\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 175, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/175\hfil Three-point third-order problems] {Three-point third-order problems with a sign-changing nonlinear term} \author[J. Henderson, N. Kosmatov \hfil EJDE-2014/175\hfilneg] {Johnny Henderson, Nickolai Kosmatov} % in alphabetical order \address{Johnny Henderson \newline Department of Mathematics, Baylor University, Waco, TX 76798-7328, USA} \email{Johnny\_Henderson@baylor.edu} \address{Nickolai Kosmatov \newline Department of Mathematics and Statistics, University of Arkansas at Little Rock, Little Rock, AR 72204-1099, USA} \email{nxkosmatov@ualr.edu} \thanks{Submitted June 10, 2014. Published August 14, 2014.} \subjclass[2000]{34B15, 34B16, 34B18} \keywords{Green's function; fixed point theorem; positive solutions; \hfill\break\indent third-order boundary-value problem} \begin{abstract} In this article we study a well-known boundary value problem \begin{gather*} u'''(t) = f(t, u(t)), \quad 0 < t < 1, \\ u(0) = u'(1/2) = u''(1)=0. \end{gather*} With $u'(\eta)=0$ in place of $u'(1/2)=0$, many authors studied the existence of positive solutions of both the positone problems with $\eta \geq 1/2$ and the semi-positone problems for $\eta > 1/2$. It is well-known that the standard method successfully applied to the semi-positone problem with $\eta > 1/2$ does not work for $\eta =1/2$ in the same setting. We treat the latter as a problem with a sign-changing term rather than a semi-positone problem. We apply Krasnosel'ski\u{\i}'s fixed point theorem \cite{kra} to obtain positive solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} We study the third-order nonlinear boundary-value problem \begin{gather} u'''(t) = f(t, u(t)), \quad 0 < t < 1, \label{e}\\ u(0) = u'(1/2) = u''(1) = 0. \label{bc} \end{gather} with a sign-changing nonlinearity. Equation \eqref{e} satisfying the three-point condition \begin{equation}\label{nbc} u(0) = u'(\eta) = u''(1) = 0, \end{equation} with $\eta \geq 1/2$ has been studied by many authors \cite{gr,s,z}. We mention also relevant results in \cite{w,ip}, where, under nonlocal conditions involving Stieltjes integrals, the positone case was considered. A good theory of positive solutions for semi-positone problems with $\eta > 1/2$ is developed in \cite{lan, yao1,yao2,yhl} (and the references therein). In particular, Yao \cite{yao1} obtained a positive solution of the boundary value problem similar to \eqref{e}, \eqref{nbc}. The author assumed that the function $f : [0, 1] \times \mathbb{R}_+ \to \mathbb{R}$ satisfies the Carath\'eodory conditions and there exists a nonnegative function $h \in L_1[0, 1]$ such that $f(t, u) \geq h(t)$, $(t, u) \in [0, 1] \times \mathbb{R}_+$. Our paper is motivated by \cite{yao1} where, we believe, the idea of a non-constant lower bound $-h(t)$ for the inhomogeneous term was originally used for the boundary value problem similar to \eqref{e}, \eqref{bc}. The author refers to this type of problem as weakly semipositone. Prior to \cite{yao1}, similar semipositone problems have been solved effectively \cite{yhl} only for $f:[0,1] \times \mathbb{R} \to [-M,\infty)$ due to selection of $h \equiv M > 0$. As in \cite{yao1}, in this paper, we need only $f:[0,1] \times \mathbb{R}_+ \to \mathbb{R}$. Regardless of the choice of $h$, as a first step, one translates the semipositone problem into a positone problem using the transformations $$ u \mapsto v-u_0 \quad \mathrm{and} \quad f(\cdot, u) \mapsto f(\cdot, v-u_0)+h(\cdot), $$ where $u_0$ is a unique solution of the problem with the nonlinear term replaced with $h$. Subsequently, the positone problem is converted into an integral equation, which is shown to have one or, depending on conditions of $f$, several positive solutions. Finally, an important feature of this approach is that it requires the inequality $v(t) \geq u_0(t)$ to hold for a fixed point of the corresponding integral operator. This comparison depends on the properties of Green's function, or in particular, on the function appearing in the definition of a cone, and the solution $u_0$. The case of $\eta = 1/2$ stands alone since this type of approach used by many authors to study the case $\eta > 1/2$ does not readily apply to the case $\eta =1/2$. The difficulty arises when we attempt to obtain the inequality $v(t) \geq u_0(t)$ for $\eta =1/2$. Since problem \eqref{e}, \eqref{bc} cannot be treated as a semipositone problem, we adopt a new set of assumptions and consider a sign-changing nonlinearity. We are unaware of any results on the case $\eta = 1/2$ with a sign-changing nonlinear term. Another benefit is that we can also obtain new results for the case $\eta > 1/2$ with a sign-changing nonlinearity by employing the concept of a sign-changing lower bound $g_0$. We think that it would not be difficult to extend our results to the case of $f$ satisfying the Carath\'eodory conditions and even treat singularities as in \cite{yhl}. Here we settle for a continuous sign-changing nonlinear term. \section{Properties of Green's function} Let $g_0 \in C[0,1]$. Then the differential equation \begin{equation}\label{g0} u'''(t) = g_0(t), \quad 0 < t < 1, \end{equation} satisfying the boundary condition \eqref{bc} has a unique solution \begin{align*} u_0(t) &= \frac{1}{2}\int_0^t(t-s)^2 g_0(s) \, ds - \frac{t^2}{2} \int_0^1 g_0(s) \, ds\\ &\quad + t\Big(\frac{1}{2} \int_0^1 g_0(s) \, ds - \int_0^{1/2} \big(\frac{1}{2}-s\big) g_0(s) \, ds\Big). \end{align*} Using Green's function \begin{equation}\label{grfn} G(t,s) = \frac{1}{2}(t-s)^2 \chi_{[0,t]}(s) + \frac{1}{2}(t-t^2) - t\big(\frac{1}{2}-s\big)\chi_{[0,1/2]}(s), \end{equation} for $(t,s) \in [0,1] \times [0,1]$, we have $$ u_0(t) = \int_0^1 G(t,s) g_0(s)\, ds. $$ Let \[ G_0(s) = G(1/2,s) = \frac{s^2}{2}\chi_{[0,1/2]}(s)+\frac{1}{8}\chi_{[1/2,1]}(s), \quad s \in [0,1]. \] We revisit the important properties \cite{lan} of \eqref{grfn} used in cone-theoretic methods: \begin{equation}\label{grul} q(t) G_0(s) \leq G(t,s) \leq G_0(s), \quad (t,s) \in [0,1] \times [0,1], \end{equation} where \begin{equation}\label{q} q(t) = 4(t-t^2). \end{equation} Also, \begin{equation}\label{L} L = \max_{t \in [0,1]} \int_0^1 G(t,s) \, ds = \frac{1}{12}, \end{equation} and, for $0 < \alpha <1/2$, \begin{equation}\label{C} C = \int_{\alpha}^{1-\alpha} G_0(s) \, ds = \frac{1}{24} (2 - 4 \alpha^3 - 3 \alpha). \end{equation} Note that if \begin{equation}\label{u01} \int_t^1 g_0(t) \, dt \geq 0, \quad t \in [0,1], \end{equation} then $u_0(t)$ is concave in $[0,1]$. If, in addition, \begin{equation}\label{u02} u_0(1) = \frac{1}{2}\int_0^1(1-s)^2 g_0(s) \, ds -\int_0^{1/2} \big(\frac{1}{2}-s\big) g_0(s) \, ds \geq 0, \end{equation} then $u_0(t) \geq 0$. Note that neither \eqref{u01} nor \eqref{u02} requires $g_0(t) \geq 0$ in all of $[0,1]$. Moreover, if $g_0(t) \geq 0$ in $[0,1]$ and $g_0(t) > 0$ in some $[\alpha,\beta] \subset [0,1]$, then $u_0(1) > 0$. This represents a difficulty due to the fact that one can not achieve the inequality $q(t) \geq \mu u_0(t)$ in $[0,1]$ for any $\mu >0$ (as $q(1) =0$ while $u_0(1) > 0$). For this reason, the case $\eta = 1/2$ is forbidden in approaching \eqref{e}, \eqref{nbc} as a semipositone problem. If the identity takes place in \eqref{u02}, that is, $u_0(1)=0$ is enforced, then we are in position to compare $q(t)$ and $u_0(t)$ in the next lemma. \begin{lemma}\label{u0} Let $g_0 \in C[0,1]$ satisfy \eqref{u01} and suppose that the identity holds in \eqref{u02}. Then there exists a constant $\mu > 0$ such that \begin{equation}\label{mu1} q(t) \geq \mu u_0(t), \quad t \in [0,1]. \end{equation} \end{lemma} \begin{proof} Since the function $q - \mu u_0$ vanishes at the end-points of $[0,1]$, it suffices to obtain $\mu > 0$ such that $-q''(t) \geq -\mu u''_0(t)$ in $[0,1]$. That is, $$ 8 \geq \mu \int_t^1 g_0(s) \, ds, \quad t \in [0,1]. $$ By \eqref{u01}, there exists $0 < \tau < 1$ and $\mu > 0$ such that \begin{equation}\label{mu} \mu \int_t^1 g_0(s) \, ds \leq \mu \int_{\tau}^1 g_0(s) \, ds = 8. \end{equation} \end{proof} Suppose that the function $f$ in \eqref{e} satisfies \begin{itemize} \item[(H1)] $f \in C([0,1] \times \mathbb{R}_+,\mathbb{R})$; \item[(H2)] there exists a function $g_0 \in C[0,1]$ such that \begin{itemize} \item[(a)] $f(t,z) + g_0(t) \geq 0$ in $[0,1] \times \mathbb{R}_+$; \item[(b)] for all $t \in [0,1]$, $\int_t^1 g_0(s) \, ds \geq 0$; \item[(c)] $$ \frac{1}{2}\int_0^1(1-s)^2 g_0(s) \, ds -\int_0^{1/2} \big(\frac{1}{2}-s\big) g_0(s) \, ds = 0. $$ \end{itemize} \end{itemize} \begin{remark} \label{rmk1}\rm It is easy to find a function $g_0$ satisfying (H2) (b) and (c). For example, one can take $g_0(t) = a(2t-1)$, $a > 0$. Of course, an example of $f(t,z)$ that fits (H2) (b) is also easy to obtain. \end{remark} \begin{remark} \label{rmk2} \rm If the inequality \eqref{u02} is replaced with the strict inequality, we cannot expect Lemma \ref{u0} to hold. So, in this paper, we need the identity in (H2) (c). If, instead of $u'(1/2) = 0$, we impose $u'(\eta) = 0$ with $\eta > 1/2$, then the problem \eqref{g0}, \eqref{nbc} has a unique solution $$ u_0(t) = \frac{1}{2}\int_0^t(t-s)^2 g_0(s) \, ds - \frac{t^2}{2} \int_0^1 g_0(s) \, ds + t\Big(\eta \int_0^1 g_0(s) \, ds - \int_0^{\eta} (\eta-s) g_0(s) \, ds\Big). $$ Again, the assumption (H1) (b) guarantees that $u_0$ is concave in $[0,1]$. So, if $$ u_0(1) = \frac{1}{2}\int_0^1(1-s)^2 g_0(s) \, ds +\big(\eta - \frac{1}{2}\big) \int_0^1 g_0(s) \, ds - \int_0^{\eta} (\eta-s ) g_0(s) \, ds \geq 0, $$ then $u(t) \geq 0$ in $[0,1]$. Similarly, the analogue of $q(t)$, in this case \cite{lan}, is $$ p(t) = \frac{1}{\eta^2}(2\eta t-t^2). $$ Noting that $p(1) \neq 0$ and $u_0$, $p$ are concave concave in $[0,1]$, we can easily obtain an analogue of Lemma \ref{u0} asserting the existence of $\mu > 0$ such that $p(t) \geq \mu u_0(t)$ in $[0,1]$. This would give a more general result than in \cite[Lemma 2.1 (4)]{yhl}, which is derived for $g_0 \equiv M > 0$. This would also allow us to extend the results of \cite{yao1} concerning an analogue of \eqref{e}, \eqref{nbc}, where $h(t)$, which serves the purpose of $g_0(t)$, is assumed to be nonnegative. \end{remark} We modify the problem \eqref{e}, \eqref{bc} as follows. First, we define $$ f_p(t,z) = \begin{cases} f(t,z)+g_0(t), & (t,z) \in [0,1] \times [0,\infty),\\ f(t,0)+g_0(t), & (t,z) \in [0,1] \times (-\infty,0). \end{cases} $$ Next, we consider the equation \begin{equation}\label{me} v'''(t) = f_p(t,v(t)- u_0(t)), \quad t \in (0,1), \end{equation} under the boundary conditions \eqref{bc}. We can easily obtain the next lemma. \begin{lemma}\label{equiv1} The function $u$ is a positive solution of the boundary value problem \eqref{e}, \eqref{bc} if, and only if, the function $v = u+ u_0$ is a solution of the boundary value problem \eqref{me}, \eqref{bc} satisfying $v(t) \geq u_0(t)$ in $[0,1]$. \end{lemma} In the Banach space $\mathcal B = C[0,1]$ endowed with usual max-norm, we consider the operator \begin{equation}\label{oper} Tv(t) = \int_0^1 G(t,s)f_p(s,v(s)- u_0(s)) \, ds, \end{equation} where $G(t,s)$ is given by \eqref{grfn}. By (H1), $T: \mathcal B \to \mathcal B$ is completely continuous. Using the function $q$ defined by \eqref{q}, we introduce the cone $$ \mathcal{C} = \{v \in \mathcal B: v(t) \geq q(t)\|v\|, \; t \in [0,1]\}. $$ By \eqref{grul}, $T:\mathcal{C} \to \mathcal{C}$ and it is also easy to show that a fixed point of $T$ is a solution of \eqref{me}, \eqref{bc}. In particular, \begin{equation}\label{cconst} v(t) \geq \gamma \|v\|, \quad t \in [\tau,1-\tau], \end{equation} where $\gamma = \min_{t \in [\alpha,1-\alpha]} q(t) = 4(\alpha-\alpha^2)$, and $\kappa = \max_{t \in[\alpha,1-\alpha]} q(t) = q(1/2) = 1$. \begin{theorem}[\cite{kra}] \label{Kr} Let $\mathcal{B}$ be a Banach space and let $\mathcal{C} \subset \mathcal{B}$ be a cone in $\mathcal{B}$. Assume that $\Omega_{1}$, $\Omega_{2}$ are open with $0 \in \Omega_1$, ${\overline{\Omega}}_1\subset \Omega_2$, and let $$ T\colon \mathcal{C} \cap({\overline{\Omega}}_2 \setminus \Omega_1)\to \mathcal{C} $$ be a completely continuous operator such that either \begin{itemize} \item[(i)] $\| Tu \| \leq \| u\|$, $u \in \mathcal C \cap \partial \Omega_1$, and $\| Tu \| \geq \| u\|$, $u\in \mathcal C \cap \partial \Omega_2$, or; \item[(ii)] $\| Tu \| \geq \| u\|$, $u \in \mathcal C \cap \partial \Omega_1$, and $\| Tu \| \leq \| u\|$, $u \in \mathcal{C} \cap \partial \Omega_2$. \end{itemize} Then $T$ has a fixed point in $\mathcal C \cap ({\overline{\Omega}}_2 \setminus \Omega_1)$. \end{theorem} \section{Positive solutions} To use Theorem \ref{Kr}, following \cite{yao1} we introduce the ``height" functions $\phi,\psi: \mathbb{R}_+ \to \mathbb{R}_+$ defined by \begin{gather*} \phi(r) = \max \{f(t,z-u_0(t))+g_0(t): t \in [0,1], \, z \in [0,r]\},\\ \psi(r) = \min \{f(t,z-u_0(t))+g_0(t): t \in [\alpha,1-\alpha], \, z \in [\gamma r,r]\}. \end{gather*} Now we present our main results. \begin{theorem}\label{t1} Assume that {\rm (H1)} and {\rm (H2)} hold. Suppose that there exist $r,R > 0$ such that $\frac{1}{\mu} < r < R$, where $\mu >0$ satisfies \eqref{mu1}, \eqref{mu}, and \begin{itemize} \item[(H3)] $\phi(r) \leq 12 r$ and $\psi(R) \geq \frac{24 R}{2 - 4 \alpha^3 - 3 \alpha}$. \end{itemize} Then the boundary-value problem \eqref{e}, \eqref{bc} has at least one positive solution. \end{theorem} \begin{proof} Let $$ \Omega_1 = \{v \in B: \|v\| < r\}, \quad \Omega_2 = \{v \in B: \|v\| < R\}. $$ For $u \in \mathcal C \cap \partial \Omega_1$, we have $v(s)- u_0(s) \geq q(s) \|v\| - u_0(s) \geq (\mu r -1) u_0(s) \geq 0$, $s \in [0,1]$. This implies that \[ f_p(s,v(s)-u_0(s)) = f(s,v(s)-u_0(s)) +g_0(s), \quad s \in [0,1]. \] In particular, $$ f(s,v(s)-u_0(s))+g_0(s) \leq \phi(r), \quad s \in [0,1], \; 0 \leq v(s) \leq r. $$ Thus, by \eqref{L} and (H3), \begin{align*} \|Tv\| &= \max_{t \in [0,1]} \int_0^1 G(t,s) f_p(s,v(s)- u_0(s)) \, ds\\ &\leq \max_{t \in [0,1]} \int_0^1 G(t, s) \, ds \, \phi(r)\\ &= L \phi(r) = \frac{1}{12} \phi(r) \leq r. \end{align*} That is, $\|Tv\| \leq \|v\|$ for all $v \in \mathcal C \cap \partial \Omega_1$. Let $v \in \mathcal{C} \cap \partial \Omega_2$. Since $R > r$, we have $v(s)- u_0(s) \geq (\mu R -1) u_0(s)\geq 0$, $s \in [0,1]$. Then, for all $s \in [\alpha,1-\alpha]$, we have, recalling \eqref{cconst}, $$ R \geq v(s) \geq q(s) \|v\| \geq \gamma R. $$ Hence \[ f_p(s,v(s)-u_0(s)) = f(s,v(s)-u_0(s))+g_0(s) \geq \psi(R), \] for $s \in [\alpha,1-\alpha]$, $\gamma R \leq v(s) \leq R$. Then, by \eqref{C} and (H3), \begin{align*} \|Tv\| &= \max_{t \in [0,1]}\int_0^1 G(t,s) f_p(s,v(s)- u_0(s)) \, ds \\ & \geq \max_{t \in [0,1]}\int_{\alpha}^{1-\alpha} G(t,s) f_p(s,v(s)- u_0(s)) \, ds \\ &\geq \max_{t \in [0,1]} \int_{\alpha}^{1-\alpha} q(t) G_0(s) \, ds \, \psi(R)\\ & = \max_{t \in [0,1]} q(t) \int_{\alpha}^{1-\alpha} G_0(s) \, ds \, \psi(R)\\ & = \kappa C \psi(R) \geq R. \end{align*} That is, $\|Tv\| \geq \|v\|$ for all $v \in \mathcal C \cap \partial \Omega_2$. By Theorem \ref{Kr}, there exists $v_0 \in \mathcal C$ with $u(t)=v_0(t)-u_0(t) \geq (\mu r -1)u_0(t) \geq 0$ in $[0,1]$. By Lemma \ref{equiv1}, $u$ is a positive solution of the sign-changing problem \eqref{e}, \eqref{bc}. \end{proof} Now we give an example of the right side of \eqref{e} satisfying the assumptions of Theorem \ref{t1}. \subsection*{Example} Let $f(t,z) = 6 z^2+32(1-2t)$ for $z \geq 0$, $t \in [0,1]$. Then $f(t,z) + g_0(t) \geq 0$ with $g_0(t)=32(2t-1)$. Of course, (H1) and (H2) hold and $$ \int_t^1 g_0(t) \, ds \leq \int_{1/2}^1 g_0(t) \, ds = 8. $$ Hence we can choose $\mu=1$ and note that $\mu r > 1$, if we choose $r=2$. Then, recalling that $v(s)-u_0(s) \geq 0$, $$ f(t,v(s)-u_0(s))+ g_0(t) = 6(v(s) -u_0(s))^2 \leq 6\|v\|^2 =24 =12 r, $$ for all $v \in \mathcal C \cap \partial \Omega_1$. This shows that the first condition of (H3) is fulfilled. It is easy to see that $$ \|u_0\| \leq \max_{t \in [0,1]} \int_0^1 G(t,s) |g_0(s)| \, ds \leq \max_{t \in [0,1]} \int_0^1 G(t,s) \, ds \, \|g_0\| = \frac{8}{3}. $$ Let now $\alpha = 1/4$ so that $C = 19/384$ and $\gamma = 4(\alpha-\alpha^2) = 3/4$. Then, for all $s \in [1/4,3/4]$, $v \in \mathcal C \cap \partial \Omega_2$, where $R =13$, we have \begin{align*} f(t,v(s)-u_0(s))+ g_0(t) & = 6(v(s) -u_0(s))^2\\ & \geq 6(\gamma \|v\| -\|u_0\|)^2 \\ &= 6\big( \frac{39}{4} - \frac{8}{3} \big)^2 = \frac{7225}{24} \\ & > \frac{4992}{19} = \frac{24 R}{2 - 4 \alpha^3 - 3 \alpha}. \end{align*} The above shows that the second part of ($H_3$) is also verified. Hence a solution $v_0$ exists in the cone and $2 \leq \|v_0\| \leq 13$. The next result can be shown along the similar lines. \begin{theorem}\label{t2} Assume that {\rm (H1)} and {\rm (H2)} hold. Suppose that there exist $r,R > 0$ such that $\frac{1}{\mu} < r < R$, where $\mu >0$ satisfies \eqref{mu1}, \eqref{mu}, and \begin{itemize} \item[(H4)] $\phi(R) \leq 12 R$ and $\psi(r) \geq \frac{24 r}{2 - 4 \alpha^3 - 3 \alpha}$. \end{itemize} Then the boundary0value problem \eqref{e}, \eqref{bc} has at least one positive solution. \end{theorem} In conclusion of this paper presents a multiplicity result for \eqref{e}, \eqref{bc} which now is considered as a nonlinear eigenvalue problem. That is, \begin{equation}\label{ee} u'''(t) = \lambda f(t, u(t)), \quad 0 < t < 1, \end{equation} subject to \eqref{bc}. The result including the assumptions and the method of proof echoes that of Ma \cite{ma}, where a fourth order semipositone boundary-value problem with dependence on the first derivative was studied. The presence of the parameter $\lambda > 0$ provides an additional control on the growth of the right side. We introduce a new set of assumptions as follows: \begin{itemize} \item[(M1)] there exists an interval $[\alpha, 1-\alpha] \subset (0,1)$ such that \[ \lim_{u \to \infty} \frac{f(t,u)}{u} = \infty, \] uniformly in $[\alpha, 1-\alpha]$; \item[(M2)] $f(t,0) > 0$, $t \in [0,1]$. \end{itemize} Our next result is a multiplicity criterion. \begin{theorem}\label{m} Assume that \textup{(H1)}, \textup{(H2)}, \textup{(M1)}, \textup{(M2)} hold. Then the boundary-value problem \eqref{ee}, \eqref{bc} has at least two positive solutions provided $\lambda > 0$ is small enough. \end{theorem} \begin{proof} We will construct open nonempty subsets $\Omega_i = \{v \in \mathcal{C}: \|v\| = R_i\}$, $i=1,\dots,4$. Now, we consider the operator $$ Tv(t) = \lambda \int_0^1 G(t,s)f_p(s,v(s)- \lambda u_0(s)) \, ds, $$ where $u_0$ is the solution of $u''' = g_0$ subject to \eqref{bc} and $f_p$ as above. Let the $R_1 > 0$. Then $$ \|Tv\| = \max_{t \in [0,1]} \lambda \int_0^1 G(t,s) f_p(s,v(s)- \lambda u_0(s)) \, ds \leq \lambda L \phi(R_1) \leq R_1 $$ for all $v \in \mathcal{C} \cap \partial \Omega_1$, provided \begin{equation}\label{l1} \lambda \leq \frac{L \phi(R_1)}{R_1}. \end{equation} Let $v \in \mathcal{C} \cap \partial \Omega_2$, where $R_2 > R_1$. Then, by Lemma \ref{u0} with $$ \mu \max_{t \in [0,1]}\int_{t}^1 g_0(s) \, ds = 8. $$ Note that the equation in (M2) holds with $f_p$ in place of $f$. Thus given $A > 0$, there exists $h \geq \frac{\gamma}{2}R_2$ such that $f_p(t,z) > Az$ for all $z \geq h$ and $t \in [\alpha,1-\alpha]$. For every $\lambda$ in \eqref{l1}, there exists a constant $A > 0$ such that \begin{equation}\label{A} \frac{1}{2} \lambda C \gamma A \geq 1, \end{equation} where $C$ is given by by \eqref{C}. For all $s \in [\alpha,1-\alpha]$, we have $$ v(s) - \lambda u_0(s) \geq v(s) - \frac{\lambda}{\mu}q(s) = v(s) - \frac{\lambda}{\mu R_2}v(s) \geq \frac{1}{2}v(s) \geq \frac{\gamma}{2}R_2 $$ provided \begin{equation}\label{l2} \lambda \leq \frac{\mu R_2}{2}. \end{equation} Hence $$ f_p(s,v(s)-\lambda u_0(s)) \geq A(v(s)-\lambda u_0(s)) \geq \frac{\gamma A}{2}R_2, \quad s \in [\alpha,1-\alpha]. $$ Then, by \eqref{A}), and recalling that $\kappa = 1$, \begin{align*} \|Tv\| &= \max_{t \in [0,1]} \lambda \int_0^1 G(t,s) f_p(s,v(s)- \lambda u_0(s)) \, ds \\ & \geq \lambda \max_{t \in [0,1]} \int_{\alpha}^{1-\alpha} q(t) G_0(s) \, ds \, \frac{\gamma A}{2}R_2 \\ & = \lambda \max_{t \in [0,1]} q(t) \int_{\alpha}^{1-\alpha} G_0(s) \, ds \, \frac{\gamma A}{2}R_2 \\ & = \lambda \kappa C \frac{\gamma A}{2}R_2 \geq R_2. \end{align*} That is, $\|Tv\| \geq \|v\|$ for all $v \in \mathcal C \cap \partial \Omega_2$. As in Theorem \ref{t1}, we have a solution $v_1$ such that $R_1 \leq \|v_1\| \leq R_2$ for every $$ 0 < \lambda \leq \lambda_0 = \min\big\{ \frac{R_1}{L \phi(R_1)},\frac{\mu R_2}{2}\big\}. $$ To make use of the assumption $(M_2)$, we note that there exist $a,b > 0$ such that $f(t,z) \geq b$ for all $t \in [0,1]$ and $z \in [0,a]$ and introduce a ``truncation" of $f$ given by $$ f_t(t,z) = \begin{cases} f(t,z), & (t,z) \in [0,1] \times [0,a]),\\ f(t,a), & (t,z) \in [0,1] \times (a,\infty). \end{cases} $$ Consider now \begin{equation}\label{emm} u'''(t) = \lambda f_t(t,u(t)), \quad 0 < t < 1, \end{equation} subject to \eqref{bc}. The operator, whose fixed point will be shown to be (a second) solution of \eqref{e}, \eqref{bc}, is $$ Tv(s) = \lambda \int_0^1 G(t,s) f_t(s,v(s)) \, ds. $$ Choose $R_3 < \min\{R_1,a\}$. Then, as in the proof of Theorem \ref{t1}, $$ \|Tv\| \leq \lambda L \phi(R_3), $$ where $\phi(R_3) = \max \{f(t,z): t \in [0,1], \, z \in [0,R_3]\}$. So, if \begin{equation}\label{l3} \lambda < \min \big\{\frac{R_3}{L \phi(R_3)}, \lambda_0 \big\}, \end{equation} then $\|Tv\| \leq \|v\|$ for all $v \in \mathcal C \cap \partial \Omega_3$. Choose $\lambda $ according to \eqref{l3}. Since $$ \lim_{z \to 0^+} \frac{f_t(t,z)}{z} \geq \lim_{z \to 0^+} \frac{b}{z} = \infty $$ uniformly in $[0,1]$. Hence there exists $0< R_4 < R_3$ such that $$ f_t(t,z) \geq B z, \quad t \in [0,1], \; z \in [0,R_4], $$ where $$ \lambda B D \geq 1, \quad D = \max_{t \in [0,1]} \int_0^1 G(t,s) q(s) \, ds. $$ Then, for all $v \in \mathcal C \cap \partial \Omega_4$, \begin{align*} \|Tv\| &= \max_{t \in [0,1]} \lambda \int_0^1 G(t,s) f_t(s,v(s))\, ds \\ & \geq \max_{t \in [0,1]} \lambda B \int_0^1 G(t,s) v(s) \, ds \\ & \geq \lambda B \max_{t \in [0,1]} \int_0^1 G(t,s) q(s) R_4 \, ds \\ & = \lambda B D R_4\\ & \geq \|v\|. \end{align*} Thus, there exists a positive solution $v_2$ with $R_4 \leq \|v_2\| \leq R_3$ for every $\lambda > 0$ satisfying \eqref{l3}. 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