\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 231, pp. 1--9.\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-2014/231\hfil Mixed type boundary-value problems] {Mixed type boundary-value problems of second-order differential systems with p-Laplacian} \author[ W. Ge, Y. Tian \hfil EJDE-2014/231\hfilneg] {Weigao Ge, Yu Tian} % in alphabetical order \address{Weigao Ge \newline Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081, China} \email{gew@bit.edu.cn, Phone 86-010-68911627} \address{Yu Tian \newline School of Science, Beijing University of Posts and Telecommunications, Beijing 100876, China} \email{tianyu2992@163.com} \thanks{Submitted January 6, 2014. Published October 29, 2014.} \subjclass[2000]{34B15, 35A15} \keywords{Mixed boundary value problem; p-Laplacian; duality principle} \begin{abstract} In this article we show the existence of solutions to a mixed boundary-value problem of second-order differential systems with a p-Laplacian. The associated Hamiltonian actions are indefinite and the discussion of the existence of solutions is due to the application of duality principle. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \allowdisplaybreaks \section{Introduction} Second-order differential systems that include the $p$-Laplacian appear in physical application; see for example \cite{m1}. In this article, we study mixed type boundary-value problems of the form \begin{equation}\label{1} \begin{gathered} (\varphi_p(x'))'+\nabla F(t, x)=0,\quad p\ge 2,\\ x(0)=x'(1)=0, \end{gathered} \end{equation} where $x\in\mathbb{R}^n$, $\varphi_p(x)=(\varphi_p(x_1), \dots, \varphi_p(x_n))^T$ with $\varphi_p(s)=|s|^{p-2}s$ for $s\in \mathbb{R}$, $F:[0, 1]\times \mathbb{R}^n\to \mathbb{R}$ is measurable in $t$ for all $x\in \mathbb{R}^n$ and continuously differentiable in $x$ for a.e. $t\in[0, 1]$. Also we study systems of the form \begin{equation}\label{1.1} \begin{gathered} (\psi_p(x'))'+\nabla F(t, x)=0,\,\,p\ge 2,\\ x(0)=x'(1)=0, \end{gathered} \end{equation} where $\psi_p(x)=|x|^{p-2}x$. When $\lim_{|x|\to\infty}F(t, x)=-\infty$, it is easy to obtain a solution of \eqref{1}, by using a minimizing sequence of this functional \[ \Phi(x)=\int_{0}^{1}[\Phi_p(x'(t))-F(t, x(t))]dt, \] where $\Phi_p(x'(t))=\sum_{i=1}^{n}\frac{1}{p}|x_{i}'|^p$. However, such an approach is not applicable if $\lim_{|x|\to \infty}F(t, x)=+\infty$, since $\Phi(x)$ does not admit maximum, and does not admit minimum. In such a case, for $\alpha>0$, we set $u=(u_1,u_2)=(x, -\varphi_p(\alpha x'))$ for \eqref{1}, and $u=(u_1, u_2)=(x,-\psi_p(\alpha x'))$ for \eqref{1.1}, $\alpha>0$. Then \eqref{1} becomes \[ \begin{gathered} -u_2'+\varphi_p(\alpha)\nabla F(t, u_1)=0,\\ u_1'+\frac{1}{\alpha}\varphi_q(u_2)=0,\\ u_1(0)=u_2(1)=0, \end{gathered} \] and \eqref{1.1} becomes \begin{gather*} -u_2'+\psi_p(\alpha)\nabla F(t, u_1)=0,\\ u_1'+\frac{1}{\alpha}\psi_q(u_2)=0,\\ u_1(0)=u_2(1)=0. \end{gather*} So \eqref{1} and \eqref{1.1} become \begin{equation}\label{2} \begin{gathered} J\dot{u}+\nabla G(t, u)=0,\\ u_1(0)=u_2(1)=0, \end{gathered} \end{equation} and \begin{equation}\label{2.2} \begin{gathered} J\dot{u}+\nabla H(t, u)=0,\\ u_1(0)=u_2(1)=0, \end{gathered} \end{equation} respectively, where \begin{gather*} G(t, u)=\Phi_q(u_2)+\varphi_p(\alpha)F(t,u_1) =\sum_{i=1}^{n}\frac{1}{q\alpha}|u_{2,i}|^q+\varphi_p(\alpha)F(t,u_1),\\ H(t, u)=\widetilde{\Phi}_q(u_2)+\varphi_p(\alpha)F(t,u_1) =\frac{1}{q\alpha}|u_2|^q+\varphi_p(\alpha)F(t, u_1), \end{gather*} with $u_1=(u_{1, 1}, \dots, u_{1, n}), u_2=(u_{2, 1}, \dots, u_{2, n})$, $q=\frac{p}{p-1}$, \[ J=\begin{pmatrix}0&-I_n\\ I_n&0\end{pmatrix} \] where $I_n$ is the $n\times n$ identity matrix. Then $G: [0, 1]\times \mathbb{R}^{2n}\to \mathbb{R}$ is measurable in $t$ for all $u\in \mathbb{R}^{2n}$ and continually differentiable in $u$ for a.e. $t\in[0, 1]$. Furthermore, if $F$ is strictly convex in $u_1$, then $G$ and $H$ are strictly convex in $u$. When $n=1$, different types of BVPs have been studied there is a series of results \cite{a1,a2,e1,g1}, whereas there are only a few results for the case $n\ge 2$, except periodic boundary value problems in \cite{t1,t2}. Let $X=\{u\in C([0, 1], \mathbb{R}^{2n}): u_1(0)=u_2(1)=0\}$. For $u\in X$ we construct functionals in the forms \begin{gather}\label{2.11} \Psi(u)=\int_{0}^{1}[\frac{1}{2}(J\dot{u}, u)+G(t,u)]dt, \\ \label{2.111} \mathcal{K}(u)=\int_{0}^{1} [\frac{1}{2}(J\dot{u}, u)+H(t,u)]dt. \end{gather} The Euler equations $\Psi(u)$ and $\mathcal{K}(u)$ are the differential systems in \eqref{2} and \eqref{2.2}, respectively. The boundary conditions in \eqref{2} and \eqref{2.2} are given by the definition of $X$. Let $u_k(t)=(u_{k, 1}(t), u_{k, 2}(t))=(\cos \lambda_k t\cdot c, \sin \lambda_k t\cdot c)$ with $c=(c_1, \dots, c_n)$, $\lambda_k=\frac{(2k+1)\pi}{2}$ and $|c|=1$. Then \begin{align*} \frac{1}{2}\int_{0}^{1}(J\dot{u}_k(t), u_k(t))dt &=\frac{\lambda_k}{2}\int_{0}^{1}[-\cos^2\lambda_k t|c|^2-\sin^2\lambda_k t|c|^2]dt\\ &=-\frac{1}{2}\lambda_k=-\frac{(2k+1)\pi}{4} \to\mp\infty \end{align*} as $k\to \pm \infty$. So $\Psi(u)$ and $\mathcal{K}(u)$ are neither bounded from below nor from above. Since $G(t, u)$ is continually differentiable in $u$ and strictly convex with respect to $u$, we can make Fenchel transform \begin{equation}\label{4} G^*(t, \dot{v})=\sup_{u\in \mathbb{R}^{2n}}[(\dot{v}, u)-G(t, u)]. \end{equation} By the transform theory, there is only one $u_{v}$ for $v$ such that \[ (\dot{v}, u_{v})-G(t, u_{v})=\sup_{u\in \mathbb{R}^{2n}}[(\dot{v}, u)-G(t, u)]. \] Therefore $\dot{v}=\nabla G(t, u_{v}), u_{v}=\nabla G^*(t, \dot{v})$ and \[G(t, u_{v})+G^*(t, \dot{v})=(\dot{v}, u_{v}).\] Let $u=u_{v}$, we have the relations \[ G(t, u)+G^*(t, \dot{v})=(\dot{v}, u)\]\[\dot{v}=\nabla G(t, u),\quad u=\nabla G^*(t, \dot{v}) \] and among them any one implies the others. The same is true for the relations \begin{gather*} H(t, u)+H^*(t, \dot{v})=(\dot{v}, u)\\ \dot{v}=\nabla H(t, u),\quad u=\nabla H^*(t, \dot{v}) \end{gather*} where $H^*(t, \dot{v})=\sup_{u\in \mathbb{R}^{2n}}[(\dot{v}, u)-H(t,u)]$. With the duality we aim to prove the following theorem. \begin{theorem}\label{thm1} Suppose $F(t, x)$ is measurable in $t$ for all $x\in \mathbb{R}^n$, strictly convex and lower semicontinuous (l.s.c.) in $x$ for a.e. $t\in[0, 1]$ and there are $a\in C(\mathbb{R}^n, \mathbb{R}^+), b\in L^2([0, 1], \mathbb{R}^+)$ such that \[ |\nabla F(t, x)|\le b(t)a(|x|) \] and there are $\delta>0$, $(\delta\in(0, \frac{\pi^2}{4})$ if $p=2$), and $\beta, \gamma\ge 0$ such that \[ -\beta\le F(t, x)\le \frac{\delta}{2} |x|^2+\gamma. \] Then \eqref{1} has at least one solution. \end{theorem} \begin{theorem}\label{thm2} Under the assumptions of Theorem \ref{thm1}, system \eqref{1.1} has at least one solution. \end{theorem} \section{Preliminaries} To prove our main theorems, we use the following propositions. \begin{proposition}\label{prop1} Assume $F: [0, 1]\times \mathbb{R}^n\to \mathbb{R}$ and $F(t, x)$ is strictly convex in $x$ for all $t\in [0, 1]$ and there are $\alpha>0$, $\beta(t), \gamma(t)\ge0$ such that \[ -\beta(t)\le F(t, x)\le \frac{\alpha}{2}|x|^2+\gamma(t). \] Then for $v=\nabla F(t, x)$ it holds that \[ |v|\le 2\alpha (|x|+\beta(t)+\gamma(t))+1,\quad \forall t\in[0, 1]. \] \end{proposition} \begin{proof} On the one hand, by $v=\nabla F(t, x)\Leftrightarrow F^*(t, v)=(v, x)-F(t, x)$, we have \begin{equation}\label{1aob} F^*(t, v)\le (v, x)+\beta(t),\quad \forall t\in[0, 1]. \end{equation} On the other hand, \begin{equation}\label{2aob} \begin{aligned} F^*(t,v)&=\sup_{x\in \mathbb{R}^N}[(v, x)-F(t, x)]\\ &\ge \sup_{x\in \mathbb{R}^N}[(v,x)-\frac{\alpha}{2}|x|^2-\gamma(t)] =\frac{1}{2\alpha}|v|^2-\gamma(t),\quad \forall t\in[0, 1]. \end{aligned} \end{equation} By \eqref{1aob} and \eqref{2aob}, \begin{equation}\label{aob} |v|^2\le 2\alpha[(v, x)+\beta(t)+\gamma(t)],\,\,\,\forall t\in[0,1]. \end{equation} If $|v|\le 1$, the result is obvious. If $|v|>1$, by \eqref{aob}, $|v|^2\le 2\alpha [|v||x|+\beta(t)|v|+\gamma(t)|v|]$. The result also follows. \end{proof} \begin{proposition}\label{prop2} If $u\in X=\{x\in H^1([0, 1], \mathbb{R}^{2n}): x_1(0)=x_2(1)=0\}$, then \begin{equation}\label{2.3} |u|_2^{2}\le \frac{4}{\pi^2}|\dot{u}|_2^{2}. \end{equation} \end{proposition} \begin{proof} Let $u=(u_1, u_2)$, $u_1, u_2\in \mathbb{R}^n$. From \begin{equation}\label{3.4} \begin{gathered} \dot{u}(t)=\lambda J u(t),\\ u_1(0)=u_2(1)=0, \end{gathered} \end{equation} and the expression $e^{\lambda Jt}=\cos(\lambda t) I+\sin(\lambda t)J$, we have the set of eigenvalues $\lambda_k$ of \eqref{3.4} \[ \lambda_k=\frac{(2k+1)\pi}{2},\quad k=0, \pm1,\pm2,\dots. \] Then for each $\lambda_k$, $k=0, 1, 2, 3,\dots$, \eqref{3.4} possesses $2n$-dimensional vector space \[ u_k(t)=\begin{pmatrix} \sin(\lambda_k t) C_{1,k}\\ \cos(\lambda_k t) C_{2,k} \end{pmatrix}, \] where $C_{1,k}, C_{2,k}\in \mathbb{R}^n$ are arbitrary vectors. Then $u\in X$ can be expressed as \[ u(t)=\begin{pmatrix}\sum_{k=0}^{\infty}\sin (\lambda_k t) C_{1, k}\\ \sum_{k=0}^{\infty}\cos(\lambda_k t) C_{2, k}\end{pmatrix},\quad C_{1, k}, C_{2, k}\in \mathbb{R}^n. \] Then \begin{gather*} \begin{aligned} |u|_2^{2} &=\sum_{k=0}^{\infty}\Big[\int_{0}^{1}\sin^{2}\lambda_k tdt\cdot|C_{1, k}|^2+\int_{0}^{1}\cos^2\lambda_kt dt\cdot|C_{2,k}|^2\Big]\\ &= \frac{1}{2}\sum_{k=0}^{\infty}[|C_{1, k}|^{2}+|C_{2,k}|^{2}], \end{aligned}\\ \begin{aligned} |\dot{u}|_2^{2}=\frac{1}{2}\sum_{k=0}^{\infty}\lambda_{k}^{2}(|C_{1, k}|^{2}+|C_{2, k}|^{2})\ge \frac{1}{2}\sum_{k=0}^{\infty}\frac{\pi^2}{4}(|C_{1, k}|^{2}+|C_{2, k}|^{2}), \end{aligned} \end{gather*} and hence \eqref{2.3} holds. \end{proof} \begin{proposition}\label{prop3} If $u\in X$, then \[ \int_{0}^{1}(J\dot{u}, u)dt \ge -\frac{2}{\pi}|\dot{u}|_2^{2}. \] \end{proposition} \begin{proof} The result follows directly from the calculation \begin{align*} \int_{0}^{1}(J\dot{u}, u)dt &\ge -\int_{0}^{1}|J\dot{u}|\cdot|u|dt\\ &\ge -\Big[\int_{0}^{1}|J\dot{u}|^2dt\cdot \int_{0}^{1}|u|^2dt\Big]^{1/2}\\ &= -\Big[\int_{0}^{1}|\dot{u}|^2dt\cdot \frac{4}{\pi^2}\int_{0}^{1}|\dot{u}|^2dt\Big]^{1/2}\\ &= -\frac{2}{\pi}|\dot{u}|_2^{2}. \end{align*} \end{proof} \begin{proposition}\label{prop4} Under the conditions in Theorem \ref{thm1}, we can choose a suitable $\alpha>0$ so that after the transform $u=(u_1, u_2)=(x, -\varphi_p(\alpha \dot{x}))$, the function $G(t, u)$ in BVP \eqref{2} satisfies \begin{equation}\label{5} -\xi\le G(t, u)\le \frac{l}{2}|u|^2+\eta, \end{equation} where $\xi, \eta\ge0, l\in (0,\frac{\pi}{2})$ are appropriate real numbers. \end{proposition} \begin{proof} If $p=2$, then $\delta\in (0,\pi^2/4)$. Choose $\alpha=1/\sqrt{\delta}$. One get $G(t,u)=\frac{\sqrt{\delta}}{2}|u_2|^2+\frac{1}{\sqrt{\delta}}F(t, u_1)$ and \[ -\frac{\beta}{\sqrt{\delta}}\le G(t, u)\le \frac{\sqrt{\delta}}{2}(|u_1|^{2}+|u_2|^{2})+\frac{\gamma}{\sqrt{\delta}}. \] Let $\xi=\beta/\sqrt{\delta}$, $\eta=\gamma/\sqrt{\delta}$, $l=\sqrt{\delta}$. Obviously $\xi, \eta>0$, $l\in (0, \frac{\pi}{2})$. If $p>2$, then $q\in(1, 2)$. Without loss of generality, assume that $\delta>\pi^2/4$. Let $\alpha=(\pi/4\delta)^{q-1}$, then \begin{align*} -\varphi_p(\alpha)\beta\le G(t,u) &\le \frac{n}{\alpha q}|u_2|^q+\varphi_p(\alpha)F(t, u_1)\\ &\le \frac{n}{\alpha q}|u_2|^q+\frac{\delta \varphi_p(\alpha)}{2}|u_1|^2+\varphi_p(\alpha)\gamma\\ &= \frac{n}{\alpha q}|u_2|^q+\frac{\pi}{8}|u_1|^2+\varphi_{p}(\alpha)\gamma. \end{align*} It follows from $q\in (1, 2)$ that there is $M>0$ such that \[ \frac{n}{\alpha q}|u_2|^q\le M+\frac{\pi}{8}|u_2|^2. \] Let $\xi=\varphi_p(\alpha) \beta, \eta=M+\varphi_p(\alpha)\gamma, l=\frac{\pi}{4}$. Then it holds \[ -\xi\le G(t, u)\le \frac{l}{2}|u|^2+\eta. \] \end{proof} For the rest of this article, we assume $G(t, u)$ satisfies \eqref{5}. Similarly we can prove he following result. \begin{proposition}\label{prop5} Under the conditions in Theorem \ref{thm2}, there is an $\alpha>0$ such that after the transform $u=(u_1, u_2)=(x, -\psi_p(\alpha \dot{x}))$, the function $H$ in \eqref{2.2} satisfies \begin{equation}\label{2.4} -\xi\le H(t, u)\le \frac{l}{2}|u|^2+\eta, \end{equation} where $\xi, \eta\ge0$, $l\in (0, \frac{\pi}{2})$ are some real numbers. \end{proposition} In the Clarke transform $G^*(t, \dot{v})=\sup_{u\in \mathbb{R}^{2n}}[(\dot{v}, u)-G(t, u)]$, $G^*(t, u)$ is convex in $u$. On the other hand, if \begin{gather*} G_{\varepsilon}(t, u)=\frac{\varepsilon}{2}(u,u)+G(t, u),\\ G_{\varepsilon}^{*}(t, \dot{v}) =\sup_{u\in \mathbb{R}^{2n}}[(\dot{v}, u)-G_{\varepsilon}(t, u)], \end{gather*} then $G_{\varepsilon}(t, u)$ is strictly convex in $u$ and satisfies $\lim_{|u|\to\infty}\frac{G_{\varepsilon}(t,u)}{|u|}=\infty$. Hence $G_{\varepsilon}^{*}(t, \dot{v})$ is differentiable in $\dot{v}$; i.e., $\nabla G_{\varepsilon}^{*}(t, y)$ is continuous in $y$. This time we have \begin{equation}\label{6} -\xi+\frac{\varepsilon}{2}|u|^2\le G_{\varepsilon}(t, u)\le \frac{l+\varepsilon}{2}|u|^2+\eta \end{equation} and \[ G_{\varepsilon}^{*}(t, \dot{v}) =\sup_{u\in \mathbb{R}^{2n}}[(\dot{v}, u)-G_{\varepsilon}(t, u)] \ge\sup_{u\in \mathbb{R}^{2n}}[(\dot{v},u) -\frac{l+\varepsilon}{2}|u|^2-\eta]=\frac{1}{2(l+\varepsilon)}|\dot{v}|^2-\eta. \] \begin{equation}\label{6'} G_{\varepsilon}^{*}(t, \dot{v})\le \frac{1}{2\varepsilon}|\dot{v}|^2+\xi. \end{equation} Let $\dot{v}\in \partial G_{\varepsilon}(t, u)$. One has \[ G_{\varepsilon}^{*}(t, \dot{v}) =(\dot{v}, u)-G_{\varepsilon}(t, u)\le (\dot{v}, u) -\frac{\varepsilon}{2}|u|^2+\xi \] and \[ \frac{1}{2(l+\varepsilon)}|\dot{v}|^2-\eta\le |\dot{v}||u|+\xi, \] which implies \begin{equation}\label{7} |\dot{v}|\le 1+2(l+\varepsilon)(|u|+\xi+\eta). \end{equation} Similarly for $u\in \partial G_{\varepsilon}^{*}(t, \dot{v})$, we have \begin{equation}\label{8} |u|\le 1+\frac{2}{\varepsilon}(|\dot{v}|+\xi+\eta). \end{equation} Let $\varepsilon>0$ be such that $l+\varepsilon\in (0,\frac{\pi}{2})$. Take in account the boundary-value problem \begin{equation}\label{9} \begin{gathered} J\dot{u}+\nabla G_{\varepsilon}(t, u)=0\\ u_1(0)=u_2(1)=0, \end{gathered} \end{equation} whose functional is \[ \Psi_{\varepsilon}(u)=\int_{0}^{1}[\frac{1}{2}(J\dot{u}, u)+G_{\varepsilon}(t, u)]dt. \] Let $v=-Ju$. Then\begin{align*} \Psi_{\varepsilon}(u) &=-\frac{1}{2}\int_{0}^{1}(J\dot{u}, u)dt+\int_{0}^{1}[(J\dot{u}, u)+G_{\varepsilon}(t,u)]dt\\ &=-\frac{1}{2}\int_{0}^{1}(J\dot{u}, u)dt-\int_{0}^{1}[(\dot{v}, u)-G_{\varepsilon}(t, u)]dt\\ &=-\int_{0}^{1}[\frac{1}{2}(J\dot{v}, v)+G_{\varepsilon}^{*}(t,\dot{v})]dt =:-\mathcal{K}_{\varepsilon}(v). \end{align*} \begin{proposition}\label{prop6} Under the conditions in Theorem \ref{thm1}, $\mathcal{K}_{\varepsilon}$ has one critical point $v_{\varepsilon}\in Y=\{y\in H^1([0, 1], \mathbb{R}^{2n}): y_1(1)=0, y_2(0)=0\}$, which minimize the value of $\mathcal{K}_{\varepsilon}$ and is uniformly bounded below for all $\varepsilon\in(0, \frac{\pi}{2}-l)$. Furthermore $u_{\varepsilon}=Jv_{\varepsilon}$ is a solution of BVP \eqref{9}. \end{proposition} \begin{proof} It follows from \[ G_{\varepsilon}(t, u)\le \frac{l+\varepsilon}{2}|u|^2+\eta =:G(u) \] that \[ G_{\varepsilon}^{*}(t, v)\ge G^*(\dot{v})=\sup_{u\in \mathbb{R}^{2n}} [(\dot{v}, u)-G(u)]=\frac{1}{2(l+\varepsilon)}|\dot{v}|^2-\eta \] and then \[ \mathcal{K}_{\varepsilon}(v)\ge\frac{1}{2} \big(\frac{1}{l+\varepsilon}-\frac{2}{\pi}\big) \int_{0}^{1}|\dot{v}(t)|^2dt-\int_{0}^{1}\eta(t)dt\ge\alpha_0 \|\dot{v}\|_2^{2}-\eta_0, \] where $\eta_0=\int_{0}^{1}\eta(t)dt$, $\alpha_0=\frac{1}{2}\big(\frac{1}{l+\varepsilon}-\frac{2}{\pi}\big)>0$. Obviously $\mathcal{K}_{\varepsilon}(v)\to+\infty$ as $\|\dot{v}\|_2\to\infty$ and uniformly bounded below. Let \[ \mathcal {K}_{\varepsilon1}(v)=\frac{1}{2}\int_{0}^{1}(J\dot{v}, v)dt,\quad \mathcal{K}_{\varepsilon2}(v)=\int_{0}^{1}G_{\varepsilon}^{*}(t, \dot{v})dt. \] Both $\mathcal{K}_{\varepsilon 1}$ and $\mathcal{K}_{\varepsilon2}$ are weakly lower semi-continuous (w.l.s.c.) imply $\mathcal{K}_{\varepsilon}$ is w.l.s.c. and then $\mathcal{K}_{\varepsilon}$ possesses one minimum at some point $v_{\varepsilon}\in Y$. At the same time, by $L(t, x, y)=\frac{1}{2}(Jy, x)+G_{\varepsilon}^{*}(t, y)$, we have from \eqref{6} \eqref{6'} and \eqref{8} that \begin{gather*} |L(t, x, y)|\le \frac{1}{2}|x||y|+\frac{1}{2\varepsilon}|y|^2+\xi,\\ |\nabla_x L(t, x, y)|=\frac{1}{2}|y|,\\ |\nabla_yL(t, x, y)|\le \frac{1}{2}|x|+|\nabla_y G^*(t, \dot{y})| \le \frac{1}{2}|x|+1+\frac{2}{\varepsilon}(|\dot{y}|+\xi+\eta), \end{gather*} and then $\mathcal{K}_{\varepsilon}$ is continuously differentiable on $Y$. As for all $w\in Y$, \begin{align*} \langle \mathcal{K}_{\varepsilon}'(v), w\rangle &=\int_{0}^{1}\left[\frac{1}{2}(J\dot{v}_{\varepsilon}, w)-\frac{1}{2}(Jv_{\varepsilon}, \dot{w})+(\nabla G_{\varepsilon}^{*}(t, \dot{v}_{\varepsilon}), \dot{w})\right]dt\\ &=\int_{0}^{1}(-Jv_{\varepsilon}+\nabla G_{\varepsilon}^{*}(t, \dot{v}_{\varepsilon}), \dot{w})dt =0. \end{align*} One gets $Jv_{\varepsilon}=\nabla G_{\varepsilon}^{*}(t, \dot{v}_{\varepsilon})$, i.e., $u_{\varepsilon}=\nabla G_{\varepsilon}^{*}(t, \dot{v}_{\varepsilon})$. From the duality principle, it holds \[ \dot{v}_{\varepsilon}=\nabla G_{\varepsilon}(t, u) \] and hence \[ -J\dot{u}_{\varepsilon}=\nabla G_{\varepsilon}(t, u); \] i.e., \[ J\dot{u}_{\varepsilon}+\nabla G_{\varepsilon}(t, u)=0. \] Clearly $v_{\varepsilon}\in Y$ implies $u_{\varepsilon}\in X$. \end{proof} Let $\varepsilon\in (0, \frac{\pi}{2}-l)$ and $H_{\varepsilon}(t, u)=H(t, u)+\frac{\varepsilon}{2}|u|^2$. Consider the system \begin{equation}\label{2.9} \begin{gathered} J\dot{u}+\nabla H_{\varepsilon}(t, u)=0,\\ u_1(0)=u_2(0)=0. \end{gathered} \end{equation} From $v=-Ju$ one has \begin{align*} \mathcal{K}_{\varepsilon}(u) &=\int_{0}^{1}[\frac{1}{2}(J\dot{u}, u)+H_{\varepsilon}(t, u)]dt \\ &=-\int_{0}^{1}[\frac{1}{2}(J\dot{v}, v)+H_{\varepsilon}^{*}(t, \dot{v})]dt =:-\Pi_{\varepsilon}(v), \end{align*} where $H_{\varepsilon}^{*}(t, \dot{v})=\sup_{u\in \mathbb{R}^{2n}}[(\dot{v}, u)-H_{\varepsilon}(t, u)]$. The following proposition can be proved in a similar way as Proposition \ref{prop5}. \begin{proposition} \label{prop7} Under the conditions given in Theorem \ref{thm2}, $\Pi_{\varepsilon}$ has one critical point $v_{\varepsilon}\in Y=\{y\in H^1([0, 1], \mathbb{R}^{2n}): y_1(1)=0, y_2(0)=0\}$, which minimize the value of $\mathcal{K}_{\varepsilon}$ and is uniformly bounded below for all $\varepsilon\in(0, \frac{\pi}{2}-l)$. Furthermore $u_{\varepsilon}=Jv_{\varepsilon}$ is a solution of \eqref{2.9}. \end{proposition} \section{Proof of main theorems} \begin{proof}[Proof of Theorem \ref{thm1}] In Proposition \ref{prop5} we have proven that for each $\varepsilon\in (0, \frac{\pi}{2}-l)$, BVP \eqref{9} has a solution $u_{\varepsilon}=Jv_{\varepsilon}$, and $\mathcal{K}_{\varepsilon}(v_{\varepsilon})$ is the minimum of $\mathcal{K}_{\varepsilon}$ on $Y$ with $\mathcal{K}_{\varepsilon}(v_{\varepsilon})\ge-\eta_0+\alpha_0\|\dot{v}_{\varepsilon}\|_2^{2}$. Furthermore, \[ G(t, u)\le G_{\varepsilon}(t, u) \] implies \[ G_{\varepsilon}^{*}(t, \dot{v})\le G^*(t, \dot{v}). \] So \[ \alpha_0\|\dot{v}_{\varepsilon}\|_2^{2}-\eta_0\le \mathcal{K}_{\varepsilon}(v_{\varepsilon})\le \mathcal{K}_{\varepsilon}(0)=\int_{0}^{1}G^*(t, 0)dt=c<\infty \] and then there is a $c_1>0$ such that \[ \|\dot{v}_{\varepsilon}\|_2^{2}0$ such that $\|u_{\varepsilon}\|_20$ such that \[ \|u_{\varepsilon}\|_X