\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 301, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/301\hfil A variational principle]
{A variational principle for boundary-value problems
 with non-linear boundary conditions}

\author[D. Yang \hfil EJDE-2015/301\hfilneg]
{Dianwu Yang}

\address{Dianwu Yang \newline
School of Mathematical Sciences, University of Jinan,
Jinan 250022, China}
\email{ss\_yangdw@ujn.edu.cn}

\thanks{Submitted August 15, 2015. Published December 7, 2015.}
\subjclass[2010]{34B15, 58E30}
\keywords{Boundary value problem; non-linear boundary value condition;
\hfill\break\indent variational principle}

\begin{abstract}
 In this article, we establish  a variational principle for a class of
 boundary-value problems with a suitable non-linear boundary conditions.
 As an application of the variational principle, we study the existence of
 classical solutions for boundary-value problems.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks

\section{Introduction}

By using the variational principle, boundary-value problems have been 
studied by numerous mathematicians
 (see \cite{Aver03, Bona10, Xie14, Tian11, Zhan12, Bona09, Bona11, Aver10, Sun12, 
Niet09, Tian10, Tian12, Zhan10} and references therein).
In  \cite{Aver03, Bona10}, the authors studied  equations with 
the boundary condition $u(0)=u(1)=0$. 
In  \cite{Tian11, Zhan12, Bona09}, the authors studied Sturm-Liouville 
boundary-value problems. 
In \cite{Xie14, Bona11}, the authors studied Neumann boundary-value problems.
In \cite{Aver10},  Han studied the periodic boundary-value problems.
In  \cite{Sun12, Niet09, Tian10, Tian12, Zhan10}, the authors applied
variational methods to impulsive differential equations. 
In all the references above, the boundary conditions are linear. 
In this article, we consider a boundary-value problem with non-linear boundary 
conditions:
\begin{equation} \label{eq11}
\begin{gathered}
x''=f(t,x),\quad  t\in[0,1], \\
H(x(0),x(1))=0,\\
\nabla H(x(0),x(1))J [(x'(0),-x'(1))-\nabla I(x(0),x(1))]^T=0.
\end{gathered}
\end{equation}
Here, $H$ and $I: R^2\to R$ are continuously differentiable, and
$$
J=\begin{pmatrix} 0& -1\\ 1 &0 \end{pmatrix}
$$
is the standard symplectic matrix. Also, we assume that the set 
$\mathcal{A}=\{(x,y):H(x,y)=0\}$ is nonempty.
If $H(x,y)=x^2+y^2$ and $I(x,y)=0$, problem \eqref{eq11} becomes
a Dirichlet boundary value problem. If  $H(x,y)=x-y$ and $I(x,y)=0$, 
problem \eqref{eq11} becomes a periodic boundary value problem. 
If $H(x,y)=x+y$ and $I(x,y)=0$, then problem \eqref{eq11} becomes a antiperiodic 
boundary value problem.

This article is organized as follows:
 in section 2, we construct a variational functional for \eqref{eq11}.
In section 3, we obtain sufficient conditions for  \eqref{eq11} to have a 
solution.

\section{Variational structure}

Let $W$ be the Sobolev space of functions $x: [0,1]\to R$ with a weak derivative
$x'\in L^2(0,1; R)$. The inner product on $W$ is 
\begin{equation} \label{eq21}
(x,y)=\int^1_0[x'(t)y'(t)+x(t)y(t)]dt
\end{equation}
and the corresponding norm is $\|\cdot\|$.
For each $x\in W$, there exists a real number $\xi\in(0,1)$ such that
$$
x(\xi)=\int_0^1x(t)dt.
$$
Then
\begin{equation} \label{eq22}
\begin{aligned}
|x(t)|&=|x(\xi)+\int_{\xi}^tx'(s)ds|\\
&\leq\Big(\int_0^1x^2(t)dt\Big)^{1/2}+\Big(\int_0^1(x'(t))^2dt\Big)^{1/2}
\leq \sqrt{2}\|x\|.
\end{aligned}
\end{equation}

To establish a variational principle for \eqref{eq11}, we assume that $f$ 
satisfies the  condition
\begin{itemize}
\item[(H1)] $f(t,x)$ is measurable in $t$ for each $x\in \mathbb{R}$, continuous 
in $x$ for almost every $t\in [0,1]$, and
there exists $h_k\in L^1(0,1)$ for any $k>0$ such that
$$
|f(t,x)|\leq h_k(t)
$$
for almost every $t\in[0,1]$ and all $|x|\leq k$.
\end{itemize}
Under this condition, we define the functional $\phi$ on $W$ by
\begin{equation} \label{eq23}
\phi(x)=\int_0^1[\frac{1}{2}(x'(t))^2+F(t,x(t))]dt+I(x(0),x(1))
\end{equation}
where $F(t,x)=\int_0^xf(t,u)du$. Then  $\phi$ is continuously differentiable, 
weakly lower semi-continuous and
\begin{equation} \label{eq24}
(\phi'(x),y)=\int_0^{1}[x'(t)y'(t)+f(t,x(t))y(t)]dt
+\nabla I(x(0),x(1)) (y(0),y(1))
\end{equation}
for all $y\in W$, see \cite{Mawh89}. Let $Y$ be a $C^1$-manifold defined by
$$
Y=\{x\in W: H(x(0),x(1))=0\}.
$$
Then, $Y$ is weakly closed since $W$ can be compactly imbedded in $C[0,1]$.
The following theorem is our main result.

\begin{theorem} \label{thm2.1} 
Assume that $f$ satisfies {\rm (H1)} and that the following
condition is satisfied,
\begin{itemize}
\item[(H2)] $\nabla H(x, y)\neq 0$ for each $(x, y)$ satisfying $H(x, y)=0$, 
or $\mathcal{A}$
is a discrete set.
\end{itemize}
If $x$ is a critical point of the functional $\phi$ defined by \eqref{eq23} 
on $Y$, then $x(t)$ is a solution of  \eqref{eq11}.
\end{theorem}

\begin{proof} 
For a given $u$ in $Y$, let $DY(u)$ denote the tangent space to $Y$ at $u$.
If $x$ is a critical point of the functional $\phi$ on $Y$, then for any 
$y\in DY(x)$ we have $(\phi'(x), y)=0$.
It follows from \eqref{eq24} that
\begin{equation} \label{eq25}
\int_0^{1}[x'(t)y'(t)+f(t,x(t))y(t)]dt+\nabla I(x(0),x(1))\cdot (y(0),y(1))=0.
\end{equation}
We define $\omega\in C(0,1; R)$ by
\begin{equation} \label{eq26}
\omega(t)=\int_0^tf(s, x(s))ds.
\end{equation}
By  Fubini's theorem and \eqref{eq25}, we obtain that for any $y\in DY(x)$,
\begin{equation} \label{eq27}
\begin{aligned}
&\int_0^1[x'(t)-\omega(t)]y'(t)dt\\
&=-\int_0^1f(t, x(t))y(t)dt-\nabla I(x(0),x(1))\cdot (y(0),y(1))\\
&\quad -\int_0^1y'(t)\int_0^tf(s, x(s))dsdt\\
&=-y(1)\int_0^1f(t, x(t))dt-\nabla I(x(0),x(1))\cdot (y(0),y(1)).
\end{aligned}
\end{equation}

We complete this proof by considering  two cases. 
When $\nabla H(x(0), x(1))\neq 0$, we have
\begin{equation} \label{eq28}
DY(x)=\{y\in W: \nabla H(x(0),x(1))\cdot (y(0),y(1))=0\}.
\end{equation}
In \eqref{eq27}, we can choose
\[
y(t)=\sin(2n\pi t),\quad n=1,2,\dots,
\]
and
\[
y(t)=1-\cos(2n\pi t),\quad n=1,2,\dots.
\]
It follows from \eqref{eq27} that
$$
\int_0^1[x'(t)-\omega(t)]\sin(2n\pi t)dt
=\int_0^1[x'(t)-\omega(t)]\cos(2n\pi t)dt=0, \quad n=1,2,\dots.
$$
A theorem for Fourier series implies that
\begin{equation} \label{eq29}
x'(t)-\omega(t)=x'(0)
\end{equation}
on $[0, 1]$. Thus, we have $x''(t)=f(t,x(t))$ and
\begin{equation} \label{eq210}
\int_0^1f(t, x(t))dt=x'(1)-x'(0).
\end{equation}
Integrating both sides of \eqref{eq29} over $[0, 1]$, we have
\begin{equation} \label{eq211}
x(1)-x(0)-\int_0^1(1-t)f(t, x(t))dt=x'(0).
\end{equation}
Set $y(t)=\nabla H(x(0),x(1))\cdot (t,t-1)$. It is easy to show that
$y\in DY(x)$ as $(y(0),y(1))=J\nabla H(x(0),x(1))$.
Inserting $y(t)$ into \eqref{eq25} we obtain
\begin{align*}
&\big[x(1)-x(0)-\int_0^1(1-t)f(t, x(t))dt\Big]\nabla H(x(0),x(1))\cdot (1,1)\\
&+\nabla I(x(0),x(1))J\nabla H(x(0),x(1))
+\int_0^1f(t, x(t))dt\nabla H(x(0),x(1))\cdot (1,0)=0.
\end{align*}
From \eqref{eq210} and \eqref{eq211}, the above equality implies
\[
\nabla H(x(0),x(1))J[(x'(0),-x'(1))-\nabla I(x(0),x(1))]^T=0.
\]

When the $\mathcal{A}$ is a discrete set, $(x(0), x(1))$ is a isolated point of
$\mathcal{A}$. Applying the implicit function  theorem we 
obtain $\nabla H(x(0),x(1))=0$, so that
$$
DY(x)=\{y\in W: y(0)=y(1)=0\}.
$$
It is easy to show that $x(t)$ is a solution of problem \eqref{eq11}. 
This completes the proof.
\end{proof}

\section{Solutions to boundary-value problems}

As an application of Theorem \ref{thm2.1}, we consider the existence of solutions 
for problem \eqref{eq11}.

\begin{theorem} \label{thm3.1} 
Assume that {\rm (H1), (H2)} hold, and
that the following conditions are satisfied:
\begin{itemize}
\item[(H3)] The set $\mathcal{A}$ is bounded.

\item[(H4)] There is a positive constant $l$ with $l<2$, and a positive function
$c\in L^1(0,1)$ such that
$$
F(t,x)\geq -c(t)(1+|x|^l)
$$
for almost every $t\in[0,1]$ and all $x\in \mathbb{R}$.
\end{itemize}
Then  \eqref{eq11} has at least one solution.
\end{theorem}

\begin{proof} 
Let $y$ be in $Y$. By (H3), there exists a positive number $M$ such that
$$
y^2(0)+y^2(1)\leq M^2.
$$
This implies 
\begin{equation} \label{eq31}
|y(t)|=|y(0)+\int_0^ty'(t)dt|\leq M+\int_0^1|y'(t)|dt
\leq M+(\int_0^1[y'(t)]^2dt)^{1/2}.
\end{equation}
Set
$$
M_1=\min_{x^2+y^2\leq M^2}I(x,y).
$$
Then,  from (H4), \eqref{eq23} and \eqref{eq31}, we have
\begin{align*}
\phi(y) 
&\geq\frac{1}{2}\int_0^1[y'(t)]^2dt-\int_0^{1}c(t)(1+|y(t)|^l)dt+M_1\\
&\geq\frac{1}{2}\int_0^1[y'(t)]^2dt+M_2(\int_0^1[y'(t)]^2dt)^{\frac{l}{2}}+M_3
\end{align*}
for some $M_2$ and $M_3$.  It follows that
$$
\lim_{\|y\|\to \infty}\phi(y)=+\infty,
$$
since $\|y\|\to \infty$ if and only if $\int_0^1[y'(t)]^2dt\to \infty$.
Hence, $\phi|Y$ is bounded from blow. Therefore, there exists a critical 
point of $\phi$ on $Y$.
By Theorem \ref{thm2.1}, problem \eqref{eq11} has at least one solution.
\end{proof}

\begin{theorem} \label{thm3.2} 
Assume that {\rm (H1)--(H3)} hold, and
that the following conditions are satisfied:
\begin{itemize}
\item[(H5)] There is a positive function $c\in L^1(0,1)$ such that
$$
F(t,x)\geq -c(t)(1+x^2)
$$
for almost every $t\in[0,1]$ and all $x\in \mathbb{R}$.

\item[(H6)] $2\int_0^1c(t)dt<1$.
\end{itemize}
Then  \eqref{eq11} has at least one solution.
\end{theorem}

\begin{proof} 
For each $y\in Y$,  from (H5), \eqref{eq23} and \eqref{eq31}, we obtain
\begin{align*}
\phi(y)
&\geq\frac{1}{2}\int_0^1[y'(t)]^2dt-\int_0^{1}c(t)(1+y^2(t))dt+M_1\\
&\geq(\frac{1}{2}-\int_0^{1}c(t)dt)\int_0^1[y'(t)]^2dt
 +M_4(\int_0^1[y'(t)]^2dt)^{1/2}+M_5
\end{align*}
for some $M_4$ and $M_5$. Assumption (H6) implies 
$$
\lim_{\|y\|\to \infty}\phi(y)=+\infty.
$$
Therefore, problem \eqref{eq11} has at least one solution.
\end{proof}

\begin{theorem} \label{thm3.3} 
Assume that {\rm  (H1), (H2)} hold, and
that the following conditions are satisfied:
\begin{itemize}
\item[(H7)] There is a positive function $c\in L^1(0,1)$ and positive 
constants $k_1$, $l$ with $l<2$ such that
$$
F(t,x)\geq k_1x^2-c(t)(1+|x|^l)
$$
for almost every $t\in[0,1]$ and all $x\in \mathbb{R}$.

\item[(H8)] There are positive constants $k_2$ and $k_3$ such that
 $I(x, y)\geq-k_2x^2-k_3y^2$.

\item[(H9)] $4(k_2+k_3)<\min\{1, 2k_1\}$.
\end{itemize}
Then \eqref{eq11} has at least one solution.
\end{theorem}

\begin{proof}
Assumptions (H7) and (H8), and \eqref{eq23} imply 
\begin{align*}
\phi(y)&\geq\frac{1}{2}\int_0^1[y'(t)]^2dt+k_1\int_0^1y^2(t)dt-
\int_0^{1}c(t)(1+|y(t)|^l)dt-k_2y^2(0)-k_3y^2(1)\\
&\geq(\frac{1}{2}\min\{1, 2k_1\}-2k_2-2k_3)\|y\|^2-\int_0^{1}c(t)(1+|y(t)|^l)dt).
\end{align*}
for each $y\in Y$. From (H9) we obtain
$$
\lim_{\|y\|\to \infty}\phi(y)=+\infty.
$$
Therefore \eqref{eq11} has at least one solution.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by the Natural Science Foundation of Shandong
Province of PR China (ZR2011AL007).


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\end{document}