\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{mathrsfs} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 210, pp. 1--18.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/210\hfil Self-adjoint BVPs for proportional derivatives] {Even-order self-adjoint boundary value problems for proportional derivatives} \author[D. R. Anderson \hfil EJDE-2017/210\hfilneg] {Douglas R. Anderson} \address{Douglas R. Anderson \newline Department of Mathematics, Concordia College, Moorhead, MN 56562, USA} \email{andersod@cord.edu} \dedicatory{Communicated by Mokhtar Kirane} \thanks{Submitted July 7, 2017. Published September 11, 2017.} \subjclass[2010]{26A24, 34A05, 49J15, 49K15} \keywords{Proportional derivatives; PD controller; Green's function; \hfill\break\indent self-adjoint boundary value problem} \begin{abstract} In this study, even order self-adjoint differential equations incorporating recently introduced proportional derivatives, and their associated self-adjoint boundary conditions, are discussed. Using quasi derivatives, a Lagrange bracket and bilinear functional are used to obtain a Lagrange identity and Green's formula; this also leads to the classification of self-adjoint boundary conditions. Next we connect the self-adjoint differential equations with the theory of Hamiltonian systems and $(n,n)$-disconjugacy. Specific formulas of Green's functions for two and four iterated proportional derivatives are also derived. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} We study the $2n$th order differential expression \begin{equation} \label{Ly} \begin{aligned} Ly(t) &= \sum_{j=0}^n \left(-D^{\alpha}\right)^{j} \big[p_{j}\left(D^{\alpha}\right)^{j}y\big](t) \\ &= \left(-D^{\alpha}\right)^{n}\left[p_n\left(D^{\alpha}\right)^{n}y\right](t) + \dots - \left(D^{\alpha}\right)^{3}\big[p_{3}\left(D^{\alpha}\right)^{3}y\big](t) \\ &\quad + \left(D^{\alpha}\right)^{2}\big[p_{2}\left(D^{\alpha}\right)^{2}y\big](t) - D^{\alpha}\left[p_{1}D^{\alpha}y\right](t) + p_0(t)y(t), \end{aligned} \end{equation} for continuous functions $p_i$ with $p_n\ne 0$, and show that it is formally self adjoint with respect to the inner product \[ \langle y,z \rangle = \int_a^b y(t)z(t)e^2_0(b,t)d_{\alpha}t, \quad d_{\alpha}t:=\frac{dt}{\kappa_0(t)}; \] that is, the identity \[ \langle Ly, z\rangle = \langle y, Lz \rangle \] holds provided that $y$ and $z$ satisfy some appropriate self-adjoint boundary conditions at $a$ and $b$. Here $D^{\alpha}$ is a proportional derivative operator \cite{and,au,cobber} modeled after a proportional-derivative controller (PD controller) \cite{ding}. This proportional derivative $D^\alpha$ of order $\alpha\in[0,1]$, where $D^0$ is the identity operator, and $D^1$ is the classical differential operator, will be used to explore corresponding higher-order linear self-adjoint equations of the form \eqref{Ly}. We will refer to an equation with $2n$ iterations of $D^{\alpha}$ as $2n$th-order equations. \begin{remark}\cite{and,au}\rm In control theory, a PD controller for controller output $u$ at time $t$ with two tuning parameters has the algorithm \[ u(t) = \kappa_pE(t) + \kappa_d\frac{d}{dt}E(t) , \] where $\kappa_p$ is the proportional gain, $\kappa_d$ is the derivative gain, and $E$ the is input deviation, or the error between the state variable and the process variable; see \cite{ding}, for example. This is the impetus for the next definition. \end{remark} \begin{definition}[A Class of Proportional Derivatives \cite{and,au}] \rm Let $\alpha\in[0,1]$, $\mathcal{I}\subseteq\mathbb{R}$, and let the functions $\kappa_0,\kappa_1:[0,1]\times\mathcal{I}\to [0,\infty)$ be continuous such that \begin{equation} \begin{gathered} \lim_{\alpha\to 0^+} \kappa_1(\alpha,t) = 1, \quad \lim_{\alpha\to 0^+} \kappa_0(\alpha,t) = 0, \quad \forall\;t\in\mathcal{I}, \\ \lim_{\alpha\to 1^-} \kappa_1(\alpha,t) = 0, \quad \lim_{\alpha\to 1^-} \kappa_0(\alpha,t) = 1, \quad \forall\;t\in\mathcal{I}, \\ \kappa_1(\alpha,t)\ne 0, \alpha\in[0,1), \quad \kappa_0(\alpha,t)\ne 0, \alpha\in(0,1], \quad \forall\;t\in\mathcal{I}. \end{gathered} \label{kappaconditions} \end{equation} Define the proportional differential operator $D^{\alpha}$ via \begin{equation}\label{derivdef} D^{\alpha} f(t) = \kappa_1(\alpha,t) f(t) + \kappa_0(\alpha,t) f'(t), \quad t\in\mathcal{I} \end{equation} provided the right-hand side exists at $t$, where $f':=\frac{d}{dt}f$. \end{definition} \begin{remark}[\cite{and,au}] \rm For the operator given in \eqref{derivdef}, $\kappa_1$ is a type of proportional gain $\kappa_p$, $\kappa_0$ is a type of derivative gain $\kappa_d$, $f$ is the error, and $u=D^{\alpha}f$ is the controller output. To illustrate, one could take $\kappa_1\equiv \cos\left(\alpha\pi/2\right)$ and $\kappa_0\equiv \sin\left(\alpha\pi/2\right)$, or $\kappa_1\equiv (1-\alpha)\omega^{\alpha}$ and $\kappa_0\equiv \alpha\omega^{1-\alpha}$ for any $\omega\in(0,\infty)$; or, $\kappa_1=(1-\alpha)|t|^{\alpha}$ and $\kappa_0=\alpha |t|^{1-\alpha}$ on $\mathcal{I}=\mathbb{R}\backslash\{0\}$, so that \[ D^{\alpha} f(t) = (1-\alpha) |t|^{\alpha} f(t) + \alpha |t|^{1-\alpha} f'(t). \] If $\kappa_1$ and $\kappa_0$ are constant with respect to the independent variable, then $D^\beta D^\alpha = D^\alpha D^\beta$, but $D^\beta D^\alpha \ne D^\alpha D^\beta$ for $\alpha,\beta\in[0,1]$ in general; see also \cite{zulf}. By \eqref{kappaconditions} and \eqref{derivdef}, \[ \lim_{\alpha\to 0^+}D^{\alpha}f = D^0f = f \quad\text{and}\quad \lim_{\alpha\to 1^-}D^{\alpha}f = D^1f = f'. \] \end{remark} Throughout the discussion to follow we will need a vital definition \cite[Definition 1.6]{au}, which establishes a type of exponential function for derivative \eqref{derivdef}. \begin{definition}[Proportional Exponential Function \cite{and,au}]\rm Let $\alpha\in(0,1]$, the points $s,t\in\mathbb{R}$ with $s\le t$, and let the function $p:[s,t]\to \mathbb{R}$ be continuous. Let $\kappa_0,\kappa_1:[0,1]\times\mathbb{R}\to [0,\infty)$ be continuous and satisfy \eqref{kappaconditions}, with $p/\kappa_0$ and $\kappa_1/\kappa_0$ Riemann integrable on $[s,t]$. Then the conformable exponential function with respect to $D^{\alpha}$ in \eqref{derivdef} is defined to be \begin{equation}\label{epts} e_p(t,s):= e^{\int_s^t \frac{p(\tau)-\kappa_1(\alpha,\tau)}{\kappa_0(\alpha,\tau)} d\tau}, \quad e_0(t,s) = e^{-\int_s^t \frac{\kappa_1(\alpha,\tau)}{\kappa_0(\alpha,\tau)} d\tau}, \end{equation} and satisfies \begin{equation}\label{expderiv} D^{\alpha} e_p(t,s) = p(t) e_p(t,s), \quad D^{\alpha} e_0(t,s) = 0. \end{equation} \end{definition} The following fundamental theorem, given in \cite[Theorem 2.4]{and} and \cite[Lemma 1.9 (ii)]{au}, relates the proportional derivative and the proportional integral using the above proportional exponential function. \begin{theorem}[Fundamental Theorem of Integral Calculus]\label{ftc} Let $\alpha\in(0,1]$. Suppose $f:[a,b]\to \mathbb{R}$ is differentiable on $[a,b]$ and $f'$ is integrable on $[a,b]$. Then \[ \int_a^b D^{\alpha}[f(t)]e_0(b,t)d_{\alpha}t = f(b) - f(a)e_0(b,a), \] where $d_{\alpha}t:=dt/\kappa_0(t)$. \end{theorem} \begin{remark}\rm As in \cite{cobber}, consider \eqref{derivdef} with $\kappa_1=(1-\alpha)$ and $\kappa_0=\alpha$, so that \[ D^{\alpha} f(t) = (1-\alpha)f(t) + \alpha f'(t). \] Then using the FTC, Theorem \ref{ftc}, as motivation and simplifying $e_0(t,\tau)$ via \eqref{epts}, define this special case of the proportional integral of $f$ as \begin{equation}\label{propint} {}_{a}\mathcal{I}_t^{\alpha}f(t):=\frac{1}{\alpha} \int_a^t f(\tau)e^{-\frac{1-\alpha}{\alpha}(t-\tau)}d\tau. \end{equation} In two recent papers \cite{caputo,caputo2}, Caputo and Fabrizio introduce a new fractional time derivative of the form \[ \mathscr{D}_t^{(\alpha)}f(t) = \frac{1}{1-\alpha} \int_a^t f'(\tau)e^{-\frac{\alpha}{1-\alpha}(t-\tau)}d\tau, \] with related fractional time integral \[ {}_{a}\mathscr{I}_{t}^{\alpha}f(t) = \frac{1}{\alpha}\int_a^t f(\tau)e^{-\frac{1-\alpha}{\alpha}(t-\tau)}d\tau. \] Note that we then have the relationships \[ \mathscr{D}_t^{(\alpha)}f(t) = {}_{a}\mathcal{I}_t^{1-\alpha}f'(t) \quad\text{and}\quad {}_{a}\mathscr{I}_{t}^{\alpha}f(t) ={}_{a}\mathcal{I}_t^{\alpha}f(t) \] using \eqref{propint}; further research needs to be done on connecting the results of \cite{caputo,caputo2} with those to follow. \end{remark} \section{Self-adjoint proportional equations} For the theory of higher order differential equations refer to \cite{cod, He, KelPet, Nai, Reid}. Consider the $2n$th-order proportional differential expression \eqref{Ly}, in which the coefficient functions $p_j:\mathcal{I}\to\mathbb{R}$ are continuous for $0\le j\le n$ and $p_n(t)\neq 0 $ for all $t\in\mathcal{I}$. \begin{definition} \rm Let $\mathbb{D}$ be the linear set of all functions $y:\mathcal{I}\to\mathbb{R}$ such that the function \[ \left(D^{\alpha}\right)^{j}\big[p_{j}\left(D^{\alpha}\right)^{j}y\big] \] is defined on $\mathcal{I}$ and is continuous for $0\le j\le n$. \end{definition} For each $y\in\mathbb{D}$ the expression $Ly$ is defined and presents a continuous function on $\mathcal{I}$. \begin{definition}[Quasi-Derivatives] \rm As in the traditional case when $\alpha=1$ (see \cite[pp. 49]{Nai}), we introduce the functions $y^{[j]}$, $0\le j\le 2n$, as the quasi-derivatives of $y$ related to the expression $Ly$. Given $y\in\mathbb{D}$, set \begin{gather} y^{[j]} = \left(D^{\alpha}\right)^{j}y, \quad 0\le j\le n-1, \quad y^{[0]} = \left(D^{\alpha}\right)^{0}y = y, \label{yj} \\ y^{[n]} = p_n \left(D^{\alpha}\right)^{n}y, \label{yn} \\ \begin{aligned} y^{[n+j]} &= p_{n-j} \left(D^{\alpha}\right)^{n-j}y - D^{\alpha}\big[y^{[n+j-1]}\big], \quad 1 \le j \le n-1 \\ &= \sum_{i=0}^j \left(-D^{\alpha}\right)^{j-i} \big[p_{n-i} \left(D^{\alpha}\right)^{n-i}y\big], \quad 0 \le j \le n-1, \end{aligned} \label{ynj}\\ \begin{aligned} y^{[2n]} &= p_0y - D^{\alpha}\big[y^{[2n-1]}\big] \\ &= \sum_{j=0}^n \left(-D^{\alpha}\right)^{j} \big[p_{j}\left(D^{\alpha}\right)^{j}y\big] = Ly. \end{aligned} \label{y2nLy} \end{gather} \end{definition} \begin{definition}[Lagrange Bracket] \rm Assume $y,z\in\mathbb{D}$ and $t\in\mathcal{I}$. The Lagrange bracket of $y$ and $z$ is given by \begin{equation}\label{bracket} \{y,z\}(t) = \sum_{j=1}^n \left\{y^{[j-1]}z^{[2n-j]} - y^{[2n-j]}z^{[j-1]}\right\}(t). \end{equation} \end{definition} \begin{definition}[Bilinear Functional]\rm Assume $y,z\in\mathbb{D}$ and $t\in\mathcal{I}$. The bilinear (in $y$ and $z$) functional $F$ is given by \begin{equation} \label{F1} F(y,z,t) = \sum_{j=1}^n \left(y^{[j-1]}z^{[2n-j]}\right)(t). \end{equation} \end{definition} Note that by combining \eqref{bracket} and \eqref{F1}, we have the Lagrange bracket in terms of the bilinear functional, namely \[ \{y,z\}(t) = F(y,z,t) - F(z,y,t). \] Using \eqref{yj} and \eqref{ynj} we get that \begin{equation} \label{F2} F(y,z,t) = \sum_{j=0}^{n-1} (-1)^j \left(D^{\alpha}\right)^{n-j-1}y(t) \sum_{i=0}^j (-1)^i \left(D^{\alpha}\right)^{j-i} \big[p_{n-i} \left(D^{\alpha}\right)^{n-i}z\big](t). \end{equation} \begin{lemma}\label{Fdelta} The bilinear functional $F$ in \eqref{F1} satisfies \[ e_0(t,a)D^{\alpha}\Big[\frac{F(y,z,\cdot)}{e_0(\cdot,a)}\Big](t) = \Big( -yLz + \sum_{j=0}^n p_{j}\left(D^{\alpha}\right)^{j} y\left(D^{\alpha}\right)^{j}z\Big)(t) \] for $t,a\in\mathcal{I}$. \end{lemma} \begin{proof} Differentiating both sides of \eqref{F1}, employing the quotient rule for $\alpha$-deriva\-tives, and taking into account the formulas \eqref{yn} and \eqref{y2nLy}, we get \begin{align*} e_0(t,a)D^{\alpha}\Big[\frac{F(y,z,\cdot)}{e_0(\cdot,a)}\Big](t) &= D^{\alpha}F(y,z,t)+\kappa_1(t)F(y,z,t) \\ &= \sum_{j=1}^n \left(y^{[j-1]}D^{\alpha}\big[z^{[2n-j]}\big] + z^{[2n-j]}D^{\alpha}\big[y^{[j-1]}\big]\right)(t) \\ &= \Big(y^{[0]}D^{\alpha}\big[z^{[2n-1]}\big]+\sum_{j=2}^n y^{[j-1]}D^{\alpha}\big[z^{[2n-j]}\big] \\ & \quad + z^{[n]}D^{\alpha}\big[y^{[n-1]}\big] + \sum_{j=1}^{n-1} z^{[2n-j]}D^{\alpha}\big[y^{[j-1]}\big]\Big)(t) \\ &= \Big(y(p_0z-Lz) + \sum_{j=2}^n y^{[j-1]}D^{\alpha}\big[z^{[2n-j]}\big] \\ &\quad + p_n\left(D^{\alpha}\right)^n y\left(D^{\alpha}\right)^nz + \sum_{j=2}^n z^{[2n-j+1]}D^{\alpha}\big[y^{[j-2]}\big]\Big)(t). \end{align*} Further, by \eqref{yj} we have \[ D^{\alpha}\big[y^{[j-2]}\big](t) = y^{[j-1]}(t) \quad\text{for } 2\le j\le n, \; t\in\mathcal{I}, \] and from \eqref{ynj} for $z$, replacing the $j$ by $n-j+1$, we find \[ z^{[2n-j+1]} = p_{j-1} \left(D^{\alpha}\right)^{j-1}z - D^{\alpha}\big[z^{[2n-j]}\big] \mbox{ for } 2\le j\le n. \] Consequently we obtain the desired result. \end{proof} \begin{theorem}[Lagrange Identity] If $y,z\in\mathbb{D}$, then for $t,a\in\mathcal{I}$ we have \begin{equation} \label{zLy} \left(zLy - yLz\right)(t) = e_0(t,a)D^{\alpha}\big[\frac{\{y,z\}}{e_0(\cdot,a)}\big](t), \end{equation} where $\{y,z\}$ is the Lagrange bracket of $y$ and $z$ defined by \eqref{bracket}. \end{theorem} \begin{proof} By \eqref{bracket} and \eqref{F1} we have \[ \{y,z\}(t) = F(y,z,t)-F(z,y,t); \] dividing both sides by $e_0(t,a)$, taking the $\alpha$ derivative, multiplying the result by $e_0(t,a)$ on both sides, and applying Lemma \ref{Fdelta} we obtain \eqref{zLy}. \end{proof} \begin{remark}[Green's Formula]\label{greenformula}\rm Let the numbers $a,b,t\in\mathcal{I}$ with $a t_1$, $t_1,t_2\in\mathcal{I}$. The result follows if we take $t_1=a$ and $t_2=t$. \end{proof} \begin{lemma} \label{etaeta} Assume $\eta\in\mathbb{D}[a,b]$. Then \begin{equation} \label{eta} F(\eta,\eta,b)-F(\eta,\eta,a)e^2_0(b,a) = -\langle \eta, L\eta \rangle + \sum_{j=0}^n \langle p_{j},[(D^{\alpha})^j\eta]^2 \rangle. \end{equation} \end{lemma} \begin{proof} Setting $y=z=\eta$ in Lemma \ref{Fdelta} we have \[ e_0(t,a)D^{\alpha}\big[\frac{F(\eta,\eta,\cdot)}{e_0(\cdot,a)}\big](t) = \Big(-\eta L\eta + \sum_{j=0}^n p_{j} \big[(D^{\alpha})^{j}\eta\big]^2\Big)(t) \] for $t,a\in\mathcal{I}$. If we multiply both sides by $e^2_0(b,t)d_{\alpha}t$ and then integrate from $a$ to $b$ we get the desired result. \end{proof} \begin{definition} \rm The set of admissible variations is given by \[ \mathcal{S} = \left\{\eta\in\mathbb{D}[a,b]: (D^{\alpha})^j\eta(a) = (D^{\alpha})^j\eta(b) =0, \quad 0\le j\le n-1 \right\}, \] with corresponding functional \begin{equation} \label{calF} \mathcal{F}(\eta) = \sum_{j=0}^n \langle p_{j},[(D^{\alpha})^j\eta]^2\rangle. \end{equation} \end{definition} For an admissible variation $\eta \in \mathcal{S}$, Lemma \ref{etaeta} implies that \[ \mathcal{F}(\eta) = \langle \eta, L\eta \rangle. \] The functional $\mathcal{F}$ is positive definite on the set of admissible variations $\mathcal{S}$ if $\mathcal{F}(\eta) \ge 0$ for all $\eta\in\mathcal{S}$, and $\mathcal{F}(\eta)=0$ if and only if $\eta =0$. Note that the bilinear functional $F$ in \eqref{F1} and the vector-valued functions $\vec{x}$ and $\vec{u}$ given above in \eqref{xuvecdef} satisfy the dot product equation \[ (\vec{x} \cdot \vec{u})(t) = F(y,y,t). \] We will use this in the proof of the next theorem. \begin{theorem}\label{posdefdisconj} Assume $p_j(t)\ge 0$ for $0\le j\le n$ and $t\in\mathcal{I}$, and $p_n(t)>0$ for $t\in\mathcal{I}$. Then the functional $\mathcal{F}$ is positive definite on $\mathcal{S}$ and the linear Hamiltonian system \eqref{nabham} being considered for $t\in[a,b]$ is disconjugate on $[a,b]$. In particular the self-adjoint BVP \begin{gather*} Ly(t) = 0, \quad t \in [a,b], \\ (D^{\alpha})^jy(a) = 0 = (D^{\alpha})^jy(b), \quad j=0,1,\ldots,n-1, \end{gather*} has only the trivial solution. \end{theorem} \begin{proof} Let $t\in\mathcal{I}$. From $p_j(t)\ge 0$ for $0\le j\le n$ and \eqref{calF}, it is clear that $\mathcal{F}(\eta)\ge 0$ for all $\eta\in\mathcal{S}$, and that $\mathcal{F}(0)=0$. Now suppose $\eta\in\mathcal{S}$ and $F(\eta)=0$. Then \[ 0 = \sum_{j=0}^n \langle p_{j},[(D^{\alpha})^j\eta]^2\rangle \ge \langle p_{n},[(D^{\alpha})^n\eta]^2\rangle, \] and since $p_n(t)>0$, we have that $(D^{\alpha})^n\eta(t) = 0$ for $t\in[a,b]$. Because $\eta$ is admissible, it solves the initial value problem \begin{gather*} (D^{\alpha})^n\eta(t) = 0, \quad t\in[a,b] \\ (D^{\alpha})^j\eta(a) = 0, \quad 0\le j\le n-1. \end{gather*} By uniqueness of solutions to initial value problems, $\eta$ is the trivial solution in the set of admissible functions, whence $\mathcal{F}$ is positive definite on that set. By \eqref{etaeta}, if $y$ is a solution of $Ly(t)=0$, $t\in[a,b]$, then \begin{align*} (\vec{x}\cdot\vec{u})(b) - (\vec{x}\cdot\vec{u})(a)e^2_0(b,a) &= F[y,y,b]-F[y,y,a]e^2_0(b,a) \\ &= \sum_{j=0}^n \langle p_{j},\left[(D^{\alpha})^jy\right]^2\rangle \\ &= \mathcal{F}(y). \end{align*} Note that the Hamiltonian system \eqref{nabham} is disconjugate on $[a,b]$ if and only if for a vector solution $\vec{x}$, $\vec{u}$ of \eqref{nabham}, the following is positive definite: \[ \int_a^b \left(\vec{x}^\top\mathcal{C}\vec{x} + \vec{u}^\top\mathcal{B}\vec{u}\right)(t)e^2_0(b,t) d_{\alpha}t = \sum_{j=0}^{n-1} \langle p_{j},\big(y^{[j]}\big)^2\rangle + \langle 1/p_n,\big(y^{[n]}\big)^2\rangle= \mathcal{F}(y). \] This completes the proof. \end{proof} The point $t=t_0$ is a zero of order (at least) $n$ of $y$ if \[ (D^{\alpha})^jy(t_0)=0, \quad j=0,1,\ldots,n-1. \] The equation $Ly=0$ is $(n,n)$ disconjugate on $[a,b]$ provided there is no nontrivial solution of $Ly=0$ with a zero of order (at least) $n$ in $(a,b]$ preceded by a zero of order (at least) $n$ in $[a,b]$. These ideas lead to the next conclusion. \begin{theorem} If $p_n(t)>0$ for $t\in[a,b]$, then $Ly(t)=0$ is $(n,n)$ disconjugate on $[a,b]$. \end{theorem} \begin{proof} Suppose $y$ is a solution of $Ly=0$, and without loss of generality assume $y$ has a zero of order $n$ at $a$, namely $(D^{\alpha})^jy(a)=0$, $j=0,1,\ldots,n-1$. Then from \eqref{F2} we have $F(y,y,a)=0$, and $F(y,y,t)\ge 0$ for all $t\in[a,b]$ by Theorem \ref{Fincr}. If $y$ has a zero at $t_0\in(a,b]$ of order $n$, then \[ (D^{\alpha})^jy(t_0)=0, \quad j=0,1,\ldots,n-1. \] But then $y$ is a trivial solution of $Ly=0$ by the previous theorem. \end{proof} \section{Second-order proportional equations} Analogous to the classic and time scales cases \cite{AtiGus}, in this section we find Green's function associated to second-order proportional equations. With this in mind, again consider \eqref{Ly}. Taking $n=1$, we find that \[ Ly(t) = -D^{\alpha}\left[p_1 D^{\alpha}y\right](t) + p_0(t)y(t), \quad t\in\mathcal{I}, \] and for each function $y\in\mathbb{D}$, \[ y^{[0]} = y, \quad y^{[1]} = p_1 D^{\alpha}y, \quad y^{[2]} = p_0y - D^{\alpha}\big[y^{[1]}\big]. \] Then \[ Ly = y^{[2]} \] as expected. In addition, the equation $Ly(t) = g(t)$ for $t\in\mathcal{I}$ is equivalent to the first order system \[ D^{\alpha}\vec{y}(t) = A(t)\vec{y}(t) + \vec{g}(t), \quad t \in \mathcal{I}, \] where \[ \vec{y} = \begin{pmatrix} y^{[0]} \\ y^{[1]} \end{pmatrix}, \quad \vec{g} = \begin{pmatrix} 0 \\ -g \end{pmatrix}, \quad A(t) = \begin{pmatrix} 0 & \frac{1}{p_1(t)} \\ p_0(t) & 0 \end{pmatrix}. \] The Wronskian of two solutions $y,z$, is \[ W_t(y,z) = \begin{vmatrix} y^{[0]}(t) & z^{[0]}(t) \\ y^{[1]}(t) & z^{[1]}(t) \end{vmatrix} = p_1(t)\left(yD^{\alpha}z - zD^{\alpha}y\right)(t) = \{y,z\}(t), \] the Lagrange bracket \eqref{bracket} of $y$ and $z$, giving rise to the following theorem. \begin{theorem}\label{wrone2thm} The Wronskian of any two solutions $y,z$ of $Ly(t) = 0$ satisfies \[ W_t(y,z)=e^2_0(t,a)W_a(y,z). \] \end{theorem} The following theorem presents a variation of constants formula for the nonhomogeneous equation $Ly(t) = g(t)$. \begin{theorem}[Variation of Constants] Suppose that $y_1,y_2$ form a fundamental system of solutions of the homogeneous equation $Ly(t)=0$. Then the general solution of the nonhomogeneous equation $Ly(t) = g(t)$ is given by \[ y(t) = c_1y_1(t) + c_2y_2(t) + \int_{t_0}^t \frac{y_1(t)y_2(s) - y_1(s)y_2(t)}{W_s(y_1,y_2)} g(s) d_{\alpha}s, \] where $t_0\in \mathcal{I}$ and $c_1, c_2$ are real constants. \end{theorem} \begin{proof} It suffices to show that the function \[ z(t) = \int_{t_0}^t \frac{y_1(t)y_2(s) - y_1(s)y_2(t)}{W_s(y_1,y_2)} g(s) d_{\alpha}s \] is a particular solution of the nonhomogeneous equation $Ly(t) = g(t)$. Differentiating both sides yields \[ D^{\alpha}z(t) = \int_{t_0}^t \frac{y_2(s)D^{\alpha}y_1(t) - y_1(s)D^{\alpha}y_2(t)}{W_s(y_1,y_2)} g(s) d_{\alpha}s. \] Hence \begin{align*} D^{\alpha}\left[p_1D^{\alpha}z\right](t) &= \frac{y_2(t)p_1(t)D^{\alpha}y_1(t)-y_1(t)p_1(t)D^{\alpha}y_2(t)}{W_t(y_1,y_2)} g(t) \\ &\quad +\int_{t_0}^t \frac{y_2(s)D^{\alpha}[p_1D^{\alpha}y_1](t) - y_1(s)D^{\alpha}[p_1D^{\alpha}y_2](t)}{W_s(y_1,y_2)} g(s) d_{\alpha}s \\ &= -g(t) + p_0(t)z(t), \end{align*} that is $z$ satisfies $Ly(t) = g(t)$. \end{proof} For $y\in\mathbb{D}[a,b]$ let \[ Ly(t) = -D^{\alpha}\left[p_1 D^{\alpha}y\right](t) + p_0(t)y(t), \quad t\in[a,b], \] together with the boundary conditions \begin{equation} \label{bnd1} \begin{gathered} \eta_{11}e_0(b,a)y(a) + \eta_{12}e_0(b,a)y^{[1]}(a) + \beta_{11}e_0(a,b)y(b) + \beta_{12}e_0(a,b)y^{[1]}(b) = 0, \\ \eta_{21}e_0(b,a)y(a) + \eta_{22}e_0(b,a)y^{[1]}(a) + \beta_{21}e_0(a,b)y(b) + \beta_{22}e_0(a,b)y^{[1]}(b) = 0, \end{gathered} \end{equation} where $\eta_{ij}, \beta_{ij}$ are given real numbers, $i,j=1,2$. Set \[ N = \begin{pmatrix} \eta_{11} & \eta_{12} & \beta_{11} & \beta_{12} \\ \eta_{21} & \eta_{22} & \beta_{21} & \beta_{22} \end{pmatrix}. \] We will assume that the matrix $N$ has rank 2. This means that the two boundary conditions \eqref{bnd1} are linearly independent. As before, we call the boundary conditions \eqref{bnd1} self adjoint with respect to the expression $Ly$ if \[ \langle Ly, z\rangle - \langle y, Lz \rangle = \{y,z\}(b)- e^2_0(b,a)\{y,z\}(a) \] for all functions $y,z\in\mathbb{D}[a,b]$ satisfying the boundary conditions \eqref{bnd1}. Recall that by Green's formula, the boundary conditions \eqref{bnd1} are self adjoint if and only if \[ e_0(a,b)\{y,z\}(b) = e_0(b,a)\{y,z\}(a). \] Set \[ N_1 = \begin{pmatrix} \eta_{11} & \eta_{12} \\ \eta_{21} & \eta_{22} \end{pmatrix}, \quad N_2 = \begin{pmatrix} \beta_{11} & \beta_{12} \\ \beta_{21} & \beta_{22} \end{pmatrix}. \] \begin{theorem} If $\det N_1=\det N_2$, then the boundary conditions \eqref{bnd1} are self adjoint. \end{theorem} \begin{proof} Let $y,z\in\mathbb{D}[a,b]$, be functions which satisfy boundary conditions \eqref{bnd1}. Then we have \[ e_0(b,a)N_1\begin{pmatrix} y(a) & z(a) \\ y^{[1]}(a) & z^{[1]}(a) \end{pmatrix} = e_0(a,b)N_2 \begin{pmatrix} -y(b) & -z(b) \\ -y^{[1]}(b) & -z^{[1]}(b) \end{pmatrix}. \] Passing to determinants we have \[ (\det N_1)e_0(b,a)\{y,z\}(a) = (\det N_2)e_0(a,b)\{y,z\}(b). \] If $\det N_1=\det N_2\neq 0$, then \[ e_0(b,a)\{y,z\}(a)=e_0(a,b)\{y,z\}(b). \] Suppose $\det N_1=\det N_2 =0$. Since $N$ has rank 2, it is clear that the boundary conditions \eqref{bnd1} are equivalent to separated boundary conditions of the form \begin{equation} \label{sep} \begin{gathered} \eta_1y(a) + \eta_2y^{[1]}(a) = 0, \quad |\eta_1|+|\eta_2| \neq 0, \\ \beta_1y(b) + \beta_2y^{[1]}(b) = 0, \quad |\beta_1|+|\beta_2| \neq 0, \end{gathered} \end{equation} where $\eta_i, \beta_i$, $i=1,2$ are real numbers. It can easily be verified that for any functions $y,z\in\mathbb{D}[a,b]$ satisfying boundary conditions \eqref{bnd1} we have \[ \{y,z\}(a) = 0 = \{y,z\}(b), \] completing the proof. \end{proof} \begin{remark} \rm As was noted above, the separated boundary conditions \eqref{sep}, in particular the boundary conditions $y(a)=y(b)=0$ are self adjoint. The ``periodic'' boundary conditions \[ e_0(b,a)y(a)=e_0(a,b)y(b), \quad e_0(b,a)y^{[1]}(a) = e_0(a,b)y^{[1]}(b) \] which are non-separated, are also self adjoint. \end{remark} We will now construct Green's function for the self-adjoint (separated) BVP \begin{gather} -D^{\alpha}\left[p_1 D^{\alpha}y\right](t) + p_0(t)y(t) = g(t) \label{bvp1} \\ \eta y(a) - \beta y^{[1]}(a) = 0, \quad \gamma y(b) + \delta y^{[1]}(b) =0, \label{cond} \end{gather} where $\eta, \beta, \gamma, \delta$ are real numbers such that $|\eta| + |\beta| \neq 0 $, $|\gamma|+|\delta| \neq 0$. \begin{remark}\rm The minus sign on the left hand side of \eqref{bvp1}, as well as in the first boundary condition of \eqref{cond}, is taken so that the positivity of Green's function can be formulated in terms of $p_1(t)>0$, $p_0(t)\ge 0$, for $\eta, \beta, \gamma, \delta \ge 0$. \end{remark} Denote by $\phi$ and $\psi$ the solutions of the corresponding homogeneous equation \begin{equation} \label{bvp0} -D^{\alpha}\left[p_1 D^{\alpha}y\right](t) + p_0(t)y(t) = 0, \quad t\in[a,b], \end{equation} under the initial conditions \begin{gather} \phi(a) = \beta, \quad \phi^{[1]}(a)=\eta, \label{cond1} \\ \psi(b) = \delta, \quad \psi^{[1]}(b) = -\gamma, \label{cond2} \end{gather} so that $\phi$ and $\psi$ satisfy the first and second boundary conditions in \eqref{cond}, respectively. From Theorem \ref{wrone2thm} we have that the Wronskian of $\phi$ and $\psi$ satisfies \[ W_t(\phi, \psi) = \phi(t)\psi^{[1]}(t) - \phi^{[1]}(t)\psi(t) = e^2_0(t,a) W_a(\phi, \psi); \] evaluating this expression at $t=a$, $t=b$, and using the boundary conditions \eqref{cond1}, \eqref{cond2} yields \[ W_a(\phi, \psi) = \beta\psi^{[1]}(a) - \eta\psi(a) = \frac{-\gamma\phi(b) - \delta\phi^{[1]}(b)}{e^2_0(b,a)}. \] Additionally, $W_a(\phi, \psi)\neq 0$ if and only if the homogeneous equation \eqref{bvp0} has only the trivial solution satisfying the boundary conditions \eqref{cond}. \begin{theorem}\label{2ptbvpthm} If $W_a(\phi, \psi)\neq 0$, then the nonhomogeneous BVP \eqref{bvp1}, \eqref{cond}, has a unique solution $y$ for which the formula \[ y(t) = \int_a^b G(t,s)g(s)d_{\alpha}s, \quad t\in[a,b] \] holds, where the function $G(t,s)$ is given by \[ G(t,s) = \frac{-1}{W_s(\phi,\psi)} \begin{cases} \phi(t)\psi(s): & a\le t\le s\le b, \\ \phi(s)\psi(t): & a\le s\le t\le b, \end{cases} \] and this $G(t,s)$ is Green's function of the BVP \eqref{bvp1}, \eqref{cond}. Furthermore the Green function satisfies the property $e_0(s,t)G(t,s) = e_0(t,s)G(s,t)$ for all $t,s\in[a,b]$. \end{theorem} \begin{proof} Since $W_a(\phi, \psi)\neq 0$, the solutions $\phi$ and $\psi$ of the homogeneous equation \eqref{bvp0} are linearly independent. Thus the general solution of the nonhomogeneous equation \eqref{bvp1} has the variation of constants form \begin{equation}\label{y1} y(t) = c_1\phi(t) + c_2\psi(t) + \int_a^t \frac{\phi(t)\psi(s)-\phi(s)\psi(t)}{W_s(\phi, \psi)}g(s) d_{\alpha}s, \end{equation} where $c_1$ and $c_2$ are real constants. We now construct $c_1$ and $c_2$ so that the function $y$ satisfies the boundary conditions \eqref{bnd1}. Using \eqref{y1} we have \begin{equation} \label{y2} y^{[1]}(t) = c_1\phi^{[1]}(t) + c_2\psi^{[1]}(t) + \int_a^t \frac{\phi^{[1]}(t)\psi(s)-\phi(s)\psi^{[1]}(t)}{W_s(\phi, \psi)}g(s) d_{\alpha}s. \end{equation} Consequently, \begin{gather*} y(a) = c_1\phi(a) + c_2\psi(a) = c_1\beta + c_2\psi(a), \\ y^{[1]}(a) = c_1\phi^{[1]}(a) + c_2\psi^{[1]}(a) = c_1\eta + c_2\psi^{[1]}(a). \end{gather*} Substituting these values of $y(a)$ and $y^{[1]}(a)$ into the first condition of \eqref{cond} we have \[ c_2\left(\eta\psi(a) - \beta\psi^{[1]}(a)\right) = 0. \] On the other hand, using the definition of $W_a(\phi, \psi)$, \[ \eta\psi(a)-\beta\psi^{[1]}(a) = -W_a(\phi, \psi) \neq 0. \] Consequently $c_2 = 0$, and \eqref{y1}, \eqref{y2}, take the form \begin{gather*} y(t) = c_1\phi(t) + \int_a^t \frac{\phi(t)\psi(s)-\phi(s)\psi(t)} {W_s(\phi, \psi)}g(s) d_{\alpha}s, \\ y^{[1]}(t) = c_1\phi^{[1]}(t) + \int_a^t \frac{\phi^{[1]}(t)\psi(s)-\phi(s)\psi^{[1]}(t)}{W_s(\phi, \psi)}g(s) d_{\alpha}s, \end{gather*} respectively. Hence \begin{gather*} y(b) = c_1\phi(b) + \int_a^b \frac{\phi(b)\psi(s)-\phi(s)\psi(b)}{W_s(\phi, \psi)}g(s) d_{\alpha}s, \\ y^{[1]}(b) = c_1\phi^{[1]}(b) + \int_a^b \frac{\phi^{[1]}(b)\psi(s)-\phi(s)\psi^{[1]}(b)}{W_s(\phi, \psi)}g(s) d_{\alpha}s. \end{gather*} Substituting these values into the second condition of \eqref{cond} yields \[ c_1\left(\gamma\phi(b) + \delta\phi^{[1]}(b)\right) + \int_a^b \frac{\left(\gamma\phi(b) + \delta\phi^{[1]}(b)\right)} {W_s(\phi, \psi)}\psi(s)g(s)d_{\alpha}s = 0. \] Again using the definition of $W_a(\phi,\psi)$, \[ \gamma\phi(b) + \delta\phi^{[1]}(b) = -e^2_0(b,a)W_a(\phi, \psi) \neq 0. \] Hence \[ c_1 = -\int_a^b \frac{\psi(s)}{W_s(\phi, \psi)}g(s)d_{\alpha}s. \] Thus $y$ has the desired form, and $G(t,s)$ satisfies $e^2_0(s,a)G(t,s)=e^2_0(t,a)G(s,t)$; this is equivalent to $e_0(s,t)G(t,s)=e_0(t,s)G(s,t)$, completing the proof. \end{proof} \begin{corollary}[Green's Function for the Two-Point Problem]\label{oldcor6} If \[ d:=\beta\gamma+\eta\delta+\eta\gamma\int_a^b\frac{d_{\alpha}\tau}{p_1(\tau)}\neq 0, \] then the nonhomogeneous BVP \eqref{bvp1}, \eqref{cond} with $p_0\equiv 0$ has a unique solution $y$ for which the formula \[ y(t) = \int_a^b G(t,s)g(s)d_{\alpha}s, \quad t\in[a,b] \] holds, where the function $G(t,s)$ is given by \[ G(t,s) = \frac{e_0(t,s)}{d}\begin{cases} \big[\beta+\eta\int_a^t\frac{d_{\alpha}\tau}{p_1(\tau)}\big] \big[\delta+\gamma\int_s^b\frac{d_{\alpha}\tau}{p_1(\tau)}\big]: & a\le t\le s\le b, \\[4pt] \big[\beta+\eta\int_a^s\frac{d_{\alpha}\tau}{p_1(\tau)}\big] \big[\delta+\gamma\int_t^b\frac{d_{\alpha}\tau}{p_1(\tau)}\big]: & a\le s\le t\le b. \end{cases} \] This $G(t,s)$ is Green's function of the BVP \eqref{bvp1}, \eqref{cond} with $p_0\equiv 0$. \end{corollary} \begin{proof} Assume \[ d:=\beta\gamma+\eta\delta+\eta\gamma\int_a^b\frac{d_{\alpha}\tau}{p_1(\tau)}\neq 0. \] Note that \[ \phi(t)=\eta e_0(t,a)\int_a^t \frac{d_{\alpha}\tau}{p_1(\tau)} +\beta e_0(t,a), \quad \psi(t) =\gamma e_0(t,b) \int_t^b \frac{d_{\alpha}\tau}{p_1(\tau)}+\delta e_0(t,b) \] satisfy \eqref{bvp0} with $p_0\equiv 0$, along with conditions \eqref{cond1} and \eqref{cond2}. The result then follows from Theorem \ref{2ptbvpthm}. \end{proof} \begin{corollary}[Green's Function for the Conjugate Problem] \index{Green's function!conjugate problem} Green's function for the conjugate boundary value problem \begin{equation}\label{cojubvp} -D^{\alpha}\left[p D^{\alpha}y\right](t)=0,\quad y(a)=y(b)=0 \end{equation} is given by \[ G(t,s) = \frac{e_0(t,s)}{\int_a^{b} \frac{1}{p(\tau)}d_{\alpha}\tau} \begin{cases} \int_a^t \frac{1}{p(\tau)} d_{\alpha}\tau \int_s^b \frac{1}{p(\tau)} d_{\alpha}\tau &: a\le t\le s\le b, \\[4pt] \int_a^s \frac{1}{p(\tau)} d_{\alpha}\tau \int_t^b \frac{1}{p(\tau)} d_{\alpha}\tau &: a\le s\le t\le b. \end{cases} \] \end{corollary} \begin{proof} By Theorem \ref{posdefdisconj}, the BVP \eqref{cojubvp} has only the trivial solution. Due to the boundary conditions $y(a)=y(b)=0$, we see that $\eta=\gamma=1$ and $\beta=\delta=0$ in \eqref{cond1} and \eqref{cond2}. The result then follows from Corollary \ref{oldcor6}. \end{proof} \begin{corollary}[Green's Function for the Focal Problem]\label{ggff} Green's function for the focal boundary value problem \begin{equation}\label{fokalbvp} -D^{\alpha}\left[p D^{\alpha}y\right](t)=0, \quad y(a) = D^{\alpha}y(b)=0 \end{equation} is given by \[ G(t,s) = e_0(t,s)\begin{cases} \int_a^t\frac{1}{p(\tau)} d_{\alpha}\tau &: a\le t\le s\le b, \\[4pt] \int_a^{s}\frac{1}{p(\tau)} d_{\alpha}\tau &: a\le s\le t\le b. \end{cases} \] \end{corollary} \begin{proof} The boundary conditions imply $\eta=\delta=1$ and $\beta=\gamma=0$ in \eqref{cond1} and \eqref{cond2}. The result again follows from Corollary \ref{oldcor6}. \end{proof} \section{Fourth-order proportional equations} In equation \eqref{Ly} let $n=2$, and consider the fourth order expression \begin{equation} \label{Ly1} Ly(t) = (D^{\alpha})^2\left[p_2 (D^{\alpha})^2y\right](t) - D^{\alpha}\big[p_1D^{\alpha}y\big](t) + p_0(t)y(t). \end{equation} For $y \in \mathbb{D}$ we have by definition \begin{gather*} y^{[0]} = y, \quad y^{[1]} = D^{\alpha}y, \quad y^{[2]} = p_2(D^{\alpha})^2y, \\ y^{[3]} = p_1D^{\alpha}y - D^{\alpha}\big[y^{[2]}\big], \quad y^{[4]} = p_0y - D^{\alpha}\big[y^{[3]}\big]. \end{gather*} It follows that \[ Ly = y^{[4]}. \] In this case, for $y,z \in \mathbb{D}$ the Lagrange bracket of $y$ and $z$ is \[ \{y,z\}(t) = y(t)z^{[3]}(t) - y^{[3]}(t)z(t) + y^{[1]}(t)z^{[2]}(t) - y^{[2]}(t)z^{[1]}(t), \] and the Lagrange identity \[ \left(zLy - yLz\right)(t) = e_0(t,a)D^{\alpha} \big[\frac{\{y,z\}}{e_0(\cdot,a)}\big](t) \] holds. Using the same techniques as in previous sections, for each function $y \in \mathbb{D}$ we have the following system of relations at $t \in \mathcal{I}$, \begin{gather*} D^{\alpha}\big[y^{[0]}\big] = y^{[1]}, \quad D^{\alpha}\big[y^{[1]}\big] = \frac{y^{[2]}}{p_2}, \\ D^{\alpha}\big[y^{[2]}\big] = p_1 y^{[1]} - y^{[3]}, \quad D^{\alpha}\big[y^{[3]}\big] = p_0 y - Ly. \end{gather*} Thus the equation $Ly(t) = g(t)$ for $t\in\mathcal{I}$ where $g:\mathcal{I}\to\mathbb{R}$ is a continuous function is equivalent to the first order system \[ D^{\alpha}\vec{y}(t) = A(t)\vec{y}(t) + \vec{g}(t), \quad t\in\mathcal{I}, \] where \[ \vec{y} = \begin{pmatrix} y^{[0]} \\ y^{[1]} \\ y^{[2]} \\ y^{[3]} \end{pmatrix}, \quad \vec{g} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ -g \end{pmatrix}, \quad A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{p_2} & 0 \\ 0 & p_1 & 0 & -1 \\ p_0 & 0 & 0 & 0 \end{pmatrix}. \] Together with the expression \eqref{Ly1}, take boundary conditions of the form \begin{equation} \label{bnd4} e_0(b,a)\sum_{j=1}^4 \eta_{ij}y^{[j-1]}(a) + e_0(a,b)\sum_{j=1}^4 \beta_{ij}y^{[j-1]}(b)=0, \quad 1 \le i \le 4. \end{equation} These boundary conditions are self adjoint if and only if \begin{align*} 0 &= e_0(a,b)\Big\{y(b)z^{[3]}(b) - y^{[3]}(b)z(b) + y^{[1]}(b)z^{[2]}(b) - y^{[2]}(b)z^{[1]}(b)\Big\} \\ &\quad -e_0(b,a)\Big\{y(a)z^{[3]}(a) + y^{[3]}(a)z(a) - y^{[1]}(a)z^{[2]}(a) + y^{[2]}(a)z^{[1]}(a)\Big\} \end{align*} for all $y,z \in \mathbb{D}_{[a,b]}$. As is the case when $\alpha=1$, it follows that by joining any one of the four types of conditions \begin{itemize} \item[(i)] $y(a) = y^{[1]}(a) = 0,$ \item[(ii)] $y^{[1]}(a) = y^{[3]}(a) = 0$, \item[(iii)] $y(a) = y^{[2]}(a) = 0$, \item[(iv)] $y^{[2]}(a) = y^{[3]}(a) = 0$, \end{itemize} with any one of the four types of conditions \begin{itemize} \item[(i)] $y(b) = y^{[1]}(b) = 0,$ \item[(ii)] $y^{[1]}(b) = y^{[3]}(b) = 0$, \item[(iii)] $y(b) = y^{[2]}(b) = 0$, \item[(iv)] $y^{[2]}(b) = y^{[3]}(b) = 0$, \end{itemize} yields the sixteen types of self-adjoint boundary conditions. The ``periodic'' boundary conditions \begin{gather*} e_0(b,a)y(a) = e_0(a,b)y(b), \quad e_0(b,a)y^{[1]}(a) = e_0(a,b)y^{[1]}(b), \\ e_0(b,a)y^{[2]}(a) = e_0(a,b)y^{[2]}(b), \quad e_0(b,a)y^{[3]}(a) = e_0(a,b)y^{[3]}(b), \end{gather*} are also self adjoint. \begin{example} \rm The Green function $G(t,s)$ for \[ (D^{\alpha})^2[p(D^{\alpha})^2y](t), \quad t\in[a,b], \] with the boundary conditions \[ y(a) = y^{[1]}(a) = y^{[2]}(b) = y^{[3]}(b) = 0 \] is given by \[ G(t,s)=\begin{cases} e_0(t,s)\int_a^t \big(\int_a^\tau \frac{h_1(s,\xi)}{p(\xi)} d_{\alpha}\xi \big) d_{\alpha}\tau &: a\le t\le s\le b, \\[4pt] e_0(t,s)\int_a^s \big(\int_a^\tau \frac{h_1(t,\xi)}{p(\xi)} d_{\alpha}\xi \big) d_{\alpha}\tau &: a\le s\le t\le b, \end{cases} \] where $h_1(v,\xi):=\int_{\xi}^v 1 \, d_{\alpha}w$. \end{example} \begin{thebibliography}{00} \bibitem{AlBoRi} C. Ahlbrandt, M. Bohner, J. Ridenhour; Hamiltonian systems on time scales, \emph{J. Math. Anal. Appl.}, 250 (2000), 561--578. \bibitem{and} D. R. 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