\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 04, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/04\hfil Nonlinear scalar two-point boundary-value problems] {Nonlinear scalar two-point boundary-value problems on time scales} \author[R. I. B. Kalhorn, J. Rodr\'iguez\hfil EJDE-2010/04\hfilneg] {Rebecca I. B. Kalhorn, Jes\'us Rodr\'iguez} % in alphabetical order \address{Rebecca I. B. Kalhorn \newline Department of Mathematics\\ North Carolina State University\\ Box 8205\\ Raleigh, NC 7695-8205, USA} \email{rkalhorn@gmail.com} \address{Jes\'us Rodr\'iguez \newline Department of Mathematics\\ North Carolina State University, Box 8205\\ Raleigh, NC 7695-8205, USA} \email{rodrigu@ncsu.edu} \thanks{Submitted March 17, 2009. Published January 6, 2010.} \subjclass[2000]{39B99, 39A10} \keywords{Boundary value problems; time scales; Schauder fixed point theorem} \begin{abstract} We establish sufficient conditions for the solvability of scalar nonlinear boundary-value problems on time scales. Our attention will be focused on problems where the solution space for the corresponding linear homogeneous boundary-value problem is nontrivial. As a consequence of our results we are able to provide easily verifiable conditions for the existence of periodic behavior for dynamic equations on time scales. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \section{Introduction} This paper is devoted to the study of scalar nonlinear boundary-value problems on time scales. We examine the problem $$\label{eqn:scalar_eq} u^{\Delta^n}(t)+a_{n-1}(t)u^{\Delta^{n-1}}(t) +\dots+a_0(t)u(t)=q(t)+g(u(t)),\quad t\in[a,b]_\mathbb{T}$$ subject to $$\label{eqn:scalar_bc} \sum_{j=1}^{n}b_{ij}u^{\Delta^{j-1}}(a)+\sum_{j=1}^{n}d_{ij}u^{\Delta^{j-1}}(b)=0,$$ for $i=1,2,\dots,n$. Throughout this paper we will assume that $\mathbb{T}$ is a time scale and $[a,b]_\mathbb{T}\subset\mathbb{T}^{\kappa^n}$ where $[a,b]_\mathbb{T}$ will denote $\{t\in\mathbb{T}:a\leq t\leq b\}$. The functions $a_0, a_1, \dots, a_{n-1}$ and $q$ are real-valued, rd-continuous functions defined on $\mathbb{T}$. The nonlinear term $g$ is continuous, real-valued, and defined on $\mathbb{R}$. We will assume the solution space for the corresponding homogeneous boundary-value problem, namely, $$\label{eqn:scalar_homo_eq} u^{\Delta^n}(t)+a_{n-1}(t)u^{\Delta^{n-1}}(t)+\dots+a_0(t)u(t)=0,\quad t\in[a,b]_\mathbb{T}$$ subject to $$\label{eqn:scalar_homo_bc} \sum_{j=1}^{n}b_{ij}u^{\Delta^{j-1}}(a) +\sum_{j=1}^{n}d_{ij}u^{\Delta^{j-1}}(b)=0,\quad\text{for } i=1,2,\dots,n,$$ has dimension 1. Let $A(t)$ be the $n\times n$ matrix-valued function given by $$A(t)=\begin{bmatrix} 0 & 1 & 0 &\dots & 0\\0 & 0 & 1 & & 0\\ \vdots & \vdots & &\ddots & \vdots\\ 0 & 0 & 0 &\dots & 1\\ -a_0(t) & -a_1(t) & -a_2(t) & \dots & -a_{n-1}(t) \end{bmatrix}.$$ Clearly $A$ is rd-continuous, and we assume $A$ is also regressive. Let the matrices $B$ and $D$ be defined by $B=(b_{ij})$ and $D=(d_{ij})$. It should be observed that linear independence of the boundary conditions is equivalent to the matrix $[B|D]$ having full rank. To analyze the boundary-value problem \eqref{eqn:scalar_eq}--\eqref{eqn:scalar_bc} we will look at the equivalent $n\times n$ system, $$\label{eqn:scalar_system_eq} x^\Delta(t) = A(t)x(t)+h(t)+f(x(t)),\quad t\in [a,b]_\mathbb{T}$$ subject to $$\label{eqn:scalar_system_bc} Bx(a)+Dx(b)=0$$ where $$[f(x)]_i=\begin{cases} 0&\text{for }i=1,2,\dots n-1\\g([x]_1)&\text{for }i=n \end{cases}$$ and $$[h(t)]_i=\begin{cases} 0&\text{for }i=1,2,\dots n-1\\ q(t)&\text{for }i=n \end{cases}.$$ Note that the solution space of $$\label{eqn:scalar_system_homo_eq} x^\Delta(t) = A(t)x(t),\quad t\in [a,b]_\mathbb{T}$$ subject to $$\label{eqn:scalar_system_homo_bc} Bx(a)+Dx(b)=0$$ has dimension one as a result of the assumption on (\ref{eqn:scalar_homo_eq})--(\ref{eqn:scalar_homo_bc}). Through use of the Lyapunov-Schmidt Procedure conditions will be established to guarantee the existence of solutions to the boundary-value problem (\ref{eqn:scalar_system_eq})--(\ref{eqn:scalar_system_bc}) and thus \eqref{eqn:scalar_eq}--\eqref{eqn:scalar_bc}. We will pay particular attention to second-order equations subject to periodic boundary conditions. We obtain results which significantly extend previous work by Etheridge and Rodr\'{i}guez concerning the periodic behavior of nonlinear discrete dynamical systems\cite{eth_periodic}. \section{Preliminaries} The notation and preliminary results presented here are a straightforward generalization of previous work in differential equations and discrete time systems \cite{eth_periodic, paddy_continuous, paddy_scalar, paddy_discrete, eth_perturbed, eth_scalar, rod_resonant}. We provide references concerning general information on time scales\cite{time_scales, survey, bohner} as well as boundary-value problems\cite{henderson, stehlik}. Let $$X=\{x\in C[a,b]_\mathbb{T}:Bx(a)+Dx(b)=0\},$$ and $$Y=C_{\textup{rd}}[a,b]_\mathbb{T}$$ where $C_\text{rd}[a,b]_\mathbb{T}$ denotes the space of rd-continuous $\mathbb{R}^n$-valued maps on $[a,b]_\mathbb{T}$, and $C[a,b]_\mathbb{T}$ denotes the subspace of $C_\text{rd}[a,b]_\mathbb{T}$ where the maps are continuous. $|\cdot|$ will denote the Euclidean norm on $\mathbb{R}^n$. The operator norm will be used for matrices, and the supremum norm will be used for $x\in Y\cup X$, that is, $$\|x\|=\sup_{t\in[a,b]_\mathbb{T}}|x(t)|.$$ It is clear that $X$ and $Y$ are Banach spaces with this norm. We define the norm of a product space, $V_1\times V_2\times\dots\times V_m$, by $$\|(v_1,v_2,\dots,v_m)\|=\sum_{i=1}^m\|v_i\|_i$$ where $\|\cdot\|_i$ denotes the norm on $V_i$. We define the operator $L:D(L)\to Y$ where $D(L)=X\cap C^1_{\textup{rd}}([a,b]_{\mathbb{T}} \to\mathbb{R}^n)$ by $$(Lx)(t)=x^\Delta(t)-A(t)x(t),\quad t\in[a,b]_\mathbb{T}$$ and the operator $F:X\to Y$ by $$(Fx)(t)=f(x(t)),\quad t\in[a,b]_\mathbb{T}.$$ Clearly $x$ is a solution to (\ref{eqn:scalar_system_eq})--(\ref{eqn:scalar_system_bc}) if and only if $Lx=h+Fx$. $\Phi$ will denote the fundamental matrix solution for $x^\Delta(t)=A(t)x(t)$, $t\in[a,b]_\mathbb{T}$ where $\Phi(a)=I$. \begin{proposition} \label{prop:dimensions_same} The solution space for the homogeneous boundary-value problem \eqref{eqn:scalar_system_homo_eq}--\eqref{eqn:scalar_system_homo_bc} and the kernel of $(B+D\Phi(b))$ have the same dimension. \end{proposition} \begin{proof} The the solution space of \eqref{eqn:scalar_system_homo_eq}--\eqref{eqn:scalar_system_homo_bc} and kernel of $L$ have the same dimension. $x\in \ker (L)$ if and only if $x^\Delta(t)=A(t)x(t)$, $t\in[a,b]_\mathbb{T}$ and $x$ satisfies the boundary conditions. This is true if and only if there is a $c$ in $\mathbb{R}^n$ such that $x(t)=\Phi(t) c$ for all $t\in[a,b]_\mathbb{T}$ and $Bc+D\Phi(b)c=0$. It follows that the kernel of $L$ and the kernel of ($B+D\Phi(b)$) have the same dimension. \end{proof} Let $d$ be a unit vector which spans the kernel of $(B+D\Phi(b))$. Define $S: [a,b]_\mathbb{T}\to \mathbb{R}^n$ by $$S(t)=\Phi(t)d.$$ The following result is obvious. \begin{corollary} label{coro1} The kernel of $L$ consists of $x$ such that $x(t)=S(t)\alpha$ for some real number $\alpha$. \end{corollary} \section{Main Result} We will now construct projections onto the kernel and image of $L$ in order to use the Lyapunov-Schmidt Procedure \cite{chow, eth_periodic}. Define $P:X\to X$ by $$(Px)(t)=S(t)d^Tx(a),\quad t\in[a,b]_\mathbb{T}.$$ \begin{proposition} \label{prop2} $P$ is a projection onto the kernel of $L$. \end{proposition} \begin{proof} The fact that $P$ is a bounded linear map is self-evident. The fact that $P$ is idempotent can be shown through direct computation. It remains to be shown that $\mathop{\rm Im}(P) = \ker(L)$. Let $x\in X$. $(Px)(t)=S(t)d^Tx(a)=S(t)\alpha$ where $\alpha =d^Tx(a)$. Therefore $\mathop{\rm Im}(P)\subset \ker(L)$. Let $x\in \ker(L)$. There exists a $\beta\in\mathbb{R}$ such that $x(t)=S(t)\beta$. $(Px)(t)=S(t)d^Tx(a)=S(t)d^TS(a)\beta=S(t)\beta= x(t)$. Therefore $\ker(L) \subset \mathop{\rm Im}(P)$. \end{proof} Let $k$ be a vector that spans the kernel of $((B+D\Phi(b))^T)$. Define the map $\Psi: [a,b]_\mathbb{T}\to \mathbb{R}^n$ by $$\Psi(t)=[D\Phi(b)\Phi^{-1}(\sigma(t))]^Tk,\quad t\in[a,b]_\mathbb{T}.$$ \begin{proposition} \label{prop3} $y$ is in the image of $L$ if and only if $\int^b_ay^T(\tau)\Psi(\tau)\Delta\tau=0$. \end{proposition} \begin{proof} Using the variation of constants formula \cite{time_scales} and the boundary conditions it is clear that $y\in \mathop{\rm Im}(L)$ if and only if there exists $x\in X$ such that $(B+D\Phi(b))x(a)+D\int_a^b\Phi(b)\Phi^{-1} (\sigma(\tau))y(\tau)\Delta\tau=0$, which is equivalent to $$-x^T(a)(B+D\Phi(b))^T =\Big[\int_a^bD\Phi(b)\Phi^{-1}(\sigma(\tau))y(\tau)\Big]^T\Delta\tau.$$ This holds if and only if $\int_a^b\big[D\Phi(b)\Phi^{-1}(\sigma(\tau))y(\tau)\big]^T\Delta\tau \beta=0$ where $\beta$ is an element of the kernel of $(B+D\Phi(b))^T$ and therefore must be a multiple of $k$. Therefore, $\int_a^by^T(\tau)\Psi(\tau)\Delta\tau=0$. \end{proof} Define the operator $W$ from $Y$ into $Y$ by $$(Wy)(t)=\Psi(t)\Big[\int^b_a|\Psi(\tau)|^2\Delta\tau\Big]^{-1} \int^b_a\Psi^T(\tau)y(\tau)\Delta\tau, \quad t\in[a,b]_\mathbb{T}.$$ \begin{proposition} \label{prop4} $E$, defined by $E=I-W$, is a projection onto the image of $L$. \end{proposition} \begin{proof} First we will show that $E$ is a projection. Since $W$ is a bounded linear map $E$ is also a bounded map. To prove $E^2=E$ it will be sufficient to show that $W^2=W$. Let $y\in Y$. \begin{align*} &(W(Wy))(t) \\ &=W\Big(\Psi(\cdot)\Big[\int^b_a|\Psi(\tau)|^2\Delta\tau\Big]^{-1} \int^b_a\Psi^T(\tau)y(\tau)\Delta\tau\Big)(t),\quad t\in[a,b]_\mathbb{T}\\ &=\Psi(t)\Big[\int^b_a|\Psi(\tau)|^2\Delta\tau\Big]^{-1} \int^b_a\Psi^T(\tau)\Psi(\tau)\Delta\tau \Big[\int^b_a|\Psi(\nu)|^2\Delta\nu\Big]^{-1} \int^b_a\Psi^T(\nu)y(\nu)\Delta\nu \\ &=\Psi(t)\Big[\int^b_a|\Psi(\nu)|^2\Delta\nu\Big]^{-1} \int^b_a\Psi^T(\nu)y(\nu)\Delta\nu =(Wy)(t). \end{align*} Finally we will prove that $\mathop{\rm Im}(E) = \mathop{\rm Im}(L)$. It is clear that $Ey\in \mathop{\rm Im}(E)$. \begin{align*} &\int_a^b\Psi^T(\tau)(Ey)(\tau)\Delta\tau\\ &=\int_a^b\Psi^T(\tau)(y-Wy)(\tau)\Delta\tau\\ &=\int_a^b\Psi^T(\tau)y(\tau)\Delta\tau- \int_a^b\Psi^T(\tau)\Psi(\tau)\Delta\tau \Big[\int^b_a|\Psi(\nu)|^2\Delta\nu\Big]^{-1} \int^b_a\Psi^T(\nu)y(\nu)\Delta\nu =0. \end{align*} Therefore $Ey\in \mathop{\rm Im}(L)$, and $\mathop{\rm Im}(E) \subset \mathop{\rm Im}(L)$. Now suppose $y\in \mathop{\rm Im}(L)$. $$(Ey)(t)=y(t)-\Psi(t)\Big[\int_a^b|\Psi(\tau)|^2\Delta\tau\Big]^{-1} \int_a^b\Psi^T(\tau)y(\tau)\Delta\tau =y(t),$$ for all $t\in[a,b]_\mathbb{T}$. Therefore $y\in \mathop{\rm Im}(E)$, and $\mathop{\rm Im}(L) \subset \mathop{\rm Im}(E)$. \end{proof} By constructing the projections $P$ and $E$ we are now able to analyze the existence of solutions to (\ref{eqn:scalar_system_eq})--(\ref{eqn:scalar_system_bc}) using the classic Lyapunov-Schmidt Procedure. We provide a self-contained presentation of our approach, but offer references \cite{chow, hale, rod_resonant, rod_galerkin,sweet} for a more general formulation and for applications to differential and difference equations. We can utilize the fact that $P$ and $E$ are projections and write $$X= \mathop{\rm Im}(P) \oplus \mathop{\rm Im}(I-P)\quad\text{and}\quad Y= \mathop{\rm Im}(I-E) \oplus \mathop{\rm Im}(E).$$ For all $x\in X$ there exists $u\in \ker(L)$ and $v\in \mathop{\rm Im}(I-P)$ such that $x=u+v$. It is clear that $L:\mathop{\rm Im}(I-P)\cap D(L)\to \mathop{\rm Im}(L)$ is a bijection, and therefore there exists a bounded linear map $M: \mathop{\rm Im}(L)\to \mathop{\rm Im}(I-P)\cap D(L)$ such that $$LMy=y,\forall y\in \mathop{\rm Im}(L)\quad\text{and}\quad MLx=v,\forall x\in X.$$ Define $H_1:\mathbb{R}\times\mathop{\rm Im}(I-P)\to\mathbb{R}$ by $$H_1(\alpha,v)=\alpha-\int_a^bg([\alpha S(\tau)+Mh(\tau) +MEF(S\alpha+v)(\tau)]_1)[\Psi(\tau)]_n\Delta\tau,$$ Define $H_2:\mathbb{R}\times\mathop{\rm Im}(I-P)\to \mathop{\rm Im}(I-P)$ by $$H_2(\alpha,v)=Mh+MEF(S\alpha+v).$$ Define $H:\mathbb{R}\times\mathop{\rm Im}(I-P)\to\mathbb{R} \times\mathop{\rm Im}(I-P)$ by $$H(\alpha,v)=(H_1(\alpha,v),H_2(\alpha,v)).$$ \begin{proposition}\label{H} $Lx=h+Fx$ if and only if there exists $(\alpha,v)\in\mathbb{R}\times\mathop{\rm Im}(I-P)$ such that $H(\alpha,v)=(\alpha,v)$. \end{proposition} \begin{proof} Let $x\in X$. There exist $\alpha\in\mathbb{R}$ and $v\in$ $\mathop{\rm Im}(I-P)$ such that $x=S\alpha+v$ and \begin{align*} Lx=h+Fx &\Longleftrightarrow \left\{\begin{array}{l}E[Lx-h-Fx]=0\\(I-E)[Lx-h-Fx]=0\end{array}\right.\\ &\Longleftrightarrow \left\{\begin{array}{l}Lv-h-EF(x)=0\\(I-E)F(x)=0\end{array}\right.\\ &\Longleftrightarrow \left\{\begin{array}{l}v=Mh+MEF(S\alpha+v)\\ \int_a^bg([\alpha S(\tau)+Mh(\tau) +MEF(S\alpha+v)(\tau)]_1)[\Psi(\tau)]_n\Delta\tau=0\end{array}\right.\\ &\Longleftrightarrow H(\alpha,v)=(\alpha,v). \end{align*} \end{proof} Define $g(\pm\infty)$ as follows, provided the corresponding limits exist, $$\lim_{x\to\pm\infty}g(x)=g(\pm\infty).$$ \begin{proposition}\label{scalar_convergence} Assume $g$ is continuous, $g(\infty)$ and $g(-\infty)$ exist, $[S(t)]_1$ is of one sign, and $g(\infty)g(-\infty)\int_a^b[\Psi(\tau)]_n\Delta\tau\neq0$. Then $$\int_a^bg([\pm\alpha S(\tau)+Mh(\tau) +MEFx(\tau)]_1)[\Psi(\tau)]_n\Delta\tau\to g(\pm\infty)\int_a^b[\Psi(\tau)]_n\Delta\tau$$ as $\alpha\to\infty$. \end{proposition} \begin{proof} We will show that $$\int_a^bg([\alpha S(\tau)+Mh(\tau)+MEFx(\tau)]_1) [\Psi(\tau)]_n\Delta\tau\to g(\infty)\int_a^b[\Psi(\tau)]_n\Delta\tau$$ as $\alpha\to\infty$. The proof for the corresponding result with the opposite sign follows an analogous argument. Let $\epsilon>0$. Since $Mh$ and $MEF$ are bounded on $[a,b]_\mathbb{T}$ and $S$ achieves its minimum on the set there exists $\alpha_0>0$ such that for all $\alpha>\alpha_0$ $$|g(\infty)-g([\alpha S(t)+Mh(t)+MEFx(t)]_1)|<\epsilon.$$ Let $\alpha>\alpha_0$. Then \begin{align*} &\Big|g(\infty)\int_a^b[\Psi(\tau)]_n\Delta\tau - \int_a^bg([\alpha S(\tau)+Mh(\tau)+MEFx(\tau)]_1)[\Psi(\tau)]_n\Delta\tau\Big|\\ &\leq\int_a^b\left|g(\infty)-g([\alpha S(\tau)+Mh(\tau) +MEFx(\tau)]_1)[\Psi(\tau)]_n\right|\Delta\tau\\ &\leq \epsilon\|\Psi\|(b-a). \end{align*} Therefore, $\int_a^bg([\pm\alpha S(\tau)+Mh(\tau) +MEFx(\tau)]_1)[\Psi(\tau)]_n\Delta\tau\to g(\pm\infty)\int_a^b[\Psi(\tau)]_n\Delta\tau$ as $\alpha\to\infty$. \end{proof} \begin{theorem}\label{thm:main} Suppose that the kernel of $(B+D\Phi(b))$ is one dimensional. If \begin{enumerate} \item[(i)] $[S(t)]_1$ is of one sign for all $t\in[a,b]_\mathbb{T}$ , \item[(ii)] $g:\mathbb{R}\to\mathbb{R}$ is continuous, \item[(iii)] $g(\infty)$ and $g(-\infty)$ exist, \item[(iv)] $g(\infty)g(-\infty)\big|\int_a^b[\Psi(\tau)]_n\Delta\tau\big|<0$, and \item[(v)] $\int^b_ah^T(\tau)\Psi(\tau)\Delta\tau=0$ \end{enumerate} then there is at least one solution to the boundary-value problem \eqref{eqn:scalar_eq}--\eqref{eqn:scalar_bc}. \end{theorem} \begin{proof} For simplicity we will assume that $g(\infty)>g(-\infty)$ and $\int_a^b[\Psi(\tau)]_n\Delta\tau>0$. Let $r=\sup_{z\in\mathbb{R}}|g(z)|$. Using Proposition~\ref{scalar_convergence} there is an $\alpha_0>0$ such that for $\alpha>\alpha_0$ \begin{gather*} \int_a^bg([S(\tau)\alpha+Mh(\tau)+MEF(S\alpha+v)(\tau)]_1) [\Psi(\tau)]_n\Delta\tau>0,\\ \int_a^bg([S(\tau)(-\alpha)+Mh(\tau)+MEF(S\alpha+v)(\tau)]_1) [\Psi(\tau)]_n\Delta\tau<0 \end{gather*} for $v\in \mathop{\rm Im}(I-P)$. We now use Schauder's Fixed Point Theorem to prove the existence of a solution to (\ref{eqn:scalar_system_eq})--(\ref{eqn:scalar_system_bc}). Let $$\mathcal{B}=\{(v,\alpha):\|v\|\leq\|Mh\|+\|ME\|r, \quad\text{and}\quad |\alpha|\leq\delta$$ where $\delta=\alpha_0+r(b-a)\|\Psi\|\}$. Note that $$\big|\int_a^bg([S(\tau)(-\alpha)+Mh(\tau)+MEF(S\alpha+v) (\tau)]_1)[\Psi(\tau)]_n\Delta\tau\big|\leq r(b-a)\|\Psi\|.$$ For $\alpha\in[0,\delta]$, we have \begin{gather*} -\delta\leq-r(b-a)\|\Psi\|\leq H_1(\alpha,v)\leq\alpha\leq\delta,\\ -\delta\leq-\alpha\leq H_1(-\alpha,v)\leq r(b-a)\|\Psi\|\leq\delta. \end{gather*} Now let $(v,\alpha)\in\mathcal{B}$. Then $$\|H_2(v,\alpha)\|=\|Mh+MEF(S\alpha+v)\|\leq\|Mh\|+\|ME\|r.$$ Since $H(\mathcal{B})\subset\mathcal{B}$ by the Schauder fixed point theorem there is at least one fixed point of $H$ in $\mathcal{B}$. If $(\hat{\alpha},\hat{v})$ is this fixed point, then $\hat{v}=Mh+MEF\hat{v}$ and $\int_a^bg([\hat{\alpha}S(\tau)+Mh(\tau) +MEF(\hat{\alpha}S+\hat{v})(\tau)]_1)[\Psi(\tau)]_n=0$. By Proposition \ref{H}, $L(\hat{\alpha}S+\hat{v})=h+F(\hat{\alpha}S+\hat{v})$, and therefore the boundary-value problem (\ref{eqn:scalar_system_eq})--(\ref{eqn:scalar_system_bc}) has at least one solution. Thus \eqref{eqn:scalar_eq}--\eqref{eqn:scalar_bc} has at least one solution. \end{proof} \section{Periodic Boundary Conditions} In this section we establish the existence of solutions to periodic boundary-value problems. We consider $$\label{eqn:periodic_eq} u^{\Delta\Delta}(t)+\beta u^\Delta(t)+\gamma u(t)=q(t)+g(u(t)) \quad t\in[a,b]_\mathbb{T}$$ subject to $$\label{eqn:periodic_bc} u(a)-u(a+T)=0\quad\text{and}\quad u^\Delta(a)-u^\Delta(a+T)=0$$ where $[a,a+T]_\mathbb{T}\subset\mathbb{T}^{\kappa^2}$ and $\beta,\gamma\in\mathbb{R}$ where $\gamma\mu-\beta$ is regressive. We will assume that the solution space of $$\label{eqn:periodic_eq_hom} u^{\Delta\Delta}(t)+\beta u^\Delta(t)+\gamma u(t)=0\quad t\in[a,a+T]_\mathbb{T}$$ subject to $$\label{eqn:periodic_bc_hom} u(a)-u(a+T)=0\quad\text{and}\quad u^\Delta(a)-u^\Delta(a+T)=0$$ is one-dimensional. Let $$A=\begin{bmatrix}0&1\\-\gamma&-\beta\end{bmatrix}.$$ It is easily verified that the kernel of $(I-\Phi(b))$ is one dimensional if and only if $A$ has at least one zero eigenvalue. First suppose $A$ has real distinct eigenvalues, zero and $\lambda$. Now the solution to the corresponding homogeneous problem is $u(t)=c_1+c_2e_\lambda(t,a)$, where $e_\lambda(\cdot,a)$ denotes the time scale exponential function \cite{time_scales}. If we impose the boundary conditions we find that the solution space of this scalar homogeneous boundary-value problem is spanned by $u(t)=1$ for $t\in[a,a+T]_\mathbb{T}$. Consequently the constant function $[1,0]^T$ spans $\ker(L)$. Now suppose $A$ has a repeated eigenvalue of zero. The solution to the corresponding homogeneous problem is $u(t)=c_1+c_2t$. If we impose the boundary conditions we find that the solution space of this scalar homogeneous boundary-value problem is spanned by $u(t)=1$ for $t\in[a,a+T]_\mathbb{T}$. Consequently the constant function $[1,0]^T$ spans the $\ker(L)$ in this case as well. We can now say that the solutions to the corresponding homogeneous boundary-value problem of \eqref{eqn:periodic_eq}--\eqref{eqn:periodic_bc} are real multiples of $[1,0]^T$. Therefore, $[S(t)]_1$ is of one sign for all $t\in[a,a+T]_\mathbb{T}$. \begin{theorem}\label{thm:periodic} If $$u^{\Delta\Delta}(t)+\beta u^\Delta(t)+\gamma u(t) =q(t)\quad t\in[a,a+T]_\mathbb{T}$$ subject to $$u(a)-u(a+T)=0\quad\text{and}\quad u^\Delta(a)-u^\Delta(a+T)=0$$ has a solution and $g(\infty)$ and $g(-\infty)$ exist where $g(\infty)g(-\infty)<0$ then there is at least one solution to equation \eqref{eqn:periodic_eq}--\eqref{eqn:periodic_bc}. \end{theorem} The proof of this theorem follows from Theorem \ref{thm:main}. It is easy to verify that the most significant results in Etheridge and Rodr\'{i}guez \cite{eth_periodic} are a direct consequence of Theorem \ref{thm:periodic}. \begin{corollary} \label{coro2} Suppose the conditions in Theorem \ref{thm:periodic} are satisfied. If \begin{enumerate} \item[(i)] $q$ is periodic with period $T$ \item[(ii)] $\mathbb{T}$ is a periodic time scale with period $T$, meaning if $t\in\mathbb{T}$ then $t+T\in\mathbb{T}$ \end{enumerate} then there exists at least one periodic solution to equation \eqref{eqn:periodic_eq}--\eqref{eqn:periodic_bc}. \end{corollary} \begin{proof} Let $x$ be a solution to \eqref{eqn:periodic_eq}--\eqref{eqn:periodic_bc}. Since $g$ is bounded and $q$ is periodic it is clear that the solution $x$ exists on all of $\mathbb{T}$. Let $x(t+T)=y(t)$. $y$ satisfies the dynamic equation \eqref{eqn:periodic_eq}, $y(a)=x(a+T)=x(a)$, and $y^\Delta(a)=x^\Delta(a+T)=x^\Delta(a)$. Therefore by uniqueness $x(t)=x(t+T)$. \end{proof} \section{Example} In this section we examine the following second-order nonlinear boundary-value problem on several time scales. consider $$\label{eqn:example_eq} u^{\Delta\Delta}(t)+\beta u^\Delta(t)+\gamma u(t)=g(u(t))\quad t\in[a,b]_\mathbb{T}$$ subject to $$\label{eqn:example_bc} B\left[\begin{array}{c}u(a)\\ u^\Delta(a)\end{array}\right] +D\left[\begin{array}{c}u(b)\\ u^\Delta(b) \end{array}\right]=0$$ where $\beta,\gamma\in\mathbb{R}$ and $\gamma\mu-\beta$ is regressive, $[a,b]_\mathbb{T}\in\mathbb{T}^{\kappa^2}$, $B$ and $D$ are $2\times2$ real matrices, and $g:\mathbb{R}\to\mathbb{R}$ is continuous. The scalar boundary-value problem (\ref{eqn:example_eq})--(\ref{eqn:example_bc}) is equivalent to the $2\times 2$ system $$\label{eqn:example_system_eq} x^\Delta(t)=Ax(t)+f(x(t))\quad t\in[a,b]_\mathbb{T}$$ subject to $$\label{eqn:example_system_bc} Bx(a)+Dx(b)=0$$ where $$A=\begin{bmatrix}0&1\\ -\gamma&-\beta\end{bmatrix},\quad f(x)=\begin{bmatrix}0\\ g(x_1)\end{bmatrix}, \quad x=\begin{bmatrix}u\\ u^\Delta \end{bmatrix}\,.$$ Suppose $d$ is the vector that spans the kernel of $(B+D\Phi(b))$ and $A$ has real, distinct eigenvalues, $\lambda_1$ and $\lambda_2$, where $\lambda_1>\lambda_2$ and both are positively regressive; i.e., $1+\lambda_k\mu>0$. Further assume that the eigenpairs for $A$ are given by $(\lambda_1,v)$ and $(\lambda_2,w)$. Let $$\hat{\Phi}(t)=\begin{bmatrix} v_1e_{\lambda_1}(t,a)&w_1e_{\lambda_2}(t,a)\\ v_2e_{\lambda_1}(t,a)&w_2e_{\lambda_2}(t,a) \end{bmatrix}.$$ It is clear that $$S(t)=\hat{\Phi}(t)\hat{\Phi}^{-1}(a)d =\hat{\Phi}(t)\begin{bmatrix}d_1\\d_2\end{bmatrix} =\begin{bmatrix} v_1d_1e_{\lambda_1}(t,a)+w_1d_2e_{\lambda_2}(t,a)\\ v_2d_1e_{\lambda_1}(t,a)+w_2d_2e_{\lambda_2}(t,a) \end{bmatrix}.$$ We will provide conditions under which $S_1$ will be of one sign. It is clear that if $v_1$, $w_1$, $d_1$, or $d_2$ are zero then $S_1(t)$ is either identically zero or of one sign. Now we investigate the case when $v_1$, $w_1$, $d_1$, and $d_2$ are all nonzero. $S_1(t)$ will be of one sign on $[a,b]_\mathbb{T}$ if and only if $v_1d_1e_{\lambda_1}(t,a)+w_1d_2e_{\lambda_2}(t,a)$ is of one sign for all $t\in[a,b]_\mathbb{T}$. This holds when either $$\frac{e_{\lambda_1}(t,a)}{e_{\lambda_2}(t,a)}>-\frac{w_1d_2}{v_1d_1}, \quad\text{for all }t\in[a,b]_\mathbb{T}$$ or $$\frac{e_{\lambda_1}(t,a)}{e_{\lambda_2}(t,a)}<-\frac{w_1d_2}{v_1d_1},\quad \text{for all }t\in[a,b]_\mathbb{T}.$$ It is easy to see that ${\dfrac{e_{\lambda_1}(t,a)}{e_{\lambda_2}(t,a)}>1}$ for any time scale. To obtain further results we consider specific time scales. The first time scale we will discuss is given by $$\mathbb{T}_1=\big\{[1-\frac{1}{2^{2n}},1-\frac{1}{2^{2n+1}}]: n=0,1,2,\dots\big\}\cup\{1\}.$$ For simplicity we assume that $a=0$ and $b=1$. $$e_{\lambda_k}(t,0) = \exp\Big\{\lambda_k\Big[t-\sum_{i=0}^{l-1}\frac{1}{2^{2i+2}}\Big]\Big\} \prod_{i=0}^{l-1}(1+\frac{1}{2^{2i+2}}\lambda_k)$$ where $t\in\big[1-\frac{1}{2^{2l}},1-\frac{1}{2^{2l+1}}\big]$ and $k=1,2$. Let $t\in[1-\frac{1}{2^{2l}},1-\frac{1}{2^{2l+1}}]$ where $l\in\mathbb{Z}^+\cup\{0\}$. Observe that \begin{align*} 1<\frac{e_{\lambda_1}(t,0)}{e_{\lambda_2}(t,0)} &=\exp \Big\{(\lambda_1-\lambda_2) \Big[t-\sum_{i=0}^{l-1}\frac{1}{2^{2i+2}}\Big]\Big\}\prod_{i=0}^{l-1} \frac{(1+\frac{1}{2^{2i+2}}\lambda_1)}{(1+\frac{1}{2^{2i+2}}\lambda_2)}\\ &<\exp \Big\{(\lambda_1-\lambda_2) \Big[1-\sum_{i=0}^\infty\frac{1}{2^{2i+2}}\Big]\Big\} \big(\frac{1+\lambda_1}{1+\lambda_2}\big)^l\\ &=\exp \big\{(\lambda_1-\lambda_2)(\frac{1}{3})\big\} \big(\frac{1+\lambda_1}{1+\lambda_2}\big)^l. \end{align*} Therefore, $S_1(t)$ will be of one sign on $[0,1]$ when $$1>-\frac{w_1d_2}{v_1d_1}\quad\text{or}\quad \exp \big\{(\lambda_1-\lambda_2)\big(\frac{1}{3}\big)\big\} \big(\frac{1+\lambda_1}{1+\lambda_2}\big)^l<-\frac{w_1d_2}{v_1d_1} \quad\text{for } l=0,1,2\dots.$$ Now we consider the time scale $$\mathbb{T}_2=\{[2n,2n+1]:n=0,1,2,\dots\}.$$ Let $a=0$ and $b>0$ where $b\in[1-\frac{1}{2^{2N}},1-\frac{1}{2^{2N+1}}]$ where $N\in\mathbb{Z}^+\cup\{0\}$. $$e_{\lambda_k}(t,0)=\exp \{\lambda_k(t-l)\}(1+\lambda_k)^l$$ where $t\in[2l,2l+1]$ and $k=1,2$. Let $t\in[1-\frac{1}{2^{2l}},1-\frac{1}{2^{2l+1}}]$ where $l\in\mathbb{Z}^+\cup\{0\}$. Note that \begin{align*} \exp \{(\lambda_1-\lambda_2)(b-N)\}(\frac{1+\lambda_1}{1+\lambda_2})^N & \geq\frac{e_{\lambda_1}(t,0)}{e_{\lambda_2}(t,0)}\\ &=\exp \{(\lambda_1-\lambda_2)(t-l)\} \big(\frac{1+\lambda_1}{1+\lambda_2}\big)^l>1. \end{align*} Therefore, $S_1(t)$ will be of one sign on $[0,b]$ when $$1>-\frac{w_1d_2}{v_1d_1}\quad\text{or}\quad \exp\{(\lambda_1-\lambda_2)(b-N)\} \big(\frac{1+\lambda_1}{1+\lambda_2}\big)^N <-\frac{w_1d_2}{v_1d_1}.$$ Finally consider the time scale $$\mathbb{T}_3=\{2^n:n=0,1,2,\dots\}.$$ Let $a=1$ and $b=2^N$ where $N\in\mathbb{Z}^+$. $$e_{\lambda_k}(t,1)=\prod_{i=0}^{l-1}(1+2^i\lambda_k)$$ where $t=2^l$ and $k=1,2$. Let $t=2^l$ where $l\in\mathbb{Z}^+\cup\{0\}$. 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