\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 152, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/152\hfil Non-existence of solutions]
{Non-existence of solutions for two-point fractional and third-order\\
boundary-value problems}

\author[G. L. Karakostas \hfil EJDE-2013/152\hfilneg]
{George L. Karakostas}  % in alphabetical order

\address{George L. Karakostas \newline
 Department of Mathematics, University of Ioannina,
 451 10 Ioannina, Greece}
\email{gkarako@uoi.gr}


\thanks{Submitted October 3, 2012. Published June 28, 2013.}
\subjclass[2000]{34B15, 34A10, 34B27, 34B99}
\keywords{Third order differential equation; two-point
boundary-value problem; \hfill\break\indent 
fractional boundary condition; nonexistence of solutions}

\begin{abstract}
 In this article, we provide sufficient conditions for the
 non-existence of solutions of the boundary-value problems with fractional
 derivative of order $\alpha\in(2,3)$  in the  Riemann-Liouville sense
 \begin{gather*}
 D_{0+}^{\alpha}x(t)+\lambda a(t)f(x(t))=0,\quad t\in(0,1),\\
 x(0)=x'(0)=x'(1)=0,
 \end{gather*}
 and in the Caputo sense
 \begin{gather*}
  ^CD^{\alpha}x(t)+f(t,x(t))=0,\quad t\in(0,1),\\
 x(0)=x'(0)=0, \quad x(1)=\lambda\int_0^1x(s)ds;
 \end{gather*}
 and for the third-order differential equation
 $$
 x'''(t)+(Fx)(t)=0, \quad \text{a.e. }t\in [0,1],
 $$
 associated with three among the following six conditions
  $$
  x(0)=0,\quad x(1)=0,\quad x'(0)=0, \quad x'(1)=0,
  \quad x''(0)=0,  \quad x''(1)=0.
  $$
  Thus, fourteen boundary-value problems  at resonance and six
  boundary-value problems at non-resonanse are studied.
  Applications of the results are, also, given.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{application}[theorem]{Application}
\allowdisplaybreaks

 \section{Introduction}

The problem of existence of solutions of  fractional differential equations and
general third-order differential equations,  satisfying two-point boundary
conditions, has been extensively discussed in the literature, see, e.g.,
\cite{3, CAB, F, FL, MG,  inp, LKJ, 4, MA, 5, OR, 6,TROY, yao, yao1, 7, 8} 
and the references therein.

 In this work we are dealing with non-existence of solutions and this,
because the two problems (namely, existence and non-existence) are equally
important in the theory of differential equations.   To our knowledge, the
problem of non-existence  for such boundary-value problems has not been studied
sufficiently. Indeed, only a few results have been given on it, and not in a
systematic research. See, e.g.,  \cite{MES, glk, wcz, yang, zwg}. 

Let $I$ be the interval $[0,1]$. We are going to  give sufficient conditions for
the non-existence of solutions of  the  two boundary-value problems
\begin{gather}\label{f1}
 D_{0+}^{\alpha}x(t)+\lambda a(t)f(x(t))=0,\quad 0<t<1,\quad 2<\alpha<3,\\
\label{f2}
x(0)=x'(0)=x'(1)=0,
\end{gather} 
(see, e.g., \cite{MES}) with fractional  Riemann-Liouville type derivative, and
\begin{gather}\label{f3}
 ^CD^{\alpha}x(t)+f(t,x(t))=0,\quad 0<t<1,\quad 2<\alpha<3,\\
\label{f4} 
x(0)=x''(0)=0,\quad x(1)=\lambda\int_0^1x(s)ds
\end{gather} 
(see, e.g., \cite{CW}), with
fractional  Caputo type derivative. The same problem is, also, discussed for
the third-order differential equation of the form
\begin{equation}\label{eq}
x'''(t)+(Fx)(t)=0, \quad a.e. \quad t\in I:=[0,1],
\end{equation}
 associated with the following boundary value conditions:
\begin{gather}\label{p1} x(0)=x(1)=x'(0)=0,\\
\label{p2} x(0)= x(1)=x'(1)=0,\\
\label{p3} x(0)= x(1)= x''(0)=0,\\
\label{p4} x(0)= x(1)=x''(1)=0,\\
\label{p5} x(0)=x'(0)= x'(1)=0,\\
\label{p6} x(1)= x'(0)=x'(1)=0,\\
\label{p7} x(0)= x'(0)= x''(0)=0,\\
\label{p8} x(0)= x'(0)= x''(1)=0,\\
\label{p9} x(0)= x'(1)=x''(0)=0,\\
\label{p10} x(1)= x'(0)= x''(1)=0,\\
\label{p11} x(1)= x'(1)= x''(0)=0,\\
\label{p12} x(0)= x'(1)= x''(1)=0,\\
\label{p13} x(1)= x'(0)= x''(0)=0,\\
\label{p14} x(1)= x'(1)= x''(1)=0,\\
\label{p15} x'(0)= x'(1)= x''(0)=0,\\
\label{p16} x'(0)= x'(1)= x''(1)=0,\\
\label{p17} x(0)= x''(0)= x''(1)=0,\\
\label{p18} x(1)= x''(0)= x''(1)=0,\\
\label{p19} x'(0)= x''(0)= x''(1)=0,\\
\label{p20} x'(1)= x''(0)= x''(1)=0.
\end{gather}
 Conditions \eqref{p1}-\eqref{p14} lead to  non-resonance boundary-value
problems  and they  are examined in section \ref{nres}. On the other hand
conditions \eqref{p15}, \eqref{p16} formulate boundary-value problems  at
resonance, and they  are investigated in the last section \ref{res}.  Existence
of solutions of problems at non-resonance with this kind  of boundary
conditions, is  investigated  in a great number of works, see e.g.
  \cite{cg,du}, \cite{mls}-\cite{inp}, \cite{li}-\cite{MA}, \cite{s1}-\cite{sz},
\cite{wg}-\cite{YF}, \cite{fy}. 

 \section{Preliminaries and some general results}

 Let $C(I, \mathbb{R})$ be the (Banach) space of all continuous
functions $y: I\to\mathbb{R}$ endowed with the sup-norm $\|\cdot\|$. 
Let, also, $U$ be a subspace of $C(I, \mathbb{R})$ and
 $F:U\to C(I, \mathbb{R})$ a
continuous function. Finally, let $T$ be an operator defined on a subset of
$C(I, \mathbb{R})\times C(I, \mathbb{R})$ with range in $C(I, \mathbb{R})$. The
question refers to the existence  of solutions of the operator equation
  \begin{equation}\label{oe}
  y=T(y,F(y)). 
\end{equation} 

Assume that the operator $T$ satisfies the
condition 
$$
\|T(y,u)\|\leq K\|u\|,
$$
 for all $(y, u)$ in the domain of $T$. Let
$K_T$ be the infimum of all such numbers. For instance, if an operator depends
only on the second argument $u$ and it is defined by an integral of the form
\begin{equation}\label{v}
(Ty)(t)=\int_0^1G(t,s)y(s)ds, \quad t\in I,
\end{equation}
where the kernel $G$ is a continuous function defined on the square
 $[0,1]\times [0,1]$, then
$$
K_T=\sup_{t\in{I}}\int_0^1|G(t,s)|ds.
$$
From the definition of  $K_T$ we have the following result.

 \begin{theorem}\label{t1} 
If it holds
 \begin{equation}
 \|Fu\|<\frac{1}{K_T}\|u\|,\quad u\in C(I,\mathbb{R}),
\end{equation}
then there is no solution of equation \eqref{oe}. 
\end{theorem}
 
An equivalent result is the following.

  \begin{corollary}
 Under the previous conditions, the set 
$$
\{x\in C(I,\mathbb{R}): \|Fx\|<\frac{1}{K_T}\|x\|\}
$$ 
does not contain any solution of equation \eqref{oe}.
\end{corollary}

  The inequality in Theorem  \ref{t1} can be guaranteed if, the function $F$
satisfies 
\begin{equation}\label{l}
L_F< \frac{1}{K_T},
\end{equation} 
where
\begin{equation}\label{f}
L_F:=\sup\{\frac{\|Fx\|}{\|x\|}<+\infty: x\in U,
\quad 0<\|x\|\}.
\end{equation}

Our main purpose is to give
information about the  number $K_T$ for several operators generated from
boundary-value problems mentioned in the introduction and then to check for
the applicability of inequality \eqref{l}.

  An interesting case, which will be used in the applications given later,  is
when the response $F$ is a Nemytskii-type operator.  Indeed, assume that  $U$
is the space $C^{(2)}(I, \mathbb{R})$. Also, let $F: U\to C(I,\mathbb{R})$ be
defined via  a continuous function 
 $f:[0,1]\times\mathbb{R}^{3}\to\mathbb{R}$, by the type
$$
(Fx)(t):=f(t,x(t), x'(t),x''(t)),\quad t\in I.
$$   
Then we have the following corollary.

\begin{corollary}\label{cor}  
If  the condition 
$$
\sup_{t\in I}\sup_{u_0,u_1, u_2\in{\mathbb{R}}}
\frac{|f(t,u_0, u_1, u_2)|}{|u_0|}<\frac{1}{K_T}
$$ 
holds, then no solution of  problem  \eqref{oe} exists.
\end{corollary}

\begin{proof} 
Let $k$ be such that
$$
\sup_{t\in I}\sup_{u_0,u_1, u_2\in{\mathbb{R}}}\frac{|f(t,u_0, u_1,
u_2)|}{|u_0|}<k<\frac{1}{K_T}
$$ 
Then, the result follows from Theorem
\ref{t1} and the fact  that for any  function $x$, with continuous derivative
of second order and some points $t_1, t_2\in I$ we have
\begin{align*}
\frac{\|f(\cdot,x(\cdot),x'(\cdot),x''(\cdot))\|}{\|x(\cdot)\|}
&=\frac{|f(t_1,x(t_1), x'(t_1), x''(t_1))|}{|x(t_2)|}\\
&\leq\frac{|f(t_1,x(t_1), x'(t_1),x''(t_1))|}{|x(t_1)|}\\
&\leq\sup_{t\in I}\sup_{u_0,u_1, u_2\in{\mathbb{R}}}\frac{|f(t,u_0, u_1,
u_2)|}{|u_0|}<k.
\end{align*}
\end{proof}

\noindent\textbf{Warning:} In the sequel when an operator $T$ is generated from a boundary-value
problem (a)-(b) instead of $K_T$ we shall write $K_{(a)-(b)}$.

\section{Nonexistence for the fractional BVP \eqref{f1}-\eqref{f2}}\label{fr1}

We start with  the  fractional boundary-value problem \eqref{f1}-\eqref{f2},
studied in \cite{MES}, where (the Krasnoselskii's fixed point theorem in cones
is applied and sufficient conditions for) the existence of solutions  is
investigated.

 Here the parameter $\lambda$ is a positive real number, 
$a:(0,1)\to[0,+\infty)$ is a continuous function with $\int_0^1a(t)dt > 0$, 
and $f:[0,+\infty)\to [0,+\infty)$ is continuous. The symbol $D_{0+}^{\alpha}u$
represents the Riemann-Liouville fractional derivative of the continuous
function $u:I\to\mathbb{R}$ of order $\alpha\in(2,3)$, (see, e.g.,
\cite{WANG}), i.e. the quantity defined by
$$
D_{0+}^{\alpha}u(t)
=\frac{1}{\Gamma(n-\alpha)}\Big(\frac{d}{dt}\Big)^n
\int_0^t\frac{u(s)}{(t-s)^{\alpha-n+1}}ds,\quad
n=\lfloor{\alpha}\rfloor+1.
$$

\begin{theorem} 
Assume that $f$ satisfies the condition
\begin{equation}\label{g} 
L_f:=\sup_{|u|>0}\frac{|f(u)|}{|u|}<
\frac{\Gamma(\alpha)}{\lambda\int_0^1s(1-s)^{\alpha-2}a(s)ds}.
\end{equation}
Then  problem \eqref{f1}-\eqref{f2} does not admit solutions.
\end{theorem}

\begin{proof}
 According to \cite{MES},  the problem is equivalent to the operator equation
\eqref{oe}, where the operator $T$ has the integral form \eqref{v} with the
Green's function $G$ being defined by the type 
$$
G(t,s):=\begin{cases}
\frac{(1-s)^{\alpha-2}t^{\alpha-1}}{\Gamma(\alpha)}a(s), \quad 0\leq t\leq
s\leq 1,\\
\Big[\frac{(1-s)^{\alpha-2}t^{\alpha-1}}{\Gamma(\alpha)}-\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\Big]a(s),
\quad 0\leq s\leq t\leq 1.
\end{cases}
$$  Here $\Gamma(\alpha)$ is the gamma
function at $\alpha$. It is not hard to see that $G$ takes positive values and
it satisfies 
$$
G(t,s)\leq G(1,s), \quad (t,s)\in [0,1]\times[0,1].
$$

 Therefore, the result will follow from Theorem \ref{t1} and inequality
\eqref{l}, when we know that the quantity in the right side of relation
\eqref{g} is less than or equal to $1/K_{\eqref{f1}-\eqref{f2}}$. So, we have
to prove this fact. Indeed, we have
\begin{align*}
K_{\eqref{f1}-\eqref{f2}}
&=\lambda \sup_{t\in I}\int_0^1G(t,s)a(s)ds\\
&=\lambda \int_0^1G(1,s)a(s)ds \\
&=\lambda\frac{1}{\Gamma(\alpha)} \int_0^1\Big[(1-s)^{\alpha-2}
-(1-s)^{\alpha-1}\Big]a(s)ds\\
& = \lambda\frac{1}{\Gamma(\alpha)}
\int_0^1s(1-s)^{\alpha-2}a(s)ds,
\end{align*}
 from which the result follows.
\end{proof}

\begin{application}\rm
Consider the values 
$$
\alpha=2.7, \quad a(t):=2t+3, \quad f(u):=\frac{8u^2+u}{u+1}(4+\sin(u)),
$$ 
as they appear in \cite{MES}. Then we have
\begin{align*}
\int_0^1s(1-s)^{\alpha-2}a(s)ds
&=\int_0^1(1-s)^{\alpha-2}(2s^2+3s)ds\\
&=\int_0^1s^{\alpha-2}(2(1-s)^2+3(1-s))ds\\
&=\int_0^1s^{\alpha-2}(5-7s+2s^2)ds
=\frac{3\alpha+7}{(\alpha-1)\alpha(\alpha+1)}.
\end{align*}
and therefore it follows that
$$
\frac{\Gamma(\alpha)}{\lambda\int_0^1s(1-s)^{\alpha-2}a(s)ds}
=\frac{(\alpha-1)\Gamma(\alpha+2)}{\lambda(3\alpha+7)}
\approx\frac{26.23339972}{15.1\lambda}.
$$
Since it holds 
$$
\frac{|f(u)|}{|u|}=\frac{8u+1}{u+1}(4+\sin(u))\leq 40,
$$  
we have, also, 
$$
\frac{\|f(u(\cdot))\|}{\|u(\cdot)\|}\leq 40,
$$ 
which, due to Theorem \ref{t1},  shows that, if the parameter $\lambda$ 
is chosen such that
$$
\lambda<\frac{26.23339972}{40\times
15.1}=\frac{26.23339972}{604}=0.04343278,
$$
then the problem has no (any, and not necessarily positive) solution. This upper
bound of  the parameter $\lambda$ agrees with the value of the parameter
suggested  in \cite{MES}.
\end{application}

\section{Nonexistence for the fractional BVP \eqref{f3}-\eqref{f4}}\label{fr2}

Here we discuss non-existence for the  fractional boundary-value problem
\eqref{f3}-\eqref{f4}, studied in \cite{CW}, where, again, (the Krasnoselskii's
fixed point theorem in cones is applied and) the existence of solutions  is
investigated.

 It is assumed that
 $^CD^{\alpha}u$ is the Caputo fractional derivative of the continuous function
$u:I\to\mathbb{R}$ at the real number $\alpha\in(2,3)$ defined by
$$
^CD^{\alpha}u(t):=\frac{1}{\Gamma(n-\alpha)}\int_0^t(t-s)^{(\alpha-1)}u(s)ds,
\quad n=\lfloor{\alpha}\rfloor+1.
$$ 
Also, the function $f :[0,1]\times[0,\to[0,\infty)$ is continuous and 
$0<\lambda<2$.

\begin{theorem} 
Assume that $f$ satisfies the condition
\begin{equation}\label{g1}
L_f:=\sup_{t\in{I}}\sup_{|u|>0}\frac{|f(t,u)|}{|u|}<
\frac{(2-\lambda)(\alpha-2)\Gamma(\alpha+2)}{2\alpha(\alpha-1)}.
\end{equation}
Then  problem \eqref{f3}-\eqref{f4} does not admit solutions.
\end{theorem}

\begin{proof}
 According to \cite{CW},  the problem is equivalent to the operator equation
\eqref{oe}, where  $T$ has the integral form \eqref{v}, with the  Green's
function $G$ being given by the type 
$$
G(t,s):=\begin{cases}
\frac{2t(1-s)^{\alpha-1}(\alpha-\lambda+\lambda
s)-(2-\lambda)\alpha(t-s)^{\alpha-1}}{(2-\lambda)\Gamma(\alpha+1)}, \quad
0\leq s\leq t\leq 1,\\
\frac{2t(1-s)^{\alpha-1}(\alpha-\lambda+\lambda
s)}{(2-\lambda)\Gamma(\alpha+1)}, \quad 0\leq t\leq s\leq 1.
\end{cases}
$$
Due to \cite[Lemmas 2.3 and 2.4]{CW},  the kernel $G$ is a nonnegative
function and it satisfies the inequality
$$
G(t,s)\leq\frac{2\alpha}{\lambda(\alpha-2)}G(1,s),
$$ 
for all $s, t, \in[0,1]$ and $\lambda\in(0,2)$. 
Hence, the result will follow from Theorem \ref{t1} and
inequality \eqref{l}, when we know that the quantity in the right side of
relation \eqref{g1} is less than or equal to the $1/K_{\eqref{f3}-\eqref{f4}}$.
To prove this fact observe that
\begin{align*}
{K_{\eqref{f3}-\eqref{f4}}}
&=\sup_{t\in I}\int_0^1G(t,s)ds\\
&\leq \frac{2\alpha}{\lambda(\alpha-2)}\int_0^1G(1,s)ds
=\frac{2\alpha(\alpha-1)}{(2-\lambda)(\alpha-2)\Gamma(\alpha+2)}.
\end{align*}
This completes the proof of the theorem.
\end{proof}

\section{Nonexistence for third-order BVPs at non-resonance}\label{nres}

In this section we give information about  non-existence for the 
third-order differential equation \eqref{eq} subject to one of the 
boundary conditions
\eqref{p1}, \eqref{p2}, $\cdots$, \eqref{p14}.

\begin{theorem}
The  boundary-value problems (at non-resonance)  \eqref{eq}-{\eqref{p1}},
\eqref{eq}-{\eqref{p2}}, \dots ,\eqref{eq}-{\eqref{p14}}, do not have solutions
provided that \eqref{l} is satisfied, where the number $L_F$ is such as  the
following tables shows: 

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
&\eqref{eq}-{\eqref{p1}}&\eqref{eq}-{\eqref{p2}}&\eqref{eq}-{\eqref{p3}}
&\eqref{eq}-{\eqref{p4}}\\
\hline
$0<L_f<$& $40.5$&$40.5$& $\inf_{t\in I}{6}/{t(t^4-2t^3+1)}$
&$9\sqrt{3}$ \\
\hline
\end{tabular}
\\[4pt]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
&\eqref{eq}-{\eqref{p5}}&\eqref{eq}-{\eqref{p6}}&\eqref{eq}-{\eqref{p7}}
&\eqref{eq}-{\eqref{p8}}&\eqref{eq}-{\eqref{p9}}\\
\hline
$0<L_f<$& $12$& $12$&\hskip .2in$6$& $3$&$3$ \\
\hline
\end{tabular}
\\[4pt]
\begin{tabular}{|c|c|c|c|c|c|}
\hline
&\eqref{eq}-{\eqref{p10}}&\eqref{eq}-{\eqref{p11}}&\eqref{eq}-{\eqref{p12}}&\eqref{eq}-{\eqref{p13}}&\eqref{eq}-{\eqref{p14}}\\
\hline
$0<L_f<$& $3$& $3$& $6$& $6$&$6$ \\
\hline
\end{tabular}\end{center}
\end{theorem}

\begin{proof}  
All the boundary-value problems are equivalent to the integral
equation \eqref{oe}, where the operator $T$ is defined by \eqref{v}. Thus the
proof is implied from Theorem \ref{t1}, provided that the numbers defined in
\eqref{g1} have values as in the tabular. Therefore what we have to (and shall)
do is to obtain the Green's functions of the  integral equations and then to
calculate the corresponding numbers $1/K_T$.

 \noindent\emph{Problem \eqref{eq}-{\eqref{p1}}:}
The corresponding  Green's function is 
\[
G(t,s):=\begin{cases}
\frac{1}{2}s(1-t)[t(1-s)+t-s], &0\leq s\leq t\leq 1,\\
\frac{t^2}{2}(1-s)^2, & 0\leq t\leq s\leq 1\end{cases}
\]
and it is nonnegative.  Hence we have
\begin{align*}
{K_{\eqref{eq}-{\eqref{p1}}}}
&=\sup_{t\in I}\Big[\int_0^tG(t,s)ds+\int_t^1G(t,s)ds\Big]\\
&=\frac{1}{2}\sup_{t\in I}\Big[\int_0^ts(1-t)[t(1-s)+t-s]ds
 +\int_t^1t^2(1-s)^2ds\Big]\\
&=\dots=\sup_{t\in I}\frac{t^2}{6}(1-t)=\frac{2}{81}
=\frac{1}{40.5}.
\end{align*}


\noindent\emph{Problem \eqref{eq}-{\eqref{p2}}:}
The corresponding Green's function is 
$$
G(t,s):=\begin{cases}
\frac{-1}{2}s^2(1-t)^2, & 0\leq s\leq t\leq 1\\
\frac{-t}{2}(1-s)[(s-t)+s(1-t)], & 0\leq t\leq s\leq 1,\end{cases}$$
 which is nonpositive. Thus we have
\begin{align*}
{K_{\eqref{eq}-{\eqref{p2}}}}
&=\sup_{t\in I}\Big[\int_0^t|G(t,s)|ds+\int_t^1|G(t,s)|ds\Big]\\
&=\frac{1}{2}\sup_{t\in I}\Big[\int_0^ts^2(1-t)^2ds
 +\int_t^1t(1-s)[(s-t)+s(1-t)]ds\Big]\\
&=\cdots=\sup_{t\in I}\frac{t}{6}(t-1)^2=\frac{2}{81}=\frac{1}{40.5}.
\end{align*}


\noindent\emph{Problem \eqref{eq}-{\eqref{p3}}:}
The corresponding Green's function is
$$
G(t,s):=\begin{cases}
\frac{1}{2}(1-t)(t-s^2), & 0\leq s\leq t\leq 1\\
\frac{t}{2}(1-s)^2,& 0\leq t\leq s\leq 1,
\end{cases}
$$
 which is nonnegative. Thus we have
\begin{align*}
{K_{\eqref{eq}-{\eqref{p3}}}}
&=\sup_{t\in I}\Big[\int_0^tG(t,s)ds+\int_t^1G(t,s)ds\Big]\\
&=\frac{1}{2}\sup_{t\in I}\Big[\int_0^t(1-t)(t-s^2)ds+\int_t^1t(1-s)^2ds\Big]\\
&=\cdots=\sup_{t\in I}\frac{t}{6}(t^4-2t^3+1)=\frac{1}{\inf_{t\in
I}\frac{6}{t(t^4-2t^3+1)}}\approx\frac{1}{14.309267}.
\end{align*}



\noindent\emph{Problem \eqref{eq}-{\eqref{p4}}:}
The corresponding Green's function is 
\[
G(t,s):=\begin{cases}
\frac{-s^2}{2}(1-t) & s\leq t,\\
\frac{-t}{2}[(s-t)+s(1-s)], & t\leq s
\end{cases}
\]
 and it  is nonpositive.  Then we have  
\[
{K_{\eqref{eq}-{\eqref{p4}}}}=\sup_{t\in
I}\frac{t}{6}(1-t)(2-t)=\frac{\sqrt{3}}{27}
=\frac{1}{9\sqrt{3}}.
\]


\noindent\emph{Problem  \eqref{eq}-{\eqref{p5}}:}
The corresponding Green's function is 
\[
G(t,s):=\begin{cases}
\frac{1}{2}[ts(1-t)+s(t-s)], & 0\leq s\leq t\leq 1,\\
\frac{t^2}{2}(1-s), & 0\leq  t\leq s\leq 1.
\end{cases}\]
The latter is nonnegative.  Hence we obtain
\[
{K_{\eqref{eq}-{\eqref{p5}}}}
=\sup_{t\in I}\Big[\frac{t^2}{4}-\frac{t^3}{6}\Big]
=\frac{1}{12}.
\]

\noindent\emph{Problem \eqref{eq}-{\eqref{p6}}:}
 The corresponding Green's function  is 
\[
G(t,s):=\begin{cases}
\frac{-s}{2}(1-t)^2, & s\leq t,\\
\frac{-1}{2}(1-s)(s-t^2), & t\leq s
\end{cases}
\]
 and it is nonpositive.  Hence it follows that
\[
{K_{\eqref{eq}-{\eqref{p6}}}}
=\sup_{t\in I}\frac{1}{12}(1-3t^2+2t^3)=\frac{1}{12}.
\]

\noindent\emph{Problem \eqref{eq}-{\eqref{p7}}: }
The corresponding Green's function is 
$$
G(t,s):=\frac{-(t-s)^2}{2}\chi_{[0,t]}(s),
$$ 
which is nonpositive.  Thus we have 
$$
{K_{\eqref{eq}-{\eqref{p7}}}}=\sup_{t\in
I}\int_0^t\frac{(t-s)^2}{2}ds=\int_0^1\frac{s^2}{2}ds=\frac{1}{6}.
$$

  \noindent\emph{Problem \eqref{eq}-{\eqref{p8}}:}
  The corresponding Green's function is  
\[
G(t,s):=\begin{cases}
\frac{1}{2}[ts+s(t-s)], &s\leq t,\\
\frac{t^2}{2}, & t\leq s,
\end{cases}
\]
 which is nonnegative. Hence we obtain
\[
{K_{\eqref{eq}-{\eqref{p8}}}}
=\sup_{t\in I}\frac{t^2}{6}(3-t)=\frac{1}{3}.
\]



\noindent\emph{Problem \eqref{eq}-{\eqref{p9}}:}
The corresponding Green's function is 
$$
G(t,s):=\begin{cases}
\frac{1}{2}(t(1-t)+t-s+s(1-s)), & 0\leq s\leq t\leq 1\\
t(1-s),& 0\leq t\leq s\leq 1,
\end{cases}
$$
 which is nonnegative.  Thus we have
\[
{K_{\eqref{eq}-{\eqref{p9}}}}
=\sup_{t\in I}\frac{t}{6}(3-t^2)=\frac{1}{3}.
\]

\noindent\emph{Problem \eqref{eq}-{\eqref{p10}}:}
The corresponding Green's function $G$ is 
\[
G(t,s):=\begin{cases}
-s(1-t), & s\leq t,\\
\frac{-1}{2}[2s-s^2-t^2], & t\leq s,\end{cases}
\] 
and it is nonpositive.  This gives
\[
{K_{\eqref{eq}-{\eqref{p10}}}}=\sup_{t\in
I}\frac{1}{6}(2-3t^2+t^3)=\frac{1}{3}.
\]

 \noindent\emph{Problem \eqref{eq}-{\eqref{p11}}:}
The corresponding Green's function is 
$$
G(t,s):=\begin{cases}
\frac{-1}{2}(1-t)^2, & 0\leq s\leq t\leq 1\\
\frac{-1}{2}(1-s)[(s-t)+(1-t)],& 0\leq t\leq s\leq 1,
\end{cases}
$$
 which is nonpositive. Thus we obtain
\[
{K_{\eqref{eq}-{\eqref{p11}}}}
=\sup_{t\in I}\frac{1}{6}[2-3t+t^3]=\frac{1}{3}.
\]

 \noindent\emph{Problem \eqref{eq}-{\eqref{p12}}:}
The corresponding Green's function is 
$$
G(t,s):=\begin{cases}
\frac{-1}{2}s^2, & 0\leq s\leq t\leq 1\\
\frac{-1}{2}[ts+t(s-t)], & 0\leq t\leq s\leq 1,
\end{cases}
$$
 which is nonpositive. Then we obtain
\[
{K_{\eqref{eq}-{\eqref{p12}}}}=\sup_{t\in
I}\frac{1}{6}[3t-3t^2+t^3]=\frac{1}{6}.
\]

 \noindent\emph{Problem \eqref{eq}-{\eqref{p13}}:}
The corresponding Green's function is 
\[
G(t,s):=\begin{cases}
\frac{1-t}{2}(1+t-2s), & 0\leq s\leq t\leq 1\\
\frac{1}{2}(1-s)^2,& 0\leq t\leq s\leq 1,
\end{cases}
\]
 which is nonnegative. Hence, it follows that
\[
{K_{\eqref{eq}-{\eqref{p13}}}}=\sup_{t\in
I}\frac{1}{6}[1-t^3]=\frac{1}{6}.
\]

 \noindent\emph{Problem \eqref{eq}-{\eqref{p14}}:}
This is a terminal value problem with Green's function given by
$$
G(t,s):=\frac{(t-s)^2}{2}\chi_{[t,1]}(s),
$$ which  is nonnegative. Thus we
have
 $$
{K_{\eqref{eq}-{\eqref{p14}}}}=\sup_{t\in
I}\int_t^1\frac{(t-s)^2}{2}ds=\frac{1}{6}.
$$
The proof is complete.
 \end{proof}


\begin{application}\rm 
Consider the differential equation
\begin{equation}\label{e0}
x'''+(\sigma+\frac{\cos(x'+x'')}{4(1+x^2)})x=0.\end{equation}
Here we have 
$$
f(u_1, u_2, u_3)=(\sigma+\frac{\cos(u_2+u_3)}{4(1+u_1^2)})u_1
$$
and therefore
$$
\sup_{u_1,u_2, u_3}\frac{f(u_1, u_2, u_3)}{u_1}=\sigma+\frac{1}{4}.
$$ 
Taking into account Corollary \ref{cor}, the previous theorem and 
Theorem \ref{t1}, we can, easily, see
 that the problems \eqref{e0}-\eqref{p1}, \eqref{e0}-\eqref{p2}, $\cdots$,
\eqref{e0}-\eqref{p14} do not have any solution, provided that the parameter
$\sigma$ has corresponding values, as the following tables show:

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$$
&\eqref{e0}-{\eqref{p1}}&\eqref{e0}-{\eqref{p2}}&\eqref{e0}-{\eqref{p3}}&\eqref{e0}-{\eqref{p4}}&\eqref{e0}-{\eqref{p5}}\\
\hline
$0<\sigma$&$<40.25$&$<40.25$&$<14.0602$&$<15.3385$&$<11.75$\\
\hline
\end{tabular}
\\[4pt]
 \begin{tabular}{|c|c|c|c|c|c|}
\hline
&\eqref{e0}-{\eqref{p6}}&\eqref{e0}-{\eqref{p7}}&\eqref{e0}-{\eqref{p8}}&\eqref{e0}-{\eqref{p9}}&\eqref{e0}-{\eqref{p10}}
\\
\hline
$0<\sigma$&$<11.75$&$<5.75$&$<2.75$&$<2.75$&$<2.75$\\
\hline
\end{tabular}
\\[4pt]
 \begin{tabular}{|c|c|c|c|c|c|}
\hline
&\eqref{e0}-{\eqref{p11}}&\eqref{e0}-{\eqref{p12}}&\eqref{e0}-{\eqref{p13}}&\eqref{e0}-{\eqref{p14}}
\\
\hline
$0<\sigma$&$<2.75$&$<5.75$&$<5.75$&$<5.75$\\
\hline
\end{tabular}
\end{center}
The upper bounds of $\sigma$ for problem \eqref{e0}-{\eqref{p3}} and
\eqref{e0}-{\eqref{p4}} are approximate values.
\end{application}


\section{Nonexistence for third-order BVPs  at resonance}\label{res}

In this section we shall discuss the problems \eqref{qq}-{\eqref{p15}} -
\eqref{qq}-{\eqref{p20}} where  \eqref{qq} is  equation
\begin{equation}\label{qq} 
x'''+f(x(t))=0,\quad \text{a. e. } t\in I.
\end{equation} 
Due to the boundary conditions {\eqref{p15}}-{\eqref{p20}},
these problems are at resonance and, so, they do not have integral equivalent
forms, as in the preceding cases. Hence we need to apply a suitable technique,
where, we need to assume some rather simple conditions on the function $f$. Of
course, this will restrict the family of responses. Indeed, in the sequel we
shall assume  that the response $f$ satisfies the conditions:
\begin{itemize}
\item[(1)] $f$ is a continuously differentiable real valued function.


\item[(2)]  There are positive real numbers $L_f>\delta$ such that
\begin{equation}\label{a}
 |f(u)|\leq L_f|u|\quad\text{and}\quad |f'(u)|\geq\delta,\quad
u\in\mathbb{R}.
\end{equation}
\end{itemize}
 Clearly, the latter implies that either
$f'(u)\geq \delta$, for all $u$, or $f'(u)\leq -\delta$, for all
$u$.
Our main theorem reads as follows.

 \begin{theorem}\label{t2}
The  boundary-value problems \eqref{qq}-{\eqref{p15}}, \dots ,
\eqref{qq}-{\eqref{p20}} (at resonance) do not admit solutions when the
parameters $L, \delta$ satisfy the relations as in the following tables:

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
&\hskip .2in \eqref{qq}-{\eqref{p15}}&
\eqref{qq}-{\eqref{p16}}&\eqref{qq}-{\eqref{p17}} \\
\hline
$\delta<L_f<$&$-5\delta+\sqrt{25\delta^2+60\delta}$&$\frac{1}{11}(-10\delta+2\sqrt{25\delta^2+165\delta})$&$-\delta+\sqrt{\delta^2+12\delta}$\\
\hline
\end{tabular}
\\[4pt]
\begin{tabular}{|c|c|c|c|}
\hline
&\eqref{qq}-{\eqref{p18}}&\eqref{qq}-{\eqref{p19}}&\eqref{qq}-{\eqref{p20}}\\
\hline
$\delta<L_f<$&$\frac{1}{3}(-\delta+\sqrt{\delta^2+36\delta})$&
$-2\delta+\sqrt{4\delta^2+24\delta}$&$\frac{1}{5}(-4\delta+2
\sqrt{4\delta^2+30\delta})$\\
\hline
\end{tabular}
\end{center}
\end{theorem}

\begin{proof} These boundary-value problems can be written as operator
equations of the form \eqref{oe}, where the operator $T$ is not  linear.  So,
for each problem, we must formulate this operator and then to calculate  the
quantity $K_T$. Also, due to \eqref{a}, Theorem \ref{t1} will be applied, when
we shall show that 
\begin{equation}\label{s}
L_fK_T<1.
\end{equation}


 \noindent\emph{Problem \eqref{qq}-{\eqref{p15}}:}  Assume that
$\delta<L_f<-5\delta+\sqrt{25\delta^2+60\delta}$. Then it holds
 \begin{equation}\label{s1}
L_f[\frac{L_f}{60\delta}+\frac{1}{6}]<1,
\end{equation}
 If there is a solution $x$ of the problem, it will satisfy the boundary
conditions $x'(0)=x'(1)=x''(0)=0$.
 Integrating equation $x'''+f(x)=0$ we get $$x''(t)=-\int_0^ty(s)ds,$$  where
$y(t):=f(x(t))$. By using the boundary conditions we obtain
\begin{equation}\label{b1}
\int_0^1(1-s)y(s)ds=0
\end{equation}
and $x(t)=x(0)-(g_1y)(t), \quad t\in[0,1]$, where
$$
(g_1y)(t):=\int_0^t\frac{(t-s)^2}{2}y(s)ds, \quad t\in{I}.
$$ 
Next, define the function 
$$
(A_1y)(v):=\int_0^1(1-t)f(v-(g_1y)(t))dt,\quad v\in\mathbb{R}.
$$ 
This is a differentiable function, which, due to \eqref{b1},
satisfies the equation 
$$
(A_1y)(0)+(A_1y)'(\mu x(0))x(0)=(A_1y)(x(0))=0,
$$  
for
some $\mu\in [0,1]$. Namely we have
$$
\int_0^1(1-t)f(-(g_1y)(t))dt+x(0)\int_0^1(1-t)f'(\mu x(0)-(g_1y)(t))dt=0.
$$
Thus the initial value satisfies the relation
$$
x(0)=-b_1(x,y)\int_0^1(1-t)f(-(g_1y)(t))dt,
$$
 where
$$
b_1(x,y):=\frac{1}{\int_0^1(1-t)f'(\mu x(0)t-(g_1y)(t))dt}.
$$ 
Clearly, it holds  $|b_1(x,y)|\leq 2/\delta$.
Therefore,  the solution $x$ satisfies the integral equation $x=T(x,f(x))$,
where the operator $T$ is defined on $C(I,\mathbb{R})\times C(I,\mathbb{R}) $
by the type
$$
T(x,y)(t):=-b_1(x,y)\int_0^1(1-r)f(-(g_1y)(r))dr-(g_1y)(t), \quad t\in
[0,1].
$$
Then, from \eqref{a} we obtain
$$
|T(x,y)(t)|\leq K_{\eqref{qq}-{\eqref{p15}}}\|y\|,
$$
where
$$
K_{\eqref{qq}-{\eqref{p15}}}:=\frac{2}{\delta}L_f\int_0^1(1-r)\frac{r^3}{6}dr
+\frac{1}{6},
$$
which implies that
$$
L_fK_{\eqref{qq}-{\eqref{p15}}}< L_f[\frac{L_f}{60\delta}+\frac{1}{6}]<1,
$$
because of \eqref{a} and \eqref{s1}.


\noindent\emph{Problem \eqref{qq}-{\eqref{p16}}:} Let
$\delta<L_f<\frac{1}{11}(-10\delta+2\sqrt{25\delta^2+55\delta})$. 
Then we have
\begin{equation}\label{11}
L_f\frac{11L_f+20\delta}{60\delta}<1.
\end{equation}
 Assume that there is a  solution $x(t), \quad t\in [0,1]$ of the problem such
that $x'(0)=x'(1)=x''(1)=0$.
 Integrating equation $x'''+f(x)=0$, we get
\begin{equation}\label{r1}
x''(t)=-\int_0^ty(s)ds,
\end{equation} 
where
$y(s):=f(x(s))$. By using the boundary conditions $x''(1)=x'(1)=0$, it follows
that
\begin{equation}\label{b3}
\int_0^1sy(s)ds=0.
\end{equation}
Integrate \eqref{r1}, once again, and get $x(t)=x(0)+(g_2y)(t)$, where
$$
(g_2y)(t):=\frac{t^2}{2}\int_t^1y(s)ds+\int_0^t\frac{2ts-s^2}{2}y(s)ds,
\quad t\in{I}.
$$
Define the function 
$$
(A_2y)(v):=\int_0^1tf(v+(g_2y)(t))dt,\quad
v\in\mathbb{R},
$$ 
for which we observe that 
$$
(A_2y)(0)+(A_2y)'(\lambda x(0))x(0)=(A_2y)(x(0))=0,
$$ 
because of \eqref{b3}, for some $\lambda\in [0,1]$.
This implies that
$$
\int_0^1tf((g_2y)(t))dt+x(0)\int_0^1tf'(\lambda x(0)+(g_2y)(t))dt=0,
$$
from which  we get
$$
x(0)=-b_2(x,y)\int_0^1tf((g_2y)(t))dt,
$$ 
where
$$
b_2(x,y):=\frac{1}{\int_0^1tf'(\lambda x(0)+(g_2(y)(t))dt}
$$ 
is such that $|b_2(x,y)|\leq 2/\delta$.
Hence, for all $t\in [0,1]$, the solution $x$ satisfies the integral relation
$$
x(t)=T(x,f(x)),
$$ 
where the operator $T$ is defined by the type
$$
T(x,y):=-b_2(x,y)\int_0^1rf((g_2y)(r))dr+(g_2y)(t), \quad t\in I
$$ on
$C(I,\mathbb{R})\times C(I,\mathbb{R})$.
 From this equation and \eqref{a} it follows that
$$
|T(x,y)(t)|\leq\frac{2}{\delta}L_f\int_0^1r|(g_2y)(r)|dr+|(g_2y)(t)|
$$
and so, finally, we obtain
$$
|T(x,y)(t)|\leq K_{\eqref{qq}-{\eqref{p16}}}\|y\|,
$$
where
$$
K_{\eqref{qq}-{\eqref{p16}}}:=\frac{2}{\delta}L_f\frac{11}{120}+\frac{1}{3}.
$$
Thus we get $L_fK_{\eqref{qq}-{\eqref{p16}}}<1$, because of  \eqref{11}.

\noindent\emph{Problem \eqref{qq}-{\eqref{p17}}:}
 Assume that $\delta<L_f<-\delta+\sqrt{\delta^2+12\delta}$. Then we have
 \begin{equation}\label{14}
L_f \frac{L_f+2\delta}{12\delta}<1.
\end{equation}
Assume that $x$ is a solution of the problem.
 Hence it satisfies $x(0)=x''(0)=x''(1)=0$.
 Integrating equation $x'''+f(x)=0$, we obtain $$x''(t)=-\int_0^ty(s)ds,$$ where
$y(s):=f(x(s))$. Thus we have
\begin{equation}\label{b5}
\int_0^1y(s)ds=0.
\end{equation}
Integrate two more times and get $x(t)=x'(0)t-(g_1y)(t), \quad t\in[0,1]$,
where  
$$
(g_1y)(t):=\int_0^t\frac{(t-s)^2}{2}y(s)ds,  \quad t\in{I}.
$$
Define the function 
$$
(A_3y)(v):=\int_0^1f(vt-(g_1y)(t))dt,\quad v\in\mathbb{R}
$$ 
and we observe that it satisfies 
$$
(A_3y)(0)+(A_3y)'(\lambda x'(0))x'(0)=(A_3y)(x'(0))=0,
$$ 
because of \eqref{b5}, for some $\lambda\in[0,1]$. This implies that
$$
\int_0^1f(-(g_1y)(t))dt+x'(0)\int_0^1tf'(\lambda x'(0)t-(g_1y)(t))dt)=0.
$$
Therefore, the value $x'(0)$ satisfies
$$
x'(0)=-b_3(x,y)\int_0^1f(-(g_1y)(t))dt,
$$ 
where the coefficient $b_3(x,y)$ is
such that  $$b_3(x,y):=\frac{1}{\int_0^1tf'(\lambda x'(0)t-(g_1y)(t))dt}$$ and
moreover 
$$
|b_3(x,y)|\leq \frac{2}{\delta},
$$ because of \eqref{a}. Hence, for
all $t\in [0,1]$, the solution $x$ satisfies the operator equation
$x(t)=T(x,f(x))(t)$, where $T$ is defined on the space
$C^{(1)}(I,\mathbb{R})\times C(I,\mathbb{R})$ by the type
$$
T(x,y)(t):=-b_3(x,y)t\int_0^1f(-(g_1y)(r))dr-(g_1y)(t).
$$
 From here and \eqref{a} we obtain
$$
|T(x,y)(t)|\leq |b_3(x,y)|L\int_0^1|(g_1y)(r)|dr+|(g_1y)|(t),
$$ and so,
finally,
$$
|T(x,y)(t)|\leq K_{\eqref{qq}-{\eqref{p17}}}\|y\|,
$$
 where
$$
K_{\eqref{qq}-{\eqref{p17}}}:= \frac{L_f}{12\delta}+\frac{1}{6}.
$$ The latter
implies that $L_fK_{\eqref{qq}-{\eqref{p17}}}<1$, due to \eqref{14}.

\noindent\emph{Problem \eqref{qq}-{\eqref{p18}}:}
Let $\delta<L_f<\frac{1}{3}(-\delta+\sqrt{\delta^2+36\delta})$. Then we have
\begin{equation}\label{104}
L_f<\frac{12\delta}{3L_f+2\delta}.
\end{equation}
 Assume that there is a  solution $x(t), \quad t\in [0,1]$ of the problem such
that $x(1)=x''(0)=x''(1)=0$.
 Integrating equation $x'''+f(x)=0$, we obtain 
$$
x''(t)=-\int_0^ty(s)ds,
$$ where
$y:=f(x)$. Thus we have
\begin{equation}\label{b7}\int_0^1y(s)ds=0.\end{equation}
Taking into account the boundary conditions, we integrate and get
$$x(t)=-(1-t)x'(0)+(g_3y)(t), \quad t\in[0,1],$$
where
$$
(g_3y)(t):=\int_t^1\frac{(1-s)^2}{2}y(s)ds+\int_0^t\frac{(1-t)^2+2(t-s)(1-t)}{2}y(s)ds,
 \quad t\in{I}.
$$
Define the function
 $$
(A_4y)(v):=\int_0^1f(-(1-t)v+(g_3y)(t))dt,\quad
v\in\mathbb{R},
$$ 
for which we observe that 
$$
(A_4y)(0)+(A_4y)'(\lambda x'(0))x'(0)=(A_4y)(x'(0))=0,
$$ 
because of \eqref{b7}, for some $\lambda\in [0,1]$. This implies that
$$
\int_0^1f((g_3y)(t))dt-x'(0)\int_0^1(1-t)f'(-(1-t)\lambda
x'(0)+(g_3y)(t))dt=0
$$
and, so,
$$
x'(0)=b_4(x,y)\int_0^1f((g_3y)(t))dt,
$$ where
$$
b_4(x,y):=\frac{1}{\int_0^1(1-t)f'(-(1-t)\lambda x'(0)+(g_3y)(t))dt}.
$$  
Due to \eqref{a},  we have $|b_4(x,y)|\leq\frac{2}{\delta}$.
Hence, for all $t\in [0,1]$, the solution $x$ satisfies the relation
$$
x(t)=T(x,f(x))(t),
$$ 
where the operator $T$ is defined on
$C^{(1)}(I,\mathbb{R})\times C(I,\mathbb{R})$ by the type
$$
T(x,y)(t):=-(1-t)b_4(x,y)\int_0^1f((g_3y)(r))dr+(g_3y)(t), \quad
t\in[0,1].
$$
 Then, as in the previous cases,  due to \eqref{a}, we obtain
$$
|T(x,y)(t)|\leq K_{\eqref{qq}-{\eqref{p18}}}\|y\|,
$$ 
where
$$
K_{\eqref{qq}-{\eqref{p18}}}:= L_f(\frac{L_f}{4\delta}+\frac{1}{6}).
$$ 
This implies that $L_fK_{\eqref{qq}-{\eqref{p18}}}<1$, because of \eqref{104}.


\noindent\emph{Problem \eqref{qq}-{\eqref{p19}}:}
Assume that $\delta<L_f<-2\delta+\sqrt{4\delta^2+24\delta}$.
 Then we have
 \begin{equation}\label{17}
L_f<\frac{24\delta}{L_f+4\delta}.
\end{equation} 
Let $x(t), \quad t\in [0,1]$ be  a solution of the problem such that
$x'(0)=x''(0)=x''(1)=0$.
 Integrating equation $x'''+f(x)=0$, we obtain $x''(t)=-\int_0^ty(s)ds$, where,
again, $y(s):=f(x(s))$. Thus we have
\begin{equation}\label{b9}
\int_0^1f(x(s))ds=0.
\end{equation}
Integrate  and get $x(t)=x(0)-(g_1y)(t), \quad t\in[0,1]$, where, again,
$$
(g_1y)(t):=\int_0^t\frac{(t-s)^2}{2}y(s)ds,  \quad t\in{I}.
$$
Define the function 
$$
(A_5y)(v):=\int_0^1f(v-(g_1y)(t))dt, \quad v\in\mathbb{R},
$$ 
and observe that 
$$
(A_5y)(0)+(A_5y)'(\lambda x(0))x(0)=(A_5y)(x(0))=0,
$$ 
because of \eqref{b9}, for some $\lambda\in [0,1]$.
This implies that
$$
\int_0^1f(-(g_1y)(t))dt+x(0)\int_0^1f'(\lambda x(0)-(g_1y)(t))dt=0
$$
and so
$$
x(0)=-b_5(x,y)\int_0^1f(-(g_1y)(t))dt,
$$ 
where
$$
b_5(x,y):=\frac{1}{\int_0^1f'(\lambda x(0)-(g_1y)(t))dt},
$$  
which, obviously,  satisfies $|b_5|\leq\frac{1}{\delta}$.
 Hence, for all $t\in [0,1]$,
 the solution $x$ satisfies the relation
$$
x(t)=T(x,f(x))(t),
$$ where the operator $T$ is defined on
$C(I,\mathbb{R})\times C(I,\mathbb{R})$ by the type
$$
T(x,y)(t):=-b_5(x,y)\int_0^1f(-(g_1y)(r))dr)-(g_1y)(t).
$$
Thus, because of \eqref{a}, we have
$$
|T(x,y)(t)|\leq
|b_5(x,y)|L_f\int_0^1\int_0^{r}\frac{(r-s)^2}{2}\|y\|\,ds\,dr
+\int_0^t\frac{(t-s)^2}{2}\|y\|ds,
$$
and so
$$
|T(x,y)(t)|\leq K_{\eqref{qq}-{\eqref{p19}}}\|y\|,
$$ 
where
$$
K_{\eqref{qq}-{\eqref{p19}}}:= L_f(\frac{L_f}{24\delta}+\frac{1}{6}),
$$
because of \eqref{a}. This gives $L_fK_{\eqref{qq}-{\eqref{p19}}}<1$, because
of \eqref{17}.

\noindent\emph{Problem \eqref{qq}-{\eqref{p20}}:}
Let $0<L_f<\frac{1}{5}(-4\delta+2\sqrt{4\delta^2+30\delta})$. Then we have
\begin{equation}\label{19}
L_f\frac{5L_f+8\delta}{24\delta}<1.
\end{equation}
 Assume that there is a  solution $x(t)$, $t\in [0,1]$ of the problem such
that $x'(1)=x''(0)=x''(1)=0$.
 Integrating the equation $x'''+f(x)=0$, we obtain $x''(t)=-\int_0^ty(s)ds$, where
$y:=f(x)$. Thus we have
\begin{equation}\label{b11}
\int_0^1y(s)ds=0.
\end{equation}
Integrate, again,  and obtain $x(t)=x(0)+(g_4y)(t)$, $t\in[0,1]$, where
$$
(g_4y)(t):=t\int_t^1(1-s)y(s)ds+\int_0^t(t-\frac{t^2}{2}-\frac{s^2}{2})y(s)ds,
 \quad t\in{I}.
$$
Define the function 
$$
(A_6y)(v):=\int_0^1f(v+(g_4y)(t))dt, \quad
v\in\mathbb{R},
$$ 
for which we observe that 
$$
(A_6y)(0)+(A_6y)'(\lambda x(0))x(0)=(A_6y)(x(0))=0,
$$ 
because of \eqref{b11}, for some 
$\lambda\in [0,1]$. This implies that
$$
\int_0^1f((g_4y)(t))dt+x(0)\int_0^1f'(\lambda x(0)+(g_4y)(t))dt=0
$$
and so we have
$$
x(0)=-b_6(x,y)\int_0^1f((g_4y)(t))dt,
$$ 
where
$$
b_6(x,y):=\frac{1}{\int_0^1f'(\lambda x(0)+(g_4y)(t))dt},
$$ 
which satisfies
$|b_6(x,y)|\leq\frac{1}{\delta}$. Hence, for all $t\in [0,1]$, the solution
$x$ satisfies the relation
$$
x(t)=T(x,f(x))(t),
$$
 where $T$ is the operator defined on
$C(I,\mathbb{R})\times C(I,\mathbb{R})$ by the type
$$
T(x,y)(t):=-b_6(x,y)\int_0^1f((g_4y)(r))dr+(g_4y)(t).
$$
Due to \eqref{a}, this gives that
$$
|T(x,y)(t)|\leq |b_6(x,y)|L_f\|y\|\int_0^1\big(\frac{r}{2}
-\frac{r^3}{6}\big)dr+\|y\|(\frac{t}{2}-\frac{t^3}{6}),
$$
and so
$$
|T(x,y)(t)|\leq K_{\eqref{qq}-{\eqref{p20}}}\|y\|,
$$ 
where
$$
K_{\eqref{qq}-{\eqref{p20}}}:=L_f(\frac{5L_f}{24\delta}+\frac{1}{3}).
$$ 
The latter gives $L_f{K_{\eqref{qq}-{\eqref{p20}}}}<1$ because of \eqref{19}.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=.3\textwidth]{fig1}\quad
\includegraphics[width=.3\textwidth]{fig2}\quad
\includegraphics[width=.3\textwidth]{fig3}\\
\includegraphics[width=.3\textwidth]{fig4}\quad
\includegraphics[width=.3\textwidth]{fig5}\quad
\includegraphics[width=.3\textwidth]{fig6}
\end{center}
\caption{If the pair of parameters $(\delta, L)$ belongs to the shaded area in
the $\delta-L$ plane, then there are no solutions of the BVPs
\eqref{qq}-{\eqref{p15}}, \eqref{qq}-{\eqref{p16}}, \eqref{qq}-{\eqref{p17}},
\eqref{qq}-{\eqref{p18}}, \eqref{qq}-{\eqref{p19}}, \eqref{qq}-{\eqref{p20}},
respectively.}
\end{figure}
\end{proof}



\begin{application} \rm 
Consider the differential equation
\begin{equation}\label{aqa}
x'''+(\sigma+\frac{1}{4(1+x^2)})x=0,
\end{equation}
for a positive parameter $\sigma$. Setting $\delta:=\sigma-\frac{1}{4}$ and 
$ L:=\sigma+\frac{1}{4}$, one can easily apply the tabular in Theorem \ref{t2}
and conclude  that the problems \eqref{aqa}-{\eqref{p15}}, $\cdots$,
\eqref{aqa}-{\eqref{p20}}  do not have solutions when the parameter $\sigma$
takes corresponding values, such as  they are approximately given in the
following tables:

\begin{center}\begin{tabular}{|c|c|c|c|}
\hline
$$ & \eqref{aqa}-{\eqref{p15}}&
\eqref{aqa}-{\eqref{p16}}& \eqref{aqa}-{\eqref{p17}}\\
\hline
$\sigma$&$0.00231<\sigma<4.92951$&$0.00254<\sigma<2.3424$&$0.01223<\sigma<3.40443$\\
\hline
\end{tabular}
\\[4pt]
\begin{tabular}{|c|c|c|c|}
\hline
& \eqref{aqa}-{\eqref{p18}}&
\eqref{aqa}-{\eqref{p19}}& \eqref{aqa}-{\eqref{p20}}\\
\hline
$\sigma$&$0.04397<\sigma<1.70604$&$0.0059<\sigma<4.24411$&$0.03973<\sigma<1.21028$\\
\hline
\end{tabular}
\end{center}
\end{application}


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\end{document}