\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 102, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/102\hfil Multiple symmetric positive solutions] {Multiple symmetric positive solutions for systems of higher order boundary-value problems on time scales} \author[P. V. S. Anand, P. Murali, K. R. Prasad \hfil EJDE-2011/102\hfilneg] {Putcha. V. S. Anand, Penugurthi Murali, Kapula R. Prasad} % in alphabetical order \address{Putcha V. S. Anand \newline CR Rao Advanced Institute of Mathematics, Statistics and Computer Science\\ University of Hyderabad Campus\\ Hyderabad, 500 046, India} \email{anand\_putcha@yahoo.com} \address{Penugurthi Murali \newline Department of Applied Mathematics \\ Andhra University \\ Visakhapatnam, 530003, India} \email{murali\_uoh@yahoo.co.in} \address{Kapula Rajendra Prasad \newline Department of Applied Mathematics \\ Andhra University \\ Visakhapatnam, 530003, India} \email{rajendra92@rediffmail.com} \thanks{Submitted January 7, 2011. Published August 10, 2011.} \subjclass[2000]{39A10, 34B15, 34A40} \keywords{Boundary value problem; cone; symmetric positive solution; \hfill\break\indent symmetric time scale} \begin{abstract} In this article, we find multiple symmetric positive solutions for a system of higher order two-point boundary-value problems on time scales by determining growth conditions and applying a fixed point theorem in cones under suitable conditions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} Symmetry creates beauty in nature and in nature every thing is almost symmetric. One can observe that symmetry in the structure of fruits, the structure of human body, the revolution of planets and the structure of atoms. Due to the importance of symmetric properties in both theory and applications, the study of existence of symmetric solutions of boundary value problems gained momentum. In this paper, we address the question of the existence of at least three symmetric positive solutions for the system of dynamical equations on symmetric time scales, $$\label{e1} \begin{gathered} (-1)^n y_1^{(\Delta\nabla)^n}=f_1(t,y_1,y_2),\quad t\in [a,b]_{\mathbb{T}} \\ (-1)^m y_2^{(\Delta\nabla)^m}=f_2(t,y_1,y_2),\quad t\in [a,b]_{\mathbb{T}} \end{gathered}$$ subject to the two-point boundary conditions $$\label{e2} \begin{gathered} y_1^{(\Delta\nabla)^i}(a)=0=y_1^{(\Delta\nabla)^i}(b),\quad i=0,1,2,\dots,n-1,\\ y_2^{(\Delta\nabla)^j}(a)=0=y_2^{(\Delta\nabla)^j}(b),\quad j=0,1,2,\dots,m-1, \end{gathered}$$ where $f_i:[a,b]_{\mathbb{T}}\times\mathbb{R}^{2}\to [0,\infty)$ are continuous and $f_i(t,y_1,y_2)=f_i(a+b-t,y_1,y_2)$ for $i=1,2$, $a\in \mathbb{T}_{k}$, $b\in\mathbb{T}^{k}$ for a time scale $\mathbb{T}$, and also $\sigma(a)<\rho(b)$. By an interval time scale, we mean the intersection of a real interval with a given time scale; i.e., $$[a,b]_{\mathbb{T}}=[a,b]\cap\mathbb{T}.$$ For time scale calculus, we refer the reader to Bohner and Peterson \cite{bop,bopp}. An interval time scale $\mathbb{T}=[a,b]_{\mathbb{T}}$ is said to be a symmetric time scale if $t\in \mathbb{T}\Leftrightarrow a+b-t\in\mathbb{T}$. If $\mathbb{T}=\mathbb{R}$ or $\mathbb{T}=h\mathbb{Z},(h>0)$ then the symmetry definition is always satisfied. In addition to, the interval time scale $\mathbb{T}=[1,2]\cup\{3,4,5\}\cup[6,7]\cup\{8\}\cup[9,10]\cup\{11,12,13\} \cup[14,15]$ has the symmetrical property. But the time scale $\mathbb{T}=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{N}\}$ is not a symmetric time scale. By a \emph{symmetric solution} $(y_1,y_2)$ of the system of boundary value problem \eqref{e1}-\eqref{e2}, we mean $(y_1,y_2)$ is a solution of \eqref{e1}-\eqref{e2} and satisfies $$y_1(t)=y_1(b+a-t)~{\rm and}~y_2(t)=y_2(b+a-t), \quad t\in [a, b ]_{\mathbb{T}}.$$ The development of the theory has gained attention by many researchers; To mention a few, we list some papers, Erbe and Wang \cite{lhh}, Eloe and Henderson \cite{pwj,pwe}, Eloe, Henderson and Sheng \cite{pqj}, Henderson and Thompson \cite{jt}, Avery and Henderson \cite{ah,aah, rj}, Avery, Davis and Henderson \cite{adh}, Davis and Henderson \cite{jh}, Davis, Henderson and Wong \cite{dhw}, Anderson \cite{dra}, Henderson and Wong \cite {jhw}, and Henderson, Murali and Prasad \cite{jmp}. This article is organized as follows. In Section 2, we establish certain lemmas and inequalities on Green's function which are needed later. In Section 3, by using the cone theory techniques, we establish the existence of at least three symmetric positive solutions to \eqref{e1}-\eqref{e2}. The main tool in this paper is an application of the Avery's generalization of the Leggett-Williams fixed point theorem for operator leaving a Banach space cone invariant. \section{Green's function and bounds} In this section, we construct the Green's function for the homogeneous SBVP corresponding to \eqref{e1}-\eqref{e2}. We estimate bounds of the Green's function, and establish some lemmas, in which we prove some inequalities on the Green's function, which are needed in our main result. Let us denote the Green's function of the problem \begin{gather*} -y^{\Delta\nabla}(t)=0,\quad t\in[a,b]_{\mathbb{T}},\\ y(a)=0=y(b), \end{gather*} as $G_1(t,s)$, and it is given by $$G_1(t,s)=\begin{cases} \frac{(b-s)(t-a)}{(b-a)}, \quad t\leq s \\ \frac{(b-t)(s-a)}{(b-a)},\quad s\leq t \end{cases}$$ for all $t,s\in [a,b]_{\mathbb{T}}$. Then, we can recursively define $$\label{e3} G_j(t,s)=\int_{a}^{b}G_{j-1}(t,r)G_1(r,s)\nabla r, \quad\text{for all } t,s\in [a,b]_{\mathbb{T}},$$ for $j=2,3,\dots,p$, and $p=\max\{m,~n\}$, where $G_j(t,s)$ is the Green's function for the problem \begin{gather*} (-1)^jy^{(\Delta\nabla)^{j}}(t)=0,\quad t\in [a,b]_{\mathbb{T}},\\ y^{(\Delta\nabla)^{i}}(a)=y^{(\Delta\nabla)^{i}}(b)=0,\quad i=0,1,2,\dots, j-1, \end{gather*} and $G_j(t,s)\geq 0$ for all $t,s\in [a,b]_{\mathbb{T}}$. For details we refer to \cite{dra,jmp}. The following lemmas are needed to establish our main result. \begin{lemma}\label{lem1} Let $l\in[\frac{b-a}{8},\frac{b-a}{2}]_{\mathbb{T}}$ and $(t,s)\in [a+l,b-l]_{\mathbb{T}}\times[a,b]_{\mathbb{T}}$, $$\label{e110} |G_j(t,s)|\geq L_l^j\phi_l^{j-1}\frac{(b-s)(s-a)}{b-a},\quad\text{for } j=1,2,\dots,p,$$ where $p$ is maximum of $\{m,n\}$, $L_l=\frac{l}{b-a}$ and $\phi_l=\int_{a+l}^{b-l}\frac{(b-s)(s-a)} {b-a}\nabla s$. \end{lemma} \begin{proof} For $j=1$ the inequality \eqref{e110} holds provided that $L_l=\frac{l}{b-a}$. Next for fixed $j$, assuming that \eqref{e110} is true, from \eqref{e3} we have for $(t,s)\in[a+l,b-l]_{\mathbb{T}}\times[a,b]_{\mathbb{T}}$, \begin{align*} |G_{j+1}(t,s)| &=|\int_a^bG_j(t,r)G_1(r,s)\nabla r|\\ &\geq|\int_{a+l}^{b-l}G_j(t,r)G_1(r,s)\nabla r|\\ &\geq\int_{a+l}^{b-l}L_l^j\phi_l^{j-1}\frac{(b-r)(r-a)} {b-a}\times L_l \frac{(b-s)(s-a)}{b-a}\nabla r\\ &=L_l^{j+1}\phi_l^{j}\frac{(b-s)(s-a)}{b-a}. \end{align*} Hence, by induction the result is true for all $j\leq p-1$. \end{proof} \begin{lemma}\label{lem2} For $(t,s)\in [a,b]_{\mathbb{T}}\times[a,b]_{\mathbb{T}}$, $$\label{e111} |G_j(t,s)|\leq \phi_0^{j-1}\frac{(b-s)(s-a)}{b-a},\quad \text{for }j=1,2,\dots,p,$$ where $\phi_0=\int_{a}^{b}\frac{(b-s)(s-a)}{b-a}\nabla s$. \end{lemma} \begin{proof} For $j=1$ the inequality \eqref{e111} is obvious. Next for fixed $j$, assume that \eqref {e111} is true, then from \eqref{e3} we have \begin{align*} |G_{j+1}(t,s)|&=|\int_a^{b}G_j(t,r)G_1(r,s)\nabla r|\\ &\leq\int_{a}^{b}\phi_0^{j-1}\frac{(b-r)(r-a)} {b-a}\times \frac{(b-s)(s-a)}{b-a}\nabla r\\ &=\phi_0^j\frac{(b-s)(s-a)}{b-a}. \end{align*} Hence, by induction the result is true for all $j\leq p-1$. \end{proof} \begin{lemma}\label{lem3} Let $t_{k}=\frac{b+a}{2}$ and $t_i\in[a, \frac{b+a}{2}]_{\mathbb{T}}$, $1\leq i\leq 3$ with $t_1\leq t_2$. For $s\in [a,b]_{\mathbb{T}}$, $$\frac{G_1(t_1,s)}{G_1(t_2,s)}\geq \frac{t_1-a}{t_2-a}\quad\text{and}\quad \frac{G_1(t_{k},s)}{G_1(t_3,s)}\leq \frac{t_{k}-a}{t_3-a}.$$ \end{lemma} \begin{proof} For $t\leq s$, we have $\frac{G_1(t_1,s)}{G_1(t_2,s)}=\frac{t_1-a}{t_2-a}$. And for $s\leq t$, we have $\frac{G_1(t_1,s)}{G_1(t_2,s)}=\frac{b-t_1}{b-t_2}$. Since $t_1\leq t_2$, we get $\frac{b-t_1}{b-t_2}\geq \frac{t_1-a}{t_2-a}$. Similarly, for $t\leq s$, we have $\frac{G_1(t_{k},s)}{G_1(t_3,s)}=\frac{t_{k}-a}{t_3-a}$. And for $s\leq t$, we have $\frac{G_1(t_{k},s)}{G_1(t_3,s)}=\frac{b-t_{k}}{b-t_3}$. Since $t_1\leq t_2$, we get $\frac{b-t_{k}}{b-t_3}\leq \frac{t_{k}-a}{t_3-a}$. \end{proof} \begin{lemma}\label{lem4} For $t, s \in [a, b]_{\mathbb{T}}$, the Green's function $G_j(t, s)$ satisfies the symmetric property, $$\label{e122} G_j(t, s)=G_j(b+a-t,~ b+a-s),\quad\text{for }j=1,2,\dots,p.$$ \end{lemma} \begin{proof} By the definition of $G_j(t,s)$, $(j=2,3,\dots,p-1)$, $$G_j(t,s)=\int_a^bG_{j-1}(t,r)G_1(r,s)\nabla r, \quad \text{for~all } t,s\in [a,b]_{\mathbb{T}}.$$ Clearly, $G_1(t,s)=G_1(a+b-t,a+b-s)$. Now, the proof is by induction. For $j=2$ the inequality \eqref{e122} is obvious. Next, assume that \eqref {e122} is true, for fixed $j$ $(j=1,2,\dots,p-1)$, then from \eqref{e3} we have \begin{align*} G_{j+1}(t,s) &=\int_a^{b}G_j(t,r)G_1(r,s)\nabla r\\ &=\int_a^{b}G_j(a+b-t,a+b-r)G_1(a+b-r,a+b-s)\nabla r\\ &=\int_a^{b}G_j(a+b-t,r_1)G_1(r_1,a+b-s)\nabla r_1\\ &=G_{j+1}(a+b-t,a+b-s), \end{align*} by using a transformation $r_1=a+b-r$. \end{proof} Let $D=\{v\mid v:[a,b]_{\mathbb{T}}\to \mathbb{R} \text{is continuous function}\}$. We define the operator $F_j:D\to D$ by $$(F_jv)(t)=\int_a^{b}G_j(t,s)v(s)\nabla s,\quad t\in [a,b]_{\mathbb{T}},\textup{ for } j=1,2,\dots, p-1.$$ By the construction of $F_j$ and properties of $G_j(t, s)$, it is clear that \begin{gather*} (-1)^j(F_jv)^{(\Delta\nabla)^{j}}(t)=v(t),\quad t\in [a,b]_{\mathbb{T}},\\ (F_jv)^{(\Delta\nabla)^{i}}(a)=(F_jv)^{(\Delta\nabla)^{i}}(b)=0,\quad i=0,1,\dots,j-1. \end{gather*} \begin{lemma}\label{lem5} For $t \in [a,b]_{\mathbb{T}}$, the operator $F_j$ satisfies the symmetric property $$F_jy(t)=F_jy(b+a-t)\quad\text{for }j=1,2,\dots,p-1.$$ \end{lemma} \begin{proof} By definition of $F_j$, and using the transformation $s_1=b+a-s$, \begin{align*} F_jy(t) &=\int_{a}^{b}G_j(t,s)v(s)\nabla s \\ &=\int_{a}^{b}G_j(a+b-t,a+b-s)v(s)\nabla s\\ &=\int_{a}^{b}G_j(a+b-t,s_1)v(s_1)\nabla s_1 \\ &=F_jy(b+a-t), \end{align*} from Lemma \ref{lem4}). \end{proof} By using the above transformations and lemmas, we can reduce the SBVP \eqref{e1},\eqref{e2} into SBVP \eqref{e6}-\eqref{e7} and vice-versa. Hence, we see that SBVP \eqref{e1}-\eqref{e2} has a solution if and only if the following problem has a solution: $$\label{e6} \begin{gathered} v_1^{\Delta\nabla}+f_1(t,F_{n-1}v_1,F_{m-1}v_2)=0,\quad t\in [a,b]_{\mathbb{T}} \\ v_2^{\Delta\nabla}+f_2(t,F_{n-1}v_1,F_{m-1}v_2)=0,\quad t\in [a,b]_{\mathbb{T}}, \end{gathered}$$ with boundary conditions $$\label{e7} v_1(a)=0=v_1(b),\quad _2(a)=0=v_2(b).$$ Indeed, if $(y_1,y_2)$ is a solution of \eqref{e1}-\eqref{e2}, then $(v_1=y_1^{(\Delta\nabla)^{(n-1)}},v_2=y_2^{(\Delta\nabla)^{(m-1)}})$ is a solution of \eqref{e6}-\eqref{e7}. Conversely, if $(v_1,v_2)$ is a solution of \eqref{e6}-\eqref{e7}, then $(y_1=F_{n-1}v_1,y_2=F_{m-1}v_2)$ is a solution of \eqref{e1}-\eqref{e2}. In fact, we have the representation $$y_1(t)=\int_a^bG_{n-1}(t,s)v_1(s)\nabla s,\quad y_2(t)=\int_a^bG_{m-1}(t,s)v_2(s)\nabla s,$$ where \begin{gather*} v_1(s)=\int_a^bG_1(s,\tau)f_1(\tau,F_{n-1}v_1 ,F_{m-1}v_2)\nabla \tau,\\ v_2(s)=\int_a^bG_1(s,\tau)f_2(\tau,F_{n-1}v_1 ,F_{m-1}v_2)\nabla \tau. \end{gather*} It is also noted that a solution $(v_1,v_2)$ of \eqref{e6}-\eqref{e7} is symmetric; i. e., $$v_1(t)=v_1(b+a-t)\quad\text{and}\quad v_2(t)=v_2(b+a-t), \quad t\in [a, b ]_{\mathbb{T}},$$ and it gives rise to a symmetric solution $(y_1,y_2)$ of \eqref{e1}-\eqref{e2}. \section{Existence of multiple symmetric positive solutions} In this section, we establish the existence of at least three symmetric positive solutions for \eqref{e1}-\eqref{e2}, by using Avery's generalization of the Leggett-Williams fixed point theorem. Let $B$ be a real Banach space with cone $P$. We consider the nonnegative continuous convex functionals $\gamma,\beta,\theta$ and nonnegative continuous concave functionals $\alpha,\psi$ on $P$, for nonnegative numbers $a',b',c',d'$ and $h'$, we define the following sets \begin{gather*} P(\gamma,c')=\{y\in P: \gamma(y)a'\}\neq \emptyset$and$\alpha(Ty)>a'$for \\$y\in P(\gamma,\theta,\alpha,a',b',c')$, \item[(B2)]$\{y\in Q(\gamma,\beta,\psi,h',d',c')|\beta(y)a'$provided$y\in P(\gamma,\alpha,a',c') $with$\theta(Ty)>b' $, \item[(B4)]$\beta(Ty)\frac{b'}{k_1(k_2+1-k_1)}$for all$(t,|u_{n-1}|,|w_{m-1}|)in \begin{align*} &[a+l,b-l]_{\mathbb{T}}\times[b'L_{k_1+1}^{n-1}\phi_{k_1+1}^{n-2} (\phi_{k_1+1}-\phi_{k_2+2}),\frac{b'(t_2-a)} {t_1-a}\phi_0^{n-2}(\phi_{k_1+1}-\phi_{k_2+2})\\ &+\frac{c'(t_{k}-a)}{t_0-a}\phi_0^{n-2}(\overline{\phi}_{k_1+1} +\phi_{k_2+2}) ]\times [b'L_{k_1+1}^{m-1}\phi_{k_1+1}^{m-2}(\phi_{k_1+1}-\phi_{k_2+2}),\\ &\frac{b'(t_2-a)} {t_1-a}\phi_0^{m-2}(\phi_{k_1+1}-\phi_{k_2+2}) + \frac{c'(t_{k}-a)}{t_0-a}\phi_0^{m-2}(\overline{\phi}_{k_1+1} +\phi_{k_2+2})], \end{align*} eitheri=1$or$i=2$. \item[(A3)]$|f_i(t,u_{n-1},w_{m-1})|<\frac{c'}{(t_0-a)(b-t_0)}$for all$(t,|u_{n-1}|,|w_{m-1}|)in \begin{align*} &[a,b]_{\mathbb{T}}\times [0, \frac{c'(t_{k}-a)}{t_0-a}\phi_0^{n-1}]\times [0,\frac{c'(t_{k}-a)}{t_0-a}\phi_0^{m-1}] ,\quad i=1, 2. \end{align*} \end{itemize} Then \eqref{e1}-\eqref{e2} has at least three symmetric positive solutions. \end{theorem} \begin{proof} Define a completely continuous operatorT:C_0\to C_0$by $$\label{e114} T(v_1,v_2) :=(T_1(v_1, v_2),T_2(v_1, v_2)),$$ where $$T_{i}(v_1, v_2):=\int_a^bG_1(t,s)f_i(s,F_{n-1}v_1, F_{m-1}v_2)\nabla s,\quad\text{for } i=1,2.$$ It is obvious that a fixed point of$T$is a solution of \eqref{e6}-\eqref{e7}. We seek three fixed points$(x_1, x_2), (y_1, y_2), (z_1, z_2)\in P$of$T$. First, we show that$T$is self map on$P$. Let$(v_1,v_2)\in P$, then$T_1(v_1, v_2)(t)\geq 0$,$T_2(v_1, v_2)(t)\geq 0$for$t\in[a,b]_{\mathbb{T}}$, and$T^{\Delta\nabla}_1(v_1, v_2)(t)\leq 0$,$T^{\Delta\nabla}_2(v_1, v_2)(t)\leq 0$for$t\in[a,b]_{\mathbb{T}}$. Further$G_1(t,s)$is symmetric, it follows that$T_1(v_1, v_2)(t)=T_1(v_1, v_2)(b+a-t),~T_2(v_1, v_2)(t)=T_2(v_1, v_2)(b+a-t)$, for$t\in[a, b]_{\mathbb{T}}$. Also, noting that$T_1(v_1,v_2)(a)=0=T_1(v_1, v_2)(b)$,$T_2(v_1,v_2)(a)=0=T_2(v_1, v_2)(b)$and$\|T(v_1,v_2)\|=|T_1(v_1, v_2)(t_{k})|+|T_2(v_1,v_2)(t_{k})|, we have \begin{align*} &\min_{t\in[a+k_0,b-k_0]_{\mathbb{T}}}(|T_1(v_1, v_2)(t)|+|T_2(v_1, v_2)(t)|)\\ &=\min_{t\in[a+k_0,t_{k}]_{\mathbb{T}}} (|T_1(v_1, v_2)(t)|+|T_2(v_1, v_2)(t)|)\\ &\geq \min_{t\in[a+k_0,t_{k}]_{\mathbb{T}}}\frac{t-a}{t_{k}-a}\| T(v_1,v_2)\|\\ &=\frac{k_0}{t_{k}-a}\| T(v_1,v_2)\|. \end{align*} ThusT:P\to P$. Next, for all$(v_1,v_2)\in P$, and using \eqref{e112},\eqref{e113},$\alpha(v_1,v_2)\leq \beta(v_1,v_2)$and$\| (v_1,v_2)\|\leq\frac{t_{k}-a}{t_0-a}\gamma(v_1,v_2)$. To show that$T:\overline{P(\gamma,c')}\to \overline{P(\gamma,c')}$, let$(v_1,v_2)\in \overline{P(\gamma,c')}$and hence$\|(v_1,v_2)\|\leq\frac{t_{k}-a}{t_0-a}c'$. Using Lemma \ref{lem2} and for$t\in [a,b]_{\mathbb{T}}, \begin{align*} |F_{n-1}v_1(t)| &=|\int_a^bG_{n-1}(t,s)v_1(s)\nabla s| \\ &\leq\frac{c'(t_{k}-a)}{t_0-a}\int_a^b|G_{n-1}(t,s)|\nabla s\\ &\leq\frac{c'(t_{k}-a)}{t_0-a}\phi_0^{n-2}\int_a^b \frac{(b-s)(s-a)}{(b-a)}\nabla s\\ &=\frac{c'(t_{k}-a)}{t_0-a}\phi_0^{n-1}. \end{align*} Similarly, fort\in [a,b]_{\mathbb{T}}, we have $$|F_{m-1}v_2(t)|\leq\frac{c'(t_{k}-a)}{t_0-a}\phi_0^{m-1}.$$ By condition (A3), $$\gamma(T_1(v_1, v_2),T_2(v_1, v_2))=|T_1(v_1, v_2) (t_0)|+|T_2(v_1, v_2)(t_0)|.$$ and \begin{align*} |T_1(v_1,v_2)(t_0)| &=|\int_a^bG_1(t_0,s)f_1(s,F_{n-1}v_1,F_{m-1}v_2)\nabla s|\\ &< \frac{c'}{(t_0-a)(b-t_0)}\int_a^b|G_1(t_0,s)|\nabla s =\frac{c'}{2}. \end{align*} Similarly,|T_2(v_1, v_2)(t_0)|< c'/2$, and hence$T:\overline{P(\gamma,c')}\to\overline{P(\gamma,c')}$. It is obvious that $$\{(v_1,v_2)\in P(\gamma,\theta,\alpha,b',\frac{b'(t_2-a)}{t_1-a},c')| \alpha(v_1,v_2)>b'\}\neq \emptyset .$$ Next, let$(v_1,v_2)\in P(\gamma,\theta,\alpha,b', \frac{b'(t_2-a)}{t_1-a},c')$, denote the set$D_1=[a+k_1,a+k_2]_{\mathbb{T}}\cup[b-k_2,\sigma(b)-k_1]_{\mathbb{T}}$. It follows that \begin{gather}\label{e115} |v_1(s)|,|v_2(s)|\in [b',\frac{b'(t_2-a)}{t_1-a}],\quad s\in D_1,\\ \label{e116} |v_1(s)|,|v_2(s)|\in[0,\frac{c'(t_{k}-a)}{t_0-a}],\quad s\in[a,b]_{\mathbb{T}}\setminus D_1. \end{gather} Using \eqref{e115}, \eqref{e116}, Lemma \ref{lem1} Lemma \ref{lem2}, $$|F_{n-1}v_1(s)|=|\int_{\tau\in D_1}G_{n-1}(s, \tau)v_1(\tau)\nabla \tau+\int_{\tau\in[a,b]_{\mathbb{T}}\setminus D_1}G_{n-1}(s,\tau)v_1(\tau)\nabla \tau|$$ and $$|F_{m-1}v_2(s)|=|\int_{\tau\in D_1}G_{m-1}(s, \tau)v_2(\tau)\nabla \tau+\int_{\tau\in[a,b]_{\mathbb{T}}\setminus D_1}|G_{m-1}(s,\tau)v_2(\tau)\nabla \tau|,$$ for$s\in D_1, we have \begin{align*} &|F_{n-1}v_1(s)|\\ &\leq\frac{b'(t_2-a)}{t_1-a}\int_{\tau\in D_1}|G_{n-1}(s, \tau)|\nabla \tau+\frac{c'(t_{k}-a)}{t_0-a}\int_{\tau\in[a,b]_{\mathbb{T}}\setminus D_1}|G_{n-1}(s,\tau)|\nabla \tau,\\ &\leq\frac{b'(t_2-a)}{t_1-a}\phi_0^{n-2}(\phi_{k_1+1}-\phi_{k_2+2})+ \frac{c'(t_{k}-a)}{t_0-a}\phi_0^{n-2}(\overline{\phi}_{k_1+1} +\phi_{k_2+2}) \end{align*} and \begin{align*} |F_{n-1}v_1(s)| &\geq\int_{\tau\in D_1}|G_{n-1}(s, \tau)v_1(\tau)|\nabla \tau\\ &\geq b'\int_{\tau\in D_1}|G_{n-1}(s, \tau)|\nabla \tau \\ &\geq b'L_{k_1+1}^{n-2}\phi_{k_1+1}^{n-2}(\phi_{k_1+1}-\phi_{k_2+2}). \end{align*} Similarly, $$|F_{m-1}v_2(s)|\leq\frac{b'(t_2-a)}{t_1-a}\phi_0^{m-2} (\phi_{k_1+1}-\phi_{k_2+2})+ \frac{c'(t_{k}-a)}{t_0-a} \phi_0^{m-2}(\overline{\phi}_{k_1+1}+\phi_{k_2+2})$$ and $$|F_{m-1}v_2(s)|\geq b'L_{k_1+1}^{m-2}\phi_{k_1+1}^{m-2} (\phi_{k_1+1}-\phi_{k_2+2}),~{for~s\in D_1}.$$ Applying (A2) we obtain \begin{align*} \alpha(T_1(v_1, v_2),&T_2(v_1, v_2)) =|T_1(v_1, v_2)(t_1)| +|T_2(v_1, v_2)(t_1)|\geq|T_1(v_1, v_2)(t_1)|\\ &=|\int_a^bG_1(t_1,s)f_1(s,F_{n-1}v_1, F_{m-1}v_2)\nabla s|\\ &\geq \int_{s\in D_1}|G_1(t_1,s)f_1(s,F_{n-1}v_1, F_{m-1}v_2)|\nabla s\\ &>\frac{b'}{k_1(k_2+1-k_1)}\int_{s\in D_1}|G_1(t_1,s)|\nabla s =b'. \end{align*} Similarly,\alpha(T_1(v_1, v_2),T_2(v_1, v_2))\geq|T_2(v_1, v_2)(t_1)|and from (A2) we have $$\alpha(T_1(v_1, v_2),T_2(v_1, v_2))\geq b'.$$ Clearly, \{(v_1,v_2)\in Q(\gamma,\beta,\psi,\frac{a'(t_3-a)}{t_{k}-a},a',c') |\beta(v_1,v_2)\frac{b'(t_2-a)}{t_1-a}. Using Lemma \ref{lem3}, we obtain \begin{align*} &\alpha(T_1(v_1, v_2), T_2(v_1, v_2))\\ &=|T_1(v_1, v_2)(t_1)|+|T_2(v_1, v_2)(t_1)|\\ &=\int_a^b|\frac{G_1(t_1,s)}{G_1(t_2,s)}G_1(t_2,s)f_1(s,F_{n-1}v_1, F_{m-1}v_2)|\nabla s \\ &\quad +\int_a^b|\frac{G_1(t_1,s)}{G_1(t_2,s)}G_1(t_2,s) f_2(s,F_{n-1}v_1,F_{m-1}v_2)|\nabla s \\ &\geq\frac{t_1-a}{t_2-a}\int_a^b|G_1(t_2,s)f_1(s,F_{n-1}v_1, F_{m-1}v_2)|\nabla s\\ &\quad +\frac{t_1-a}{t_2-a}\int_a^b|G_1(t_2,s)f_2(s,F_{n-1}v_1, F_{m-1}v_2)|\nabla s\\ &=\frac{t_1-a}{t_2-a}\theta(T_1(v_1, v_2),T_2(v_1, v_2)) >b'. \end{align*} Finally, we show that (B4) holds. Let (v_1,v_2)\in Q(\gamma,\beta,a',c') with \psi(T_1(v_1, v_2),T_2(v_1, v_2))<\frac{a'(t_3-a)}{t_{k}-a}.\$ In view of Lemma \ref{lem3}, we have \begin{align*} \beta(T_1(v_1, v_2),T_2(v_1, v_2)) &=|T_1(v_1, v_2)(t_{k})|+|T_2(v_1, v_2)(t_{k})|\\ &=\int_a^b|\frac{G_1(t_{k},s)}{G_1(t_3,s)}G_1(t_3,s) f_1(s,F_{n-1}v_1,F_{m-1}v_2)|\nabla s \\ &\quad +\int_a^b|\frac{G_1(t_{k},s)}{G_1(t_3,s)}G_1(t_3,s) f_2(s,F_{n-1}v_1,F_{m-1}v_2)|\nabla s \\ &\leq\frac{t_{k}-a}{t_3-a}\int_a^b|G_1(t_3,s)f_1(s,F_{n-1}v_1, F_{m-1}v_2)|\nabla s \\ &\quad +\frac{t_{k}-a}{t_3-a}\int_a^b|G_1(t_3,s)f_2(s, F_{n-1}v_1,F_{m-1}v_2)|\nabla s\\ &=\frac{t_{k}-a}{t_3-a}\psi(T_1(v_1, v_2),T_2(v_1, v_2))