\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}. Finally, since $R_3 < R_1$,
the solutions are distinct.
\end{proof}

\subsection*{Acknowledgments}	
The article was prepared while the second author was on sabbatical leave
at Baylor University. He is grateful to Baylor University for its support
and hospitality.

\begin{thebibliography}{00} \frenchspacing

\bibitem{w} J. R. Graef, J. R. L. Webb;
 Third order boundary value problems with nonlocal boundary conditions,
\emph{Nonlinear Anal.} \textbf{71} (2009), 1542–1551.
										
\bibitem{gr} J. R. Graef, B. Yang;
 Positive solutions of a nonlinear third order eigenvalue problem,
 \emph{Dynam. Systems Appl.} \textbf{15} (2006), no. 1, 97--110.

\bibitem{ip} G. Infante, P. Pietramala;
 A third order BVP subject to nonlinear boundary conditions, 
\emph{Math. Bohem.} \textbf{135} (2010), 113–121.

\bibitem{kra} M. A. Krasnosel'ski\u{\i};
 ``Topological Methods in the Theory of Nonlinear  Integral Equations'',
 (English) Translated by A.H. Armstrong;
 A Pergamon Press Book, MacMillan, New York, 1964.

\bibitem{lan} K. Q. Lan;
 Eigenvalues of semi-positone Hammerstein integral equations and applications 
to boundary value problems,  \emph{Nonlinear Anal.} \textbf{71} (2009), 5979-5993.

\bibitem{ma} R. Ma;
 Multiple positive solutions for a semipositone fourth-order problem, 
\emph{Hiroshima Math. J.} \textbf{33} (2013), 217-227.

\bibitem{s} Y. Sun;
 Existence of triple positive solutions for a third-order three-point 
boundary-value problem,  \emph{J. Comp. Appl. Math.} \textbf{221} (2008), 194--201.

\bibitem{yao1} Q. Yao;
 Positive solutions of a weak semipositone third-order three-point boundary value
 problem,  \emph{J. Math. Res. Exp.} \textbf{30} (2010), no. 1, 173--180.

\bibitem{yao2} Q. Yao;
 Positive solutions of singular third-order three-point boundary value problems,
 \emph{J. Math. Anal. Appl.}  \textbf{354} (2009), 207--212.

\bibitem{yhl} H. Yu, L. Haiyan, Y. Liu,
Multiple positive solutions to third-order three-point singular semipositone 
boundary value problem,
\emph{Proc. Indian Acad. Sci. (Math. Sci.)} \textbf{114} (2004), no. 4, 409--422.
												
\bibitem{z} L. Zhang, B. Sun, C. Xing;
 Existence of solutions to a third-order three-point boundary value problem,
 \emph{Ann. Differential Equations} \textbf{28} (2012), no. 3, 368--372.

\end{thebibliography}

\end{document}