\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 74, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/74\hfil Initial-value problems] {Initial-value problems for first-order differential systems with general nonlocal conditions} \author[O. Nica\hfil EJDE-2012/74\hfilneg] {Octavia Nica} \address{Octavia Nica \newline Department of Mathematics, Babe\c s-Bolyai University\\ 400084 Cluj, Romania} \email{octavia.nica@math.ubbcluj.ro} \thanks{Submitted February 17, 2012. Published May 14, 2012.} \subjclass[2000]{34A34, 34A12, 45G10} \keywords{Nonlinear differential system; nonlocal initial condition; \hfill\break\indent fixed point; vector norm; matrix convergent to zero} \begin{abstract} This article concerns the existence of solutions to initial-value problems for nonlinear first-order differential systems with nonlocal conditions of functional type. The fixed point principles by Perov, Schauder and Leray-Schauder are applied to a nonlinear integral operator split into two operators, one of Fredholm type and the other of Volterra type. The novelty in this article is combining this approach with the technique that uses convergent to zero matrices and vector norms. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In this article, we study the nonlocal initial-value problem for the first-order differential system $$\begin{gathered} x'(t)=f_1(t,x(t),y(t))\\ y'(t)=f_2(t,x(t),y(t))\quad\text{a.e. on }[0,1]\\ x(0)=\alpha [x],\quad y(0)=\beta [y]. \end{gathered} \label{1}$$ Here, $f_1,f_2:[0,1]\times \mathbb{R}^2\to \mathbb{R}$ are Carath\'{e}odory functions, $\alpha ,\beta :C[0,1]\to \mathbb{R}$ are linear and continuous functionals. Nonlocal problems have been extensively discussed in the literature by different methods; see Boucherif \cite{ref4}, Boucherif-Precup \cite{ref6}, Byszewski \cite{ref7}, Byszewski-Lakshmikantham \cite{ref8}, Nica-Precup \cite{prec-nica}, Ntouyas-Tsamatos \cite{nt}, Precup \cite{ref11}, Webb-Lan \cite{webb-lan}, Webb \cite{webb}, Webb-Infante \cite{webb-infante, webb-infante1,webb-infante2} and references therein. In the recent paper, \cite{nica-func-w}, Problem \eqref{1} was studied using as main tools the fixed point principles by Perov, Schauder and Leray-Schauder, together with the technique that uses convergent to zero matrices and vector norms. Note that the $m$-point boundary condition $x(0)+\sum_{k=1}^m a_kx(t_k)=0$ is a particular case of condition $x(0)=\alpha [x]$ when $$\alpha [x]=-\sum_{k=1}^m a_kx(t_k). \label{intro_1}$$ In \cite{ref6}, the authors studied the nonlocal initial-value problem for first-order differential equations \begin{gather*} x'(t)=f(t,x(t))\quad (\text{a.e. on }[0,1]) \\ x(0)+\sum_{k=1}^m a_kx(t_k)=0, \end{gather*} assuming that $f:[0,1]\times \mathbb{R}^2\to \mathbb{R}$ is a Carath\'{e}odory function, $t_k$ are given points with $0\leq t_1\leq t_2\leq \dots \leq t_{m}<1$ and $\ a_k$, $\widetilde{a}_k$ are real numbers with $1+\sum_{k=1}^m a_k\neq 0$ and $1+\sum_{k=1}^m \widetilde{a}_k\neq 0$. The main idea there was to rewrite the problem as a fixed point problem, involving a sum of two operators, one of Fredholm type whose values depend only on the restrictions of functions to $[0,t_{m}]$, and the other one, a Volterra type operator depending on the restrictions to $[t_{m},1]$. The same strategy was adapted in \cite{prec-nica} for the first-order differential system \begin{gather*} x'(t)=f(t,x(t),y(t)) \\ y'(t)=g(t,x(t),y(t))\quad (\text{a.e. on }[0,1])\\ x(0)+\sum_{k=1}^m a_kx(t_k)=0, \quad y(0)+\sum_{k=1}^m \widetilde{a}_ky(t_k)=0. \end{gather*} In this article, the nonlocal conditions are expressed by means of linear continuous functionals on $C[0,1]$, as in the works by Webb-Lan \cite{webb-lan}, Webb \cite{webb}, Webb-Infante \cite{webb-infante}, \cite{webb-infante1}, \cite{webb-infante2}. Our main assumption on functionals $\alpha ,\beta$ extends to the general case the specific property of the particular functional \eqref{intro_1} of depending only on the points from a proper subinterval $[0,t_0]$ of $[0,1]$, namely $[0,t_{m}]$ (taking $t_0:=t_{m}$). More exactly, we require the following property: $$x|_{[0,t_0]}=y|_{[0,t_0]}\text{ implies }\alpha [x-y]=0, \text{ whenever }x,y\in C[0,1]. \label{intro_point}$$ Therefore, \eqref{intro_point} reads that the value of functional $\alpha$ on any function $x$ only depends on the restriction of $x$ to the fixed subinterval $[0,t_0]$. The key property of functional $\alpha$ satisfying \eqref{intro_point} is that $$\alpha [u]\leq \| \alpha \| \cdot | u| _{C[0,t_0]}\,, \label{intro_key}$$ for every $u\in C[0,1]$. Normally, for a given functional $\alpha :C[0,1]\to \mathbb{R},$ we have $| \alpha [g]| \leq \| \alpha \| \cdot | g| _{C[0,1]}.$ However, if $\alpha$ satisfies condition \eqref{intro_point}, then $| \alpha [g]| \leq \| \alpha \|\cdot | g| _{C[0,t_0]}.$ Indeed, for each $g\in C[0,1]$, if we let $\widetilde{g}\in C[0,1]$ be defined by $\widetilde{g}(t)=\begin{cases} g(t), &\text{if } t\in [0,t_0] \\ g(t_0),& \text{if }t\in [t_0,1], \end{cases}$ then $| \alpha [g]|=| \alpha [ \widetilde{g}]| \leq \| \alpha \| \cdot | \widetilde{g} | _{C[0,1]} =\| \alpha \| \cdot | g| _{C[0,t_0]}.$ The goal of this work is to revisite system \eqref{1} under the assumption that both functionals $\alpha$ and $\beta$ satisfy \eqref{intro_point}, using the strategy from \cite{prec-nica}. Problem \eqref{1} is equivalent to the following integral system in $\ C [0,1]^2$: \begin{gather*} x(t)=\frac{1}{1-\alpha [1]}\alpha [g_1]+\int_0^{t}f_1(s,x(s),y(s))ds \\ y(t)=\frac{1}{1-\beta [1]}\beta [g_2]+\int_0^{t}f_2(s,x(s),y(s))ds, \end{gather*} where $g_1(t):=\int_0^{t}f_1(s,x(s),y(s))ds,\quad g_2(t):=\int_0^{t}f_2(s,x(s),y(s))ds.$ This can be viewed as a fixed point problem in $C[0,1]^2$ for the completely continuous operator $T:C[0,1]^2\to C[0,1]^2$, $T=(T_1,T_2)$, where $T_1$ and $T_2$\ are given by \begin{gather*} T_1(x,y)(t)=\frac{1}{1-\alpha [1]}\alpha [g_1]+\int_0^{t}f_1(s,x(s),y(s))ds, \\ T_2(x,y)(t)=\frac{1}{1-\beta [1]}\beta [g_2]+\int_0^{t}f_2(s,x(s),y(s))ds. \end{gather*} In fact, under assumption \eqref{intro_point} on $\alpha$ and $\beta$, operators $T_1$ and $T_2$ appear as sums of two integral operators, one of Fredholm type, whose values depend only on the restrictions of functions to $[0,t_0]$, and the other one, a Volterra type operator depending on the restrictions to $[t_0,1]$, as this was pointed out in \cite{ref6}. Thus, $T_1$ can be rewritten as $T_1=T_{F_1}+T_{V_1}$, where \begin{gather*} T_{F_1}(x,y)(t)=\begin{cases} \frac{1}{1-\alpha [1]}\alpha [g_1]+\int_0^{t}f_1( s,x(s),y(s))ds,& \text{if }t0$. In the next sections, three fixed point principles will be used to prove the existence of solutions for the semilinear problem, namely the fixed point theorems by Perov, Schauder and Leray-Schauder (see \cite{ref11}). In all three cases a key role will be played by the so called convergent to zero matrices. A square matrix$M$with nonnegative elements is said to be \emph{convergent to zero} if $M^{k}\to 0\quad \text{as }k\to \infty .$ It is known that the property of being convergent to zero is equivalent to each of the following three conditions (for details see \cite{ref11,p1}): \begin{itemize} \item[(a)]$I-M$is nonsingular and$(I-M)^{-1}=I+M+M^2+\dots$, where$I$stands for the unit matrix of the same order as$M$; \item[(b)] the eigenvalues of$M$are located in the interior of e the unit disc of the complex plane; \item[(c)]$I-M$is nonsingular and$(I-M)^{-1}$has nonnegative elements. \end{itemize} The following lemma whose proof is immediate from characterization (b) of convergent to zero matrices will be used in the sequel: \begin{lemma} \label{lem1.1} If$A$is a square matrix that converges to zero and the elements of an other square matrix$B$are small enough, then$A+B$also converges to zero. \end{lemma} We finish this introductory section by recalling (see \cite{ref12,ref11}) three fundamental results which will be used in the next sections. Let$X$be a nonempty set. By a \emph{vector-valued metric} on$X$we mean a mapping$d:X\times X\to \mathbb{R}_{+}^{n}$such that \begin{itemize} \item[(i)]$d(u,v)\geq 0$for all$u,v\in X$and if$d(u,v)=0$then$u=v$; \item[(ii)]$d(u,v)=d(v,u)$for all$u,v\in X$; \item[(iii)]$d(u,v)\leq d(u,w)+d(w,v)$for all$u,v,w\in X$. \end{itemize} Here, for$x=(x_1,x_2,\dots ,x_{n})$,$y=(y_1,y_2,\dots ,y_{n})$, by$x\leq y$we mean$x_i\leq y_i$for$i=1,2,\dots ,n$. We call the pair$(X,d)$a \emph{generalized metric space}. For such a space convergence and completeness are similar to those in usual metric spaces. An operator$T:X\to X$is said to be \emph{contractive} (with respect to the vector-valued metric$d$on$X$) if there exists a convergent to zero matrix$M$such that $d(T(u),T(v))\leq Md(u,v)\quad \text{for all }u,v\in X.$ \begin{theorem}[Perov] \label{thm1.2} Let$(X,d)$be a complete generalized metric space and$T:X\to X$a contractive operator with Lipschitz matrix$M$. Then$T$has a unique fixed point$u^{\ast }$and for each$u_0\in X$we have $d(T^{k}(u_0),u^{\ast })\leq M^{k}(I-M)^{-1}d(u_0,T(u_0))\quad \text{for all }k\in \mathbf{N}.$ \end{theorem} \begin{theorem}[Schauder] \label{thm1.3} Let$X$be a Banach space,$D\subset X$a nonempty closed bounded convex set and$T:D\to D$a completely continuous operator (i.e.,$T$is continuous and$T(D)$is relatively compact). Then$T$has at least one fixed point. \end{theorem} \begin{theorem}[Leray-Schauder] \label{thm1.4} Let$(X,\|\cdot\|_{X})$be a Banach space,$R>0$and$T:\overline{B}_{R}(0;X)\to X$a completely continuous operator. If$\|u\|_{X}0, we have \begin{align*} &| T_1(x,y)(t)-T_1(\overline{x},\overline{y})(t)| \\ &\leq | \frac{1}{1-\alpha [1]}| | \alpha [g_1-\overline{g}_1]| +\int_0^{t}| f_1(s,x(s),y(s))-f_1(s,\overline{x} (s),\overline{y}(s))| ds \\ &\quad +\int_{t_0}^{t}| f_1(s,x(s),y(s)) -f_1(s,\overline{x}(s),\overline{y}(s))| ds. \end{align*} Hence, \eqref{intro_key} gives \begin{align*} &| T_1(x,y)(t)-T_1(\overline{x},\overline{y})(t)| \\ &\leq \big(\frac{\| \alpha \| }{| 1-\alpha [1]| }+1\big) \big(a_1t_0| x-\overline{x} | _{C[0,t_0]}+b_1t_0| y-\overline{y}|_{C[0,t_0]}\big)\\ &\quad +\int_{t_0}^{t}| f_1(s,x(s),y(s)) -f_1(s,\overline{x}(s),\overline{y}(s))| ds. \end{align*} The last integral can be further estimated as follows: \begin{align*} &\int_{t_0}^{t}| f_1(s,x(s),y(s)) -f_1(s,\overline{x}(s),\overline{y}(s)) | ds \\ &\leq \int_{t_0}^{t}(a_2| x(s)-\overline{x} (s)| +b_2| y(s)-\overline{y}(s)| )ds \\ &= a_2\int_{t_0}^{t}| x(s)-\overline{x}(s)| \cdot e^{-\theta (s-t_0)}\cdot e^{\theta (s-t_0)}ds \\ &\quad +b_2\int_{t_0}^{t}| y(s)-\overline{y}(s)| \cdot e^{-\theta (s-t_0)}\cdot e^{\theta (s-t_0)}ds \\ &\leq \frac{a_2}{\theta }e^{\theta (t-t_0)}\| x-\overline{x} \| _{C[t_0,1]}+\frac{b_2}{\theta }e^{\theta (t-t_0)}\| y-\overline{y}\| _{C[t_0,1]}. \end{align*} Thus \begin{align*} | T_1(x,y)(t)-T_1(\overline{x},\overline{y})(t)| &\leq A_{\alpha }a_1t_0| x-\overline{x}| _{C[0,t_0]}+A_{\alpha }b_1t_0| y-\overline{y}| _{C[0,t_0]} \\ &\quad +\frac{a_2}{\theta }e^{\theta (t-t_0)}\| x-\overline{x} \| _{C[t_0,1]}+\frac{b_2}{\theta }e^{\theta (t-t_0)}\| y-\overline{y}\| _{C[t_0,1]}. \end{align*} Dividing bye^{\theta (t-t_0)}$and taking the supremum when$t\in [t_0,1]$, we obtain $$\label{101_1} \begin{split} \| T_1(x,y)-T_1(\overline{x},\overline{y})\|_{C[t_0,1]} &\leq A_{\alpha }a_1t_0| x-\overline{x}| _{C[0,t_0]}+A_{\alpha }b_1t_0| y-\overline{y}| _{C[0,t_0]} \\ &\quad+\frac{a_2}{\theta }\| x-\overline{x}\| _{C[t_0,1]}+ \frac{b_2}{\theta }\| y-\overline{y}\| _{C[t_0,1]}. \end{split}$$ Now \eqref{100} and \eqref{101_1} imply $$\| T_1(x,y)-T_1(\overline{x},\overline{y})\| \leq (A_{\alpha }a_1t_0+\frac{a_2}{\theta })\| x- \overline{x}\| +(A_{\alpha }b_1t_0+\frac{b_2}{\theta } )\| y-\overline{y}\| . \label{103}$$ Similarly, $$\| T_2(x,y)-T_2(\overline{x},\overline{y})\| \leq (B_{\beta }A_1t_0+\frac{A_2}{\theta })\| x- \overline{x}\| +(B_{\beta }B_1t_0+\frac{B_2}{\theta } )\| y-\overline{y}\| . \label{104}$$ Using the vector norm we can put both inequalities \eqref{103}, \eqref{104} under the vector inequality $\| T(u)-T(\overline{u})\| \leq M_{\theta }\| u-\overline{u}\| ,$ where $$M_{\theta }=\begin{bmatrix} A_{\alpha }a_1t_0+\frac{a_2}{\theta } & A_{\alpha }b_1t_0+\frac{ b_2}{\theta } \\ B_{\beta }A_1t_0+\frac{A_2}{\theta } & B_{\beta }B_1t_0+\frac{B_2 }{\theta } \end{bmatrix}. \label{m0}$$ Clearly the matrix$M_{\theta }$can be represented as$M_{\theta}=M_0+M_1$, where $M_1=\begin{bmatrix} \frac{a_2}{\theta } & \frac{b_2}{\theta } \\ \frac{A_2}{\theta } & \frac{B_2}{\theta } \end{bmatrix}.$ Since$M_0$is assumed to be convergent to zero, from Lemma \ref{lem1.1} we have that$M_{\theta }$also converges to zero for large enough$\theta >0.~$The result follows now from Perov's fixed point theorem. \end{proof} \section{Nonlinearities with growth at most linear. Application of Schauder's fixed point theorem} Here we show that the existence of solutions to problem \eqref{1} follows from Schauder's fixed point theorem when$ f_1,f_2$, instead of the Lipschitz condition, satisfy the more relaxed condition of growth at most linear: \begin{gather} | f_1(t,x,y)| \leq \begin{cases} a_1| x| +b_1| y| +c_1, & \text{if }t\in [0,t_0]\\ a_2| x| +b_2| y| +c_2, & \text{if }t\in [t_0,1], \end{cases} \label{300} \\ | g(t,x,y)| \leq \begin{cases} A_1| x| +B_1| y| +C_1, & \text{if }t\in [0,t_0]\\ A_2| x| +B_2| y| +C_2, & \text{if }t\in [t_0,1]. \end{cases} \label{301} \end{gather} \begin{theorem} \label{thm3.1} If$f_1,f_2$satisfy \eqref{300}, \eqref{301} and matrix \eqref{m} converges to zero, then \eqref{1} has at least one solution. \end{theorem} \begin{proof} To apply Schauder's fixed point theorem, we look for a nonempty, bounded, closed and convex subset$B$of$C[0,1]^2$so that$T(B)\subset B$. Let$x,y$be any elements of$C[0,1]$. For$t\in [0,t_0], using \eqref{intro_point} and \eqref{intro_key}, we have $$\begin{split} | T_1(x,y)(t)| &=| \frac{1}{1-\alpha [1]}\alpha [g_1]+\int_0^{t}f_1(s,x(s),y(s))ds| \\ &\leq | \frac{1}{1-\alpha [1]}| | \alpha [g_1]| +\int_0^{t}(a_1| x(s)| +b_1| y(s)| +c_1)ds \\ &\leq \frac{\| \alpha \| }{| 1-\alpha [1]| }| g_1| _{C[0,t_0]} +a_1t_0| x| _{C[0,t_0]}+b_1t_0| y| _{C[0,t_0]}+c_1t_0. \end{split} \label{fw2}$$ Also \begin{align*} | g_1(t)| &\leq \int_0^{t}| f_1(s,x(s),y(s))| ds \\ &\leq \int_0^{t}(a_1| x(s)| +b_1|y(s)| +c_1)ds \\ &\leq a_1t_0| x| _{C[0,t_0] }+b_1t_0| y| _{C[0,t_0]}+c_1t_0, \end{align*} which gives $$| g_1| _{C[0,t_0]}\leq a_1t_0| x| _{C[0,t_0]} +b_1t_0| y| _{C[0,t_0]}+c_1t_0. \label{g2_fw}$$ From \eqref{fw2} and \eqref{g2_fw}, we obtain $$\begin{split} | T_1(x,y)| _{C[0,t_0]} &\leq (\frac{\| \alpha \| }{| 1-\alpha [1]| }+1) (a_1t_0| x| _{C[0,t_0]}+b_1t_0| y| _{C[0,t_0]})+\widetilde{c}_1 \\ &= a_1t_0A_{\alpha }| x| _{C[0,t_0]}+b_1t_0A_{\alpha }| y| _{C[0,t_0]}+ \widetilde{c}_1, \end{split} \label{302_1}$$ where\widetilde{c}_1:=c_1t_0A_{\alpha }$. For$t\in [t_0,1]$and any$\theta >0, we have \begin{align*} | T_1(x,y)(t)| &= a_1t_0A_{\alpha }| x| _{C[0,t_0]}+b_1t_0A_{\alpha }| y| _{C[0,t_0]}+\widetilde{c}_1 \\ &\quad +\int_{t_0}^{t}(a_2| x(s)| +b_2|y(s)| +c_2\text{ })ds \\ &\leq a_1t_0A_{\alpha }| x| _{C[0,t_0]}+b_1t_0A_{\alpha }| y| _{C[0,t_0]} +\widetilde{c}_1+(1-t_0)c_2 \\ &\quad +a_2\int_{t_0}^{t}| x(s)| \cdot e^{-\theta (s-t_0)}\cdot e^{\theta (s-t_0)}ds \\ &\quad +b_2\int_{t_0}^{t}| y(s)| \cdot e^{-\theta (s-t_0)}\cdot e^{\theta (s-t_0)}ds \\ &\leq a_1t_0A_{\alpha }| x| _{C[0,t_0]}+b_1t_0A_{\alpha }| y| _{C[0,t_0]}+c_0 \\ &\quad +\frac{a_2}{\theta }e^{\theta (t-t_0)}\| x\| _{C[t_0,1]}+\frac{b_2}{\theta }e^{\theta (t-t_0)}\|y\| _{C[t_0,1]}, \end{align*} where\ c_0:=\widetilde{c}_1+(1-t_0)c_2$. Dividing by$e^{\theta (t-t_0)}$and taking the supremum, it follows that $$\label{303} \begin{split} \| T_1(x,y)\| _{C[t_0,1]} &\leq a_1t_0A_{\alpha}| x| _{C[0,t_0]}+b_1t_0A_{\alpha}| y| _{C[0,t_0]} \\ &\quad +\frac{a_2}{\theta }e^{\theta (t-t_0)}\| x\| _{C[t_0,1]}+\frac{b_2}{\theta }e^{\theta (t-t_0)}\| y\| _{C[t_0,1]}+c_0. \end{split}$$ Clearly, \eqref{302_1} and \eqref{303} give $$\| T_1(x,y)\| \leq (a_1t_0A_{\alpha }+\frac{ a_2}{\theta })\| x\| +(b_1t_0A_{\alpha }+ \frac{b_2}{\theta })\| y\| +\widetilde{c}_0, \label{304}$$ where$\widetilde{c}_0=\max \left\{ \widetilde{c}_1,c_0\right\} $. Similarly, $$\| T_2(x,y)\| \leq (A_1t_0B_{\beta } +\frac{A_2}{\theta })\| x\| +(B_1t_0B_{\beta } +\frac{B_2}{\theta })\| y\| +\widetilde{C}_0, \label{305}$$ with$\widetilde{C}_0=\max \{ \widetilde{C}_1,C_0\} $, where$\widetilde{C}_1:=C_1t_0B_{\beta }$and$C_0:=\widetilde{C}_1+(1-t_0)C_2$. Now \eqref{304} and \eqref{305} can be put together as $\begin{bmatrix} \| T_1(x,y)\| \\ \| T_2(x,y)\| \end{bmatrix} \leq M_{\theta } \begin{bmatrix} \| x\| \\ \| y\| \end{bmatrix} +\begin{bmatrix} \widetilde{c}_0 \\ \widetilde{C}_0 \end{bmatrix},$ where the matrix$M_{\theta }$is given by \eqref{m0} and converges to zero for a large enough$\theta >0$. Next we look for two positive numbers$R_1,R_2$such that if$\| x\| \leq R_1,\|y\| \leq R_2$, then$\| T_1(x,y)\| \leq R_1$,$\| T_2(x,y)\| \leq R_2$. To this end it is sufficient that $$\begin{gathered} (a_1t_0A_{\alpha }+\frac{a_2}{\theta })R_1+( b_1t_0A_{\alpha }+\frac{b_2}{\theta })R_2+\widetilde{c} _0\leq R_1 \\ (A_1t_0B_{\beta }+\frac{A_2}{\theta })R_1+( B_1t_0B_{\beta }+\frac{B_2}{\theta })R_2+\widetilde{C} _0\leq R_2, \end{gathered} \label{r1r2}$$ or equivalently $M_{\theta }\begin{bmatrix} R_1 \\ R_2 \end{bmatrix} +\begin{bmatrix} \widetilde{c}_0 \\ \widetilde{C}_0 \end{bmatrix} \leq \begin{bmatrix} R_1 \\ R_2 \end{bmatrix},$ whence $\begin{bmatrix} R_1 \\ R_2 \end{bmatrix} \geq (I-M_{\theta })^{-1} \begin{bmatrix} \widetilde{c}_0 \\ \widetilde{C}_0 \end{bmatrix}.$ Note that$I-M_{\theta }$is invertible and its inverse$(I-M_{\theta })^{-1}$has nonnegative elements since$M_{\theta }$converges to zero. Thus, if$B=\{ (x,y)\in C[0,1]^2:\| x\| \leq R_1,\| y\| \leq R_2\} $, then$T(B)\subset B$and Schauder's fixed point theorem can be applied. \end{proof} \section{More general nonlinearities. Application of the Leray-Schauder principle} We now consider that nonlinearlities$f_1,f_2$satisfy more general growth conditions, namely: \begin{gather} | f_1(t,u)| \leq \begin{cases} \omega _1(t,| u| _{e}), & \text{if }t\in [0,t_0] \\ \gamma (t)\beta _1(| u| _{e}), & \text{if }t\in [t_0,1], \end{cases} \label{400} \\ | f_2(t,u)| \leq \begin{cases} \omega _2(t,| u| _{e}), & \text{if }t\in [0,t_0] \\ \gamma (t)\beta _2(| u| _{e}), & \text{if }t\in [t_0,1], \end{cases} \label{401} \end{gather} for all$u=(x,y)\in \mathbb{R}^2$, where by$| u| _{e}$we mean the Euclidean norm in$\mathbb{R}^2$. Here$\omega _1,\omega _2 $are Carath\'{e}odory functions on$[0,t_0]\times \mathbb{R}_{+}$, nondecreasing in their second argument,$\gamma \in L^{1}[t_0,1 ]$, while$\beta _1,\beta _2:\mathbb{R}_{+}\to \mathbb{R} _{+}\ $are nondecreasing and$1/\beta _1,1/\beta _2\in L_{loc}^{1}( \mathbb{R}_{+})$. \begin{theorem} \label{thm4.1} Assume that \eqref{400}, \eqref{401} hold. In addition assume that there exists a positive number$R_0$such that for$\rho=(\rho _1,\rho _2)\in (0,\infty )^2$, $$\frac{1}{\rho _1}\int_0^{t_0}\omega _1(t,| \rho | _{e})dt\geq \frac{1}{A_{\alpha }} \text{ and } \frac{1}{\rho _2}\int_0^{t_0}\omega _2(t,| \rho | _{e})dt\geq \frac{1}{B_{\beta }} \quad \text{imply}\quad | \rho | _{e}\leq R_0 \label{410}$$ and $$\int_{R^{\ast }}^{\infty }\frac{d\tau }{\beta _1(\tau )+\beta _2(\tau )} >\int_{t_0}^{1}\gamma (s)ds, \label{450}$$ where$R^{\ast }=\big[(A_{\alpha }\int_0^{t_0}\omega _1(t,R_0)dt)^2+(B_{\beta }\int_0^{t_0}\omega _2(t,R_0)dt)^2\big]^{1/2}$. Then \eqref{1} has at least one solution. \end{theorem} \begin{proof} The result will follow from the Leray-Schauder fixed point theorem once we have proved the boundedness of the set of all solutions to equation$u=\lambda T(u)$, for$\lambda \in [0,1]$. Let$u=(x,y)$be such a solution. Then, for$t\in [0,t_0]$, also using condition \eqref{intro_point} and \eqref{intro_key}, we have $$\label{402} \begin{split} | x(t)| &= | \lambda T_1(x,y)(t)| \\ &= \lambda | \frac{1}{1-\alpha [1]}\alpha [ g_1]+\int_0^{t}f_1(s,x(s),y(s))ds| \\ &\leq \frac{\| \alpha \| }{| 1-\alpha [1]| }| g_1| _{C[0,t_0]}+\int_0^{t}| f_1(s,x(s),y(s))| ds \\ &\leq (\frac{\| \alpha \| }{| 1-\alpha [1]| }+1)\int_0^{t_0}\omega _1(s,| u(s)| _{e})ds \\ &= A_{\alpha }\int_0^{t_0}\omega _1(s,| u(s)|_{e})ds. \end{split}$$ Similarly, $$| y(t)| \leq B_{\beta }\int_0^{t_0}\omega _2(s,| u(s)| _{e})ds. \label{403}$$ Let$\rho _1=| x| _{C[0,t_0]}$,$\rho _2=|y| _{C[0,t_0]}$. Then from \eqref{402}, \eqref{403}, we deduce \begin{gather*} \rho _1\leq A_{\alpha }\int_0^{t_0}\omega _1(s,|u(s)| _{e})ds \\ \rho _2\leq B_{\beta }\int_0^{t_0}\omega _2(s,|u(s)| _{e})ds. \end{gather*} By \eqref{410}, this guarantees $$| \rho | _{e}\leq R_0. \label{453}$$ Next we let$t\in [t_0,1] . Then \begin{align*} | x(t)| &= | \lambda T_1(x,y)(t)| \\ &\leq A_{\alpha }\int_0^{t_0}\omega _1(s,R_0)ds+\int_{t_0}^{t}| f_1(s,x(s),y(s))| ds \\ &\leq A_{\alpha }\int_0^{t_0}\omega _1(s,R_0)ds+\int_{t_0}^{t}\gamma (s)\beta _1(| u(s)| _{e})ds \\ &= :\phi _1(t) \end{align*} and similarly $| y(t)| \leq B_{\beta }\int_0^{t_0}\omega _2(s,R_0)ds+\int_{t_0}^{t}\gamma (s)\beta _2(|u(s)| _{e})ds = :\phi _2(t).$ Denote\psi (t):=(\phi _1^2(t)+\phi _2^2(t))^{1/2}. Then $$\begin{gathered} \phi _1' (t)=\gamma (t)\beta _1(| u(t)| _{e})\leq \gamma (t)\beta _1(\psi (t)) \\ \phi _2' (t)=\gamma (t)\beta _2(| u(t)| _{e})\leq \gamma (t)\beta _2(\psi (t))). \end{gathered} \label{ls6}$$ Consequently, \begin{align*} \psi ' (t) &= \frac{\phi _1(t)\phi _1' (t)+\phi _2(t)\phi _2' (t)}{\psi (t)} \\ &\leq \gamma (t)\cdot \frac{\phi _1(t)}{\psi (t)}\cdot \beta _1(\psi (t))+\gamma (t)\cdot \frac{\phi _2(t)}{\psi ( t)}\cdot \beta _2(\psi (t)) \\ &\leq \gamma (t)[\beta _1(\psi (t))+\beta _2(\psi (t))]. \end{align*} It follows that $\int_{t_0}^{t}\frac{\psi '(s)}{\beta _1(\psi (s))+\beta _2(\psi (s))}ds\leq \int_{t_0}^{t}\gamma (s)ds.$ Furthermore, using \eqref{450} we obtain $$\int_{\psi (t_0)}^{\psi (t)}\frac{d\tau }{\beta _1(\tau )+\beta _2(\tau ))}\leq \int_{t_0}^{t}\gamma (s)ds\leq \int_{t_0}^{1}\gamma (s)ds<\int_{R^{\ast }}^{\infty }\frac{d\tau }{\beta _1(\tau )+\beta _2(\tau )}. \label{451}$$ Note that\psi (t_0)=R^{\ast }$. Then from \eqref{451} it follows that there exists$R_1$such that $\psi (t)\leq R_1,$ for all$t\in [t_0,1]$. Then$| x(t)| \leq R_1$and$| y(t)| \leq R_1$, for all$t\in [t_0,1]$, whence $$| x| _{C[t_0,1]}\leq R_1,\quad |y| _{C[t_0,1]}\leq R_1. \label{454}$$ Let$R=\max \{R_0,R_1\}$. From \eqref{453}, \eqref{454} we have$| x| _{C[0,1]}\leq R$and$| y|_{C[0,1]}\leq R$as desired. \end{proof} \begin{remark} \rm If$\omega _1(t,\tau )=\gamma _0(t)\beta _0(\tau )$, then the first inequality in \eqref{410} implies that$\beta _0(\tau )\leq c\tau +c' $for all$\tau \in R_{+}$and some constants$c$and$c'$; i.e., the growth of$\beta _0$is at most linear. However,$\beta _1$may have a superlinear growth. Thus we may say that under the assumptions of Theorem \ref{thm4.1}, the growth of$f_1(t,u)$in$u$is at most linear for$t\in [0,t_0]$and can be superlinear for$t\in [t_0,1]$. The same can be said about$f_2(t,u)$. \end{remark} In particular, when$\alpha =\beta =0$, problem \eqref{1} becomes the classical local initial value problem $$\begin{gathered} x'=f_1(t,x,y)\\ y'=f_2(t,x,y)\quad (\text{a.e. }t\in [0,1])\\ x(0)=y(0)=0, \end{gathered} \label{1*}$$ and our assumptions reduce to the classical conditions (see \cite{ref9,op}) and Theorem \ref{thm4.1} gives the following result. \begin{corollary} Assume that $| f_1(t,u)| \leq \gamma (t)\beta _1(|u| _{e}), \quad | f_2(t,u)| \leq \gamma (t)\beta _2(| u| _{e})$ for$t\in [0,1]$and$u\in \mathbb{R}^2$, where$\gamma \in L^{1}[0,1]$, while$\beta _1,\beta _2:\mathbb{R} _{+}\to \mathbb{R}_{+}$are nondecreasing and$1/\beta _1,1/\beta _2\in L_{loc}^{1}(\mathbb{R}_{+})$. In addition assume that $\int_0^{\infty }\frac{d\tau }{\beta _1(\tau )+\beta _2(\tau )} >\int_0^{1}\gamma (s)ds.$ Then problem \eqref{1*} has at least one solution. \end{corollary} A result similar to the above corollary was given in \cite{nica-func-w}. \begin{remark} Since the trivial solution satisfies the boundary conditions, the solution given by Theorem \ref{thm4.1} might be zero. \end{remark} \section{Numerical examples} In this section, we give some numerical examples to illustrate the existence results from Sections 2 and 3. \begin{example} \label{examp5.1} \rm Consider the initial value problem $$\begin{gathered} x'(t)=0.1+\frac{1}{4}\frac{y^2(t)}{1+y^2(t)}\sin (2x(t))=:f(x,y) \\ y'(t)=0.1+\frac{2}{3}\frac{y^2(t)}{1+y^2(t)}\cos (2x(t))=:g(x,y) \\ x(0)=\int_0^{1/2}x(s)ds, \quad y(0)=\int_0^{1/2}y(s)ds, \end{gathered}\label{ex5.1}$$ for$t\in [0,40]$. \end{example} We have that $\alpha [u]=\int_0^{1/2}u(s)ds\Longrightarrow \alpha [1 ]=\frac{1}{2}\Longrightarrow \| \alpha \| =\frac{1}{2 }.$ Consequently,$t_0=1/2$,$A_{\alpha }=2=B_{\beta }$and $M_0=\begin{pmatrix} a_1 & b_1 \\ A_1 & B_1 \end{pmatrix}.$ However, \begin{gather*} \sup_{\xi ,\eta \in \mathbb{R}} | \frac{\partial f(\xi ,\eta )}{\partial x}| \leq \frac{1}{2}=a_1, \quad \sup_{\xi ,\eta \in\mathbb{R}}| \frac{\partial f(\xi ,\eta )}{\partial y} | \leq \frac{3\sqrt{3}}{32}=b_1, \\ \sup_{\xi ,\eta \in\mathbb{R}}| \frac{\partial g(\xi ,\eta )}{\partial x} | \leq \frac{4}{3}=A_1, \quad \sup_{\xi ,\eta \in\mathbb{R}}| \frac{\partial g(\xi ,\eta )}{\partial y} | \leq \frac{\sqrt{3}}{4}=B_1 \end{gather*} and then $M_0=\begin{pmatrix} \frac{1}{2} & \frac{3\sqrt{3}}{32} \\ \frac{4}{3} & \frac{\sqrt{3}}{4} \end{pmatrix}$ has the eigenvalues$\lambda _1=0$,$\lambda _2=0.9330\cdots$. From Theorem \ref{thm2.1}, problem \eqref{ex5.1} has a unique solution, see Figure \ref{fig1}. \begin{figure}[ht] \begin{center} \setlength{\unitlength}{1mm} \begin{picture}(103,60)(-6,-3) \put(0,0){\line(1,0){97}} \multiput(12.5,-.5)(12.5,0){7}{\line(0,1){1}} \put(23,-4){10} \put(48,-4){20} \put(73,-4){30} \put(98,-4){40} \put(96.6,-.89){$\rightarrow$} \put(100,2){$t$} \put(0,0){\line(0,1){50}} \multiput(-.5,10)(0,10){5}{\line(1,0){1}} \put(-6,-1){0.0} \put(-6,9){0.5} \put(-6,19){1.0} \put(-6,29){1.5} \put(-6,39){2.0} \put(-6,49){2.5} \qbezier(0,2)(9,9)(17,25) \qbezier(17,25)(35,55)(62,46) \qbezier(62,46)(81,40)(100,42) \put(75,46){$x(t)$} \qbezier(0,2)(4,5)(10,16) \qbezier(10,16)(14,22)(18,16) \qbezier(18,16)(25,5)(60,17) \qbezier(60,17)(70,22)(80,19) \qbezier(80,19)(90,16)(100,17) \put(75,22){$y(t)$} \end{picture} \end{center} \caption{The Chebpack$x,y$solutions of the problem \eqref{ex5.1}. The numerical errors for the nonlocal boundary conditions are$bc1$and$bc$2} \label{fig1} \end{figure} The exact solution is approximated by the Matlab package \emph{Chebpack} \cite{cp} and verified by the \texttt{ode45} solver of Matlab (i.e. \texttt{ode45} is applied to \eqref{ex5.1} with the initial conditions$x(0)$,$y(0) $given by Chebpack). \begin{example} \label{examp5.2} \rm Consider the initial value problem $$\label{ex5.2} \begin{gathered} x'=-0.9x-1.8\frac{xy}{2+x^2}+90:= f(x,y) \\ y'=-0.2y-1.8\frac{xy}{2+x^2}+750:= g(x,y) \\ x(0)=\int_0^{1/2}x(s)ds, \quad y(0)=\int_0^{1/2}y(s)ds, \end{gathered}$$ for$t\in [0,1]$. \end{example} We consider $M_0=\begin{pmatrix} a_1 & b_1 \\ A_1 & B_1 \end{pmatrix}.$ We have $| \frac{x}{2+x^2}| \leq \frac{\sqrt{2}}{4},$ so that the matrix $M_0=\begin{pmatrix} 0.9 & 0.6364 \\ 0 & 0.8364 \end{pmatrix}$ has the eigenvalues$\lambda _1=0.9$,$\lambda _2=0.8364$. From Theorem \ref{thm3.1}, problem \eqref{ex5.2} has at least one solution. Let us denote $bc1(x_0,y_0)=x_0-\int_0^{1/2}x(s)ds,\quad bc2(x_0,y_0)=y_0-\int_0^{1/2}y(s)ds$ where$x(s)$and$y(s)$are obtained by integrating the differential system \eqref{ex5.2} with initial conditions$x(0)=x_0$,$y(0)=y_0$. In Figure \ref{fig2}, approximated$x_0$,$y_0$, shows the numerical contour lines of$bc1(x_0,y_0)=0$(solid line) and of$bc2(x_0,y_0)=0$(dashed line). Their intersections give the initial conditions for which the solutions$x(s)$,$y(s)$approximate the nonlocal conditions from \eqref{ex5.2}. We have in that region three intersection points$1,2,3$corresponding to three different solutions, which are improved by \texttt{fsolve} from Matlab to \begin{gather*} \texttt{init1}=[11.7467173136538\quad 167.2358959061741];\\ \texttt{init2}=[3.6799740768135\quad 156.9860214200612];\\ \texttt{init3}=[0.1962071293693\quad 152.5406128950519]. \end{gather*} Taking these values as initial conditions for a Matlab solver for differential systems, we obtain the corresponding three numerical solutions of \eqref{ex5.2} represented in Figure \ref{fig2} with an accuracy about$10^{-7}$in nonlocal conditions. \begin{figure}[ht] \begin{center} \scriptsize \setlength{\unitlength}{1mm} \begin{picture}(55,25)(-6,-3) \put(0,20){\line(1,0){50}} \put(0,0){\line(1,0){50}} \multiput(5,-.5)(15,0){4}{\line(0,1){1}} \put(4.5,-3){0} \put(19.5,-3){5} \put(33.5,-3){10} \put(48.5,-3){15} \put(0,0){\line(0,1){20}} \put(-.5,10){\line(1,0){1}} \put(-5.5,-.7){150} \put(-5.5,9.3){160} \put(-5.5,19.3){170} \put(50,0){\line(0,1){20}} \qbezier(0,4)(14,0)(17,5) \qbezier(17,5)(20,20)(50,18) \put(28,14){$bc2$} \qbezier(5,20)(5.5,10)(7,0) \qbezier(17,0)(18,10)(21,20) \qbezier(38,20)(41,10)(42,0) \put(40,15.5){1} \put(19.5,7.8){2} \put(7,3.2){3} \put(42.5,7){$bc1$} \put(6.5,16){$bc1$} \end{picture} % \qquad % \begin{picture}(55,25)(-6,-3) \put(0,20){\line(1,0){50}} \put(0,0){\line(1,0){50}} \put(25,-.5){\line(0,1){1}} \put(-.5,-3){0} \put(23,-3){0.5} \put(48,-3){1.0} \put(0,0){\line(0,1){20}} \multiput(-.5,5)(0,10){2}{\line(1,0){1}} \put(-4,4.3){20} \put(-4,14.3){40} \put(50,0){\line(0,1){20}} \qbezier(0,1)(25,15)(50,19) \put(23,9.6){$x_1(t)$} \end{picture} % \\ %% \begin{picture}(55,30)(-6,-3) \put(0,20){\line(1,0){50}} \put(0,0){\line(1,0){50}} \put(25,-.5){\line(0,1){1}} \put(23,-3){0.5} \put(48,-3){1.0} \put(0,0){\line(0,1){20}} \put(-.5,10){\line(1,0){1}} \put(-3,-.7){0} \put(-3,9.7){5} \put(-4,19.7){10} \put(50,0){\line(0,1){20}} \qbezier(0,8)(30,30)(30,1) \put(30,1){\line(1,0){20}} \put(22,9.6){$x_2(t)$} \end{picture} % \qquad % \begin{picture}(55,30)(-6,-3) \put(0,20){\line(1,0){50}} \put(0,0){\line(1,0){50}} \put(25,-.5){\line(0,1){1}} \put(23,-3){0.5} \put(48,-3){1.0} \put(0,0){\line(0,1){20}} \put(-.5,14.5){\line(1,0){1}} \put(-3,-.7){0} \put(-4.5,13.7){0.5} \put(50,0){\line(0,1){20}} \qbezier(0,2)(3,25)(5,16) \qbezier(5,16)(10,2)(50,3) \put(23,7){$x_3(t)$} \end{picture} \end{center} \caption{Contour lines of$bc1(x_0,y_0)=0$and of$bc2(x_0,y_0)=0\$. 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