u_{2}(x)$, $x\in\overline{\Omega}$. From the proofs of Theorems 1-3, we can easily see that if $v=\psi_{m_{0},\infty}$ then \begin{gather*} \Delta v(x)+f(x,v(x))\geq 0,\quad x\in\Omega,\\ v(x)> 0,\quad x\in\Omega, \\ \phi(x) \geq v(x) \geq 0,\quad x\in\partial\Omega, \end{gather*} and $\Delta v \in L^{1}(\Omega)$. Therefore it follows from Lemma \ref{lm1} that $$u_{i}(x)\geq v(x),\quad x\in\overline{\Omega},\; i=1,2.$$ Moreover, by the Hopf's strong maximum principle, we have $\frac{\partial v}{\partial n}<0$ on $\partial \Omega$, hence there exists $c_{0}>0$ such that $u_{i}(x)\geq c_0d(x), \;x\in\overline{\Omega},\; i=1,2$, where $d(x)=\mathop{\rm dist}(x, \partial\Omega)$. Let $\Omega_{\varepsilon}=\{x\in\Omega : d(x) \leq \varepsilon \}$ for $\varepsilon >0$ and $U_{i}(\delta )=\{x\in\Omega : u_{i}(x)\leq \delta \}$, $i=1,2$. Since $\partial\Omega\in C^{2,\theta }$, there exist $\varepsilon \in (0, \delta)$ such that if $x\in\Omega_{\varepsilon}$, then there is a unique $y_{x}\in\partial\Omega$ such that dist$(x, y_{x})= d(x), c_0d(x)<\delta$. Thus, for some $M>0$ only depending on $\partial\Omega$, \begin{align*} \int_{\Omega_{\varepsilon}}|f\left(x, c_0d(x)\right)|dx &\leq M\int_{\partial \Omega}\int_{0}^{\varepsilon}|f(y-sn_{y}, c_0s)|ds\,dy\\ &\leq M\int_{\partial \Omega } \int_{0}^{\varepsilon}F(s, c_0s)ds\,dy\\ &\leq M^{*} < +\infty , \end{align*} where $$M^{*}=M\int_{\partial \Omega} \int_{0}^{\delta}F(s, c_0s)ds\,dy\,.$$ By the hypothesis (F5), there exists $M_{0}>0$ such that $$0\leq F(r, s)\leq F(r, t)+M_{0}\quad \mbox{for }\delta\geq s\geq t>0 \; r\in(0, \delta).$$ Therefore, \begin{align*} \int_{\Omega_{\varepsilon} \cap U_{i}(\delta )}|f(x, u_{i}(x))|dx &\leq \int_{\Omega_{\varepsilon}}|f\big(x, c_0d(x)\big)|dx + M_{0}\mathop{\rm meas}(\Omega) \\ &\leq M^{*}+M_{0}\mathop{\rm meas}(\Omega) < +\infty ,\quad i=1,2. \end{align*} Consequently, \begin{align*} \int_{\Omega}|f\big(x, u_{i}(x)\big)|dx &\leq \int_{\Omega_{\varepsilon}\cap U_{i}(\delta)}|f\big(x, u_{i}(x)\big)|dx + \int_{\Omega\backslash (\Omega_{\varepsilon }\cap U_{i}(\delta ))}| f\big(x, u_{i}(x)\big)|dx \\ &\leq M^{*}+M_{0}\mathop{\rm meas}(\Omega) +M_{i}^{**}\mathop{\rm meas}(\Omega) < +\infty , \end{align*} where $$M_{i}^{**}=\max_{x\in\overline{\Omega},\,\delta\leq s\leq \delta_{i}^{*}}|f(x, s)|, \quad \delta_{i}^{*}=\max_{x\in\overline{\Omega}}u_{i}(x),\quad i=1,2.$$ Therefore, $$\int_{\Omega}|\Delta u_{i}|dx =\int_{\Omega} |f(x,u_{i})|dx < +\infty, \quad i=1,2.$$ i.e., $\Delta u_{i}\in L^{1}(\Omega )$, $i=1,2$. 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