\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 19, 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 (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/19\hfil Third-order three-point nonlocal problems] {Third-order nonlocal problems with sign-changing nonlinearity on time scales} \author[D. R. Anderson, C. C. Tisdell\hfil EJDE-2007/19\hfilneg] {Douglas R. Anderson, Christopher C. Tisdell} % in alphabetical order \address{Douglas R. Anderson \newline Department of Mathematics and Computer Science, Concordia College, Moorhead, MN 56562, USA} \email{andersod@cord.edu} \address{Christopher C. Tisdell \newline School of Mathematics and Statistics, University of New South Wales, Sydney, UNSW 2052, Australia} \email{cct@unsw.edu.au} \thanks{Submitted October 31, 2006. Published January 27, 2007.} \thanks{Supported by grant DP0450752 from the Australian Research Council's Discovery Projects.} \subjclass[2000]{34B18, 34B27, 34B10, 39A10} \keywords{Boundary value problem; time scale; three-point; Green's function} \begin{abstract} We are concerned with the existence and form of positive solutions to a nonlinear third-order three-point nonlocal boundary-value problem on general time scales. Using Green's functions, we prove the existence of at least one positive solution using the Guo-Krasnoselskii fixed point theorem. Due to the fact that the nonlinearity is allowed to change sign in our formulation, and the novelty of the boundary conditions, these results are new for discrete, continuous, quantum and arbitrary time scales. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \section{Statement of the problem} We will develop an interval of $\lambda$ values whereby a positive solution exists for the following nonlinear, third-order, three-point, nonlocal boundary-value problem on arbitrary time scales \begin{gather} (px^{\Delta\Delta})^\nabla(t)=\lambda f(t,x(t)), \quad t\in[t_1,t_3]_\mathbb{T}, \label{bvp} \\ \alpha x(\rho(t_1))-\beta x^\Delta(\rho(t_1)) =\int_{\xi_1}^{\xi_2} a(t)x(t)\nabla t, \nonumber \\ x^\Delta(t_2)=0, \quad (px^{\Delta\Delta})(t_3) =\int_{\eta_1}^{\eta_2} b(t)(px^{\Delta\Delta})(t)\nabla t, \label{bvpbc} \end{gather} where: $p$ is a left-dense continuous, real-valued function on $\mathbb{T}$ with $p>0$; $\lambda>0$ is a real scalar; \begin{itemize} \item[(H1)] the real scalars $\alpha,\beta>0$ and the three boundary points satisfy $t_10$. \end{lemma} \begin{proof} Fix $s\in(\rho(t_1),t_2]_\mathbb{T}$. Then $u_1(\rho(t_1),s)=\frac{\beta}{\alpha}\int_{\rho(t_1)}^s\frac{\Delta r}{p(r)}>0$ and $u_1(\cdot,s)$ is increasing, so that $00$ for $s\in(\rho(t_1),t_2]_\mathbb{T}$ and $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$. Now fix $s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$. The branch $u_2$ is positive at $\rho(t_1)$, increases until $t_2$, and then decreases until $s$. We then switch to branch $v_2$, which continues to decrease, so that $v_2(t,s)\ge v_2(\sigma^2(t_3),s)$. As a function of $s$, $v_2(\sigma^2(t_3),s)$ is also decreasing, whence $v_2(t,s)\ge v_2(\sigma^2(t_3),\sigma^2(t_3))=u_2(\sigma^2(t_3))$ for $u_2$ given in \eqref{u2}. Thus $G(t_2,s)\ge G(t,s)$ for $s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$ and $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$ as well, and $G(t,s)>0$ for $s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$ and $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$ if and only if (H1) holds. \end{proof} \begin{remark}\label{rmk1.4} \rm If $\mathbb{T}=\mathbb{Z}$, $\alpha=1$, $\beta=0$, and $p(t)\equiv 1$, then the necessary and sufficient condition for the Green function to be positive is $t_2-t_1-1 \ge t_3-t_2$; see \cite{and2}. \end{remark} \begin{lemma}\label{lemma15} Assume {\rm (H1)}. For any $(t,s)\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\times(\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$, the Green function \eqref{greenf} corresponding to the problem \eqref{b1}, \eqref{b2} satisfies, using \eqref{u2}, $$\frac{u_2(t)}{u_2(t_2)} \le \frac{G(t,s)}{G(t_2,s)} \le 1.$$ \end{lemma} \begin{proof} The right-hand inequality follows from the previous lemma. For the left-hand inequality, we proceed by analyzing branches of the Green function \eqref{greenf}. For fixed $t\in[\rho(t_1),s)_\mathbb{T}$ and $s\in(t,t_2]_\mathbb{T}$, $$\frac{G(t,s)}{G(t_2,s)} = \frac{\int_{\rho(t_1)}^t\int_u^{s} \frac{\Delta r}{p(r)}\Delta u+\frac{\beta}{\alpha}\int_{\rho(t_1)}^{s} \frac{\Delta r}{p(r)}}{\int_{\rho(t_1)}^{s}\int_u^{s}\frac{\Delta r}{p(r)} \Delta u+\frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}\frac{\Delta r}{p(r)}} =:\phi(s).$$ Then \begin{align*} &\phi^{\nabla}(s)\\ &=\frac{(t-s)\left(\int_{\rho(t_1)}^t\int_u^s \frac{\Delta r}{p(r)}\Delta u+\frac{\beta}{\alpha}\int_{\rho(t_1)}^{s} \frac{\Delta r}{p(r)} \right)+ \left(t-\rho(t_1)+\frac{\beta}{\alpha}\right) \int_t^s\int_u^s\frac{\Delta r}{p(r)}\Delta u}{p^{\rho}(s) \left(\int_{\rho(t_1)}^{s}\int_u^{s}\frac{\Delta r}{p(r)}\Delta u +\frac{\beta}{\alpha}\int_{\rho(t_1)}^{s}\frac{\Delta r}{p(r)}\right) \left(\int_{\rho(t_1)}^{\rho(s)}\int_u^{\rho(s)}\frac{\Delta r}{p(r)} \Delta u+\frac{\beta}{\alpha}\int_{\rho(t_1)}^{\rho(s)}\frac{\Delta r}{p(r)} \right)}. \end{align*} The denominator of $\phi^\nabla$ is clearly positive, so consider the numerator, \begin{align*} \psi(s)&:= \psi_1(s)+\psi_2(s) \\ &:= (t-s)\int_{\rho(t_1)}^t\int_u^s\frac{\Delta r}{p(r)}\Delta u + \left(t-\rho(t_1)\right)\int_t^s\int_u^s\frac{\Delta r}{p(r)}\Delta u \\ &\quad +\frac{\beta}{\alpha}\Big((t-s)\int_{\rho(t_1)}^{s} \frac{\Delta r}{p(r)}+\int_t^s\int_u^s\frac{\Delta r}{p(r)} \Delta u \Big). \end{align*} Note that $\psi(t)=0$; the first part satisfies $$\psi_1^\nabla(s)=\frac{(t-\rho(t_1))\nu(s)}{p^\rho(s)} -\int_{\rho(t_1)}^t\int_u^s\frac{\Delta r}{p(r)}\Delta u =-\int_{\rho(t_1)}^t\int_u^{\rho(s)}\frac{\Delta r}{p(r)}\Delta u \le 0$$ for $s\in(t,t_2]_\mathbb{T}$, while the second part satisfies $$\psi_2(s)\le \frac{\beta}{\alpha}\Big((t-s)\int_{\rho(t_1)}^{s} \frac{\Delta r}{p(r)}+\int_t^s\int_t^s\frac{\Delta r}{p(r)}\Delta u \Big) =\frac{\beta}{\alpha}(t-s)\int_{\rho(t_1)}^{t}\frac{\Delta r}{p(r)} \le 0.$$ Therefore, the whole numerator satisfies $\psi(s)\le 0$, so that $\phi^\nabla\le 0$ and $\phi$ is nonincreasing as a function of $s$. Thus $\phi(s)\ge \phi(t_2)$ for $s\in(t,t_2]$; in other words, $$\frac{G(t,s)}{G(t_2,s)} \ge \frac{G(t,t_2)}{G(t_2,t_2)} = \frac{u_2(t)}{u_2(t_2)}.$$ For $s\in[\rho(t_1),t_2]_\mathbb{T}$ and $t\in[s,\sigma^2(t_3)]_\mathbb{T}$, $$\frac{G(t,s)}{G(t_2,s)}\equiv 1 \ge \frac{u_2(t)}{u_2(t_2)}.$$ If $s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$ and $t\in[\rho(t_1),s)_\mathbb{T}$, then $$\frac{G(t,s)}{G(t_2,s)} = \frac{u_2(t)}{u_2(t_2)}.$$ Finally, if $s\in[t_2,\sigma^2(t_3)]_\mathbb{T}$ and $t\in[s,\sigma^2(t_3)]_\mathbb{T}$, then $$\frac{G(t,s)}{G(t_2,s)}=\frac{u_2(t)+\int_s^t\int_{s}^u \frac{\Delta r}{p(r)}\Delta u}{u_2(t_2)} \ge \frac{u_2(t)}{u_2(t_2)}.$$ \end{proof} \section{Exploring the nonlocal problem} In this section we turn our attention to the problem $$\label{bvp2} (px^{\Delta\Delta})^\nabla(t)=\lambda y(t), \quad t\in[t_1,t_3]_\mathbb{T},$$ with nonlocal boundary conditions \eqref{bvpbc}, where $y$ is as in (H5), and $\lambda>0$. Assume (H2) and (H3), and and use \eqref{u2} to define $$\label{D} D:=u_2(t_2)\Big(1-\int_{\eta_1}^{\eta_2}b(t)\nabla t\Big) \Big(\int_{\xi_1}^{\xi_2} a(t)\nabla t-\alpha\Big) < 0.$$ \begin{lemma} Assume {\rm (H1)} through {\rm (H5)}. Then the nonhomogeneous dynamic equation \eqref{bvp2} with boundary conditions \eqref{bvpbc} has a unique solution $x^*$, where for $t \in [\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$, $$\label{form} x^*(t)=\lambda\Big(\int_{\rho(t_1)}^{t_3} G(t,s)y(s) \nabla s+A(y)u_2(t)+B(y)\left(u_2(t_2)-u_2(t)\right)\Big)$$ holds, where: $G(t,s)$ is the Green function \eqref{greenf} of the boundary-value problem \eqref{b1}, \eqref{b2}; and the functionals $A$ and $B$ are defined using \eqref{u2} by $$\label{A} A(y):=\frac{1}{D}\begin{vmatrix} \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t+u_2(t_2)\left(\alpha-\int_{\xi_1}^{\xi_2}a(t)\nabla t\right) \\ \int_{\eta_1}^{\eta_2}b(t)\left(\int_{t}^{t_3}y(s)\nabla s\right)\nabla t & \int_{\xi_1}^{\xi_2}a(t)\left(\int_{\rho(t_1)}^{t_3}G(t,s)y(s)\nabla s\right)\nabla t \end{vmatrix},$$ $$\label{B} B(y):=\frac{1}{D}\begin{vmatrix} \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t \\ \int_{\eta_1}^{\eta_2}b(t)\left(\int_{t}^{t_3}y(s)\nabla s\right)\nabla t & \int_{\xi_1}^{\xi_2}a(t)\left(\int_{\rho(t_1)}^{t_3}G(t,s)y(s)\nabla s\right)\nabla t \end{vmatrix}.$$ \end{lemma} \begin{proof} For $y$ as in (H5), we show that the function $x^*$ given in \eqref{form} is a solution of \eqref{bvp2} with conditions \eqref{bvpbc} only if $A(y)$ and $B(y)$ are given by \eqref{A} and \eqref{B}, respectively. If $x^*$ is a solution of \eqref{bvp2}, \eqref{bvpbc}, then $$x^*(t)=\lambda\int_{\rho(t_1)}^t G(t,s)y(s)\nabla s + \lambda\int_{t}^{t_3} G(t,s)y(s)\nabla s + Au_2(t) + B(u_2(t_2)-u_2(t))$$ for some constants $A$ and $B$. Taking the delta derivative with respect to $t$ yields $$x^{*\Delta}(t) = \lambda\int_{\rho(t_1)}^t G^\Delta(t,s)y(s)\nabla s + \lambda\int_{t}^{t_3} G^\Delta(t,s)y(s)\nabla s + A\int_t^{t_2}\frac{\Delta r}{p(r)} - B\int_t^{t_2}\frac{\Delta r}{p(r)};$$ since $p$ times the delta derivative of this expression is $$(px^{*\Delta\Delta})(t) = -\lambda \int_t^{t_3}y(s)\nabla s-A+B,$$ we see that \eqref{bvp2} holds. It is also clear that $x^{*\Delta}(t_2)=0$ is satisfied. To meet the other two boundary conditions in \eqref{bvpbc}, we must have at $\rho(t_1)$ that $$\label{311} \alpha Bu_2(t_2)= \int_{\xi_1}^{\xi_2}a(t) \Big(\lambda\int_{\rho(t_1)}^{t_3}G(t,s)y(s)\nabla s+Au_2(t)+ B(u_2(t_2) -u_2(t))\Big)\nabla t,$$ while at $t_3$ we have $$\label{312} -A+B = \int_{\eta_1}^{\eta_2}b(t) \Big(-\lambda\int_{t}^{t_3}y(s)\nabla s-A+B\Big)\nabla t.$$ Combining \eqref{311} and \eqref{312}, we arrive at the system of equations \begin{align*} & A\int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t + B\Big[\int_{\xi_1}^{\xi_2}a(t)\left(u_2(t_2)-u_2(t)\right)\nabla t -\alpha u_2(t_2)\Big] \\ &= -\lambda\int_{\xi_1}^{\xi_2}a(t) \Big(\int_{\rho(t_1)}^{t_3}G(t,s)y(s)\nabla s\Big)\nabla t \end{align*} and $$A\Big[\int_{\eta_1}^{\eta_2}b(t)\nabla t-1\Big] + B\Big[1-\int_{\eta_1}^{\eta_2}b(t)\nabla t\Big] = -\lambda\int_{\eta_1}^{\eta_2}b(t)\Big(\int_{t}^{t_3}y(s)\nabla s\Big) \nabla t.$$ The determinant of the coefficients of $A$ and $B$ is $D$, given by \eqref{D}, which is negative, and by elementary linear algebra we verify \eqref{A} and \eqref{B} with $\lambda$ factored out. Also note that $A(y)>B(y)>0$ since $D<0$ and $$A(y)-B(y)=\frac {\int_{\eta_1}^{\eta_2}b(t)\Big(\int_{t}^{t_3}y(s) \nabla s\Big)\nabla t}{1-\int_{\eta_1}^{\eta_2}b(t)\nabla t}.$$ \end{proof} \begin{corollary} \label{coro2.2} Assume {\rm (H1)} through {\rm (H5)}. Then the unique solution $x^*$ as in \eqref{form} of the problem \eqref{bvp2}, \eqref{bvpbc} satisfies $x^*(t)\ge 0$ for $t\in[\rho(t_1),\sigma^2(t_3)]$. \end{corollary} \begin{proof} From Lemma \ref{lemma13} we know that on the Green function \eqref{greenf} satisfies $G(t,s)\ge 0$. Assumption (H3) applied to \eqref{A} and \eqref{B} imply that $A(y)>B(y)>0$. \end{proof} \begin{lemma}\label{lemma23} Assume {\rm (H1)} through {\rm (H5)}. Then the unique solution $x^*$ as in \eqref{form} of the problem \eqref{bvp2}, \eqref{bvpbc} satisfies $$\theta\|x^*\| \le x^*(t) \le \lambda \theta \Theta \quad\text{on}\quad [\rho(t_1),\sigma^2(t_3)]_\mathbb{T},$$ where using \eqref{u2} we take $$\label{theta} \theta:=\min\left\{\frac{u_2(\rho(t_1))}{u_2(t_2)},\frac{u_2(\sigma^2(t_3))}{u_2(t_2)}\right\}\in(0,1), \quad \|x^*\|:=\max_{t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}}x^*(t)=x^*(t_2),$$ and $$\label{Theta} \Theta:=\frac{1}{\theta}u_2(t_2)\left(1 + \bar{A}\right)\int_{\rho(t_1)}^{t_3}y(s)\nabla s$$ for $$\bar{A}:=\frac{1}{D}\begin{vmatrix} \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t+u_2(t_2)\left(\alpha-\int_{\xi_1}^{\xi_2}a(t)\nabla t\right) \\ \int_{\eta_1}^{\eta_2}b(t)\nabla t & u_2(t_2)\int_{\xi_1}^{\xi_2}a(t)\nabla t \end{vmatrix}.$$ \end{lemma} \begin{proof} >From previous work, it is clear that for all $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$, $$x^*(t) \le x^*(t_2) =\lambda\Big(\int_{\rho(t_1)}^{t_3}G(t_2,s)y(s)\nabla s+A(y)u_2(t_2) \Big).$$ For $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$, from Lemma \ref{lemma13} and Lemma \ref{lemma15}, the Green function \eqref{greenf} satisfies $$\frac{G(t,s)}{G(t_2,s)} \ge \frac{u_2(t)}{u_2(t_2)} \ge \min\Big\{\frac{u_2(\rho(t_1))}{u_2(t_2)}, \frac{u_2(\sigma^2(t_3))}{u_2(t_2)}\Big\} = \theta\in(0,1)$$ by (H1) and \eqref{u2}, and \begin{align*} & x^*(t) \\ &= \lambda\Big(\int_{\rho(t_1)}^{t_3} \frac{G(t,s)}{G(t_2,s)}G(t_2,s)y(s)\nabla s+A(y)\frac{u_2(t)}{u_2(t_2)} u_2(t_2)+B(y)(u_2(t_2)-u_2(t))\Big) \\ &\ge \lambda\Big(\int_{\rho(t_1)}^{t_3} \theta G(t_2,s)y(s) \nabla s+A(y)\theta u_2(t_2)\Big) \\ & = \theta\|x^*\|. \end{align*} Consequently, $\theta\|x^*\| \le x^*(t)$ for all $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$. For $D$ in \eqref{D} and $A(y)$ in \eqref{A}, \begin{aligned} & A(y)\\ & = \frac{1}{D}\begin{vmatrix} \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t+u_2(t_2) \left(\alpha-\int_{\xi_1}^{\xi_2}a(t)\nabla t\right) \\ \int_{\eta_1}^{\eta_2}b(t)\left(\int_{t}^{t_3}y(s)\nabla s\right)\nabla t & \int_{\xi_1}^{\xi_2}a(t)\left(\int_{\rho(t_1)}^{t_3}G(t,s)y(s) \nabla s\right)\nabla t \end{vmatrix} \nonumber \\ &\le \frac{1}{D}\begin{vmatrix} \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t+u_2(t_2) \left(\alpha-\int_{\xi_1}^{\xi_2}a(t)\nabla t\right) \\ \int_{\eta_1}^{\eta_2}b(t)\nabla t & u_2(t_2) \int_{\xi_1}^{\xi_2}a(t)\nabla t \end{vmatrix} \int_{\rho(t_1)}^{t_3}y(s)\nabla s \nonumber \\ & = \bar{A}\int_{\rho(t_1)}^{t_3}y(s)\nabla s < \infty. \end{aligned} \label{AAbar} As a result, for $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$, \begin{align*} x^*(t) &\le \lambda \Big(\int_{\rho(t_1)}^{t_3}G(t_2,s)y(s)\nabla s+A(y)u_2(t_2) \Big) \\ &\le \lambda u_2(t_2)(1+\bar{A})\int_{\rho(t_1)}^{t_3} y(s)\nabla s \\ &\le \lambda\theta\Theta \end{align*} using \eqref{theta}, \eqref{Theta}, and \eqref{AAbar}. \end{proof} \section{An existence result on cones} Let $\mathcal{B}$ denote the Banach space $C[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$ with the norm $$\|x\|=\sup_{t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}}|x(t)|.$$ Define the cone $\mathcal{P}\subset\mathcal{B}$ by $$\mathcal{P}=\{x\in\mathcal{B}: x(t)\ge \theta\|x\| \;\text{on}\; [\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\},$$ where $\theta$ is given in \eqref{theta}. Consider the related boundary-value problem \begin{gather*} (px^{\Delta\Delta})^\nabla(t)=\lambda f^*(t,x(t)), \quad t\in[t_1,t_3]_\mathbb{T}, \\ \alpha x(\rho(t_1))-\beta x^\Delta(\rho(t_1))=\int_{\xi_1}^{\xi_2} a(t)x(t)\nabla t, \\ x^\Delta(t_2)=0, \quad (px^{\Delta\Delta})(t_3)=\int_{\eta_1}^{\eta_2} b(t)(px^{\Delta\Delta})(t)\nabla t, \end{gather*} where $$\label{fstar} f^*(t,x(t)):=f(t,x^\dagger(t)) + y(t), \quad x^\dagger(t):=\max\{x(t)-x^*(t),0\},$$ such that $x^*$ given in \eqref{form} is the solution of \eqref{bvp2}, \eqref{bvpbc}, and $y$ is from (H5). For any fixed $x\in\mathcal{P}$, $x^\dagger\le x\le \|x\|$ and by (H5), \begin{align*} &\int_{\rho(t_1)}^{t_3}G(t,s)f^*(s,x(s))\nabla s\\ &\le \int_{\rho(t_1)}^{t_3}G(t_2,s)\left(z(s)h(x^\dagger(s))+y(s)\right)\nabla s \\ &\le \Big(\max_{0\le\tau\le\|x\|}h(\tau)+1\Big) \int_{\rho(t_1)}^{t_3}G(t_2,s)\left(z(s)+y(s)\right)\nabla s<\infty. \end{align*} For $A$ in \eqref{A} and using \eqref{AAbar}, we have $$A(z+y) \le \bar{A}\int_{\rho(t_1)}^{t_3}(z(s)+y(s))\nabla s;$$ likewise for $B$ in \eqref{B} and using (H5), \begin{align*} &B(z+y)\\ & = \frac{1}{D}\begin{vmatrix} \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t \\ \int_{\eta_1}^{\eta_2}b(t)\big(\int_{t}^{t_3}(z(s)+y(s))\nabla s\big) \nabla t & \int_{\xi_1}^{\xi_2}a(t)\big(\int_{\rho(t_1)}^{t_3}G(t,s)(z(s)+y(s)) \nabla s\big)\nabla t \end{vmatrix} \\ &\le \frac{1}{D}\begin{vmatrix} \int_{\eta_1}^{\eta_2}b(t)\nabla t-1 & \int_{\xi_1}^{\xi_2}a(t)u_2(t)\nabla t \\ \int_{\eta_1}^{\eta_2}b(t)\nabla t & u_2(t_2)\int_{\xi_1}^{\xi_2}a(t)\nabla t \end{vmatrix}\int_{\rho(t_1)}^{t_3}(z(s)+y(s))\nabla s \\ & < \infty. \end{align*} This allows us to define for $y\in\mathcal{P}$ the operator $T:\mathcal{P}\to \mathcal{B}$ for $t\in [\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$ by $$\label{T} (Tx)(t):=\lambda\Big(\int_{\rho(t_1)}^{t_3} G(t,s)f^*(s,x(s))\nabla s + A(f^*)u_2(t) +B(f^*)(u_2(t_2)-u_2(t))\Big),$$ using \eqref{A}, \eqref{B}, and \eqref{fstar}. \begin{lemma}\label{lemma31} Assume {\rm (H1)} through {\rm (H5)}. Then $T:\mathcal{P}\to\mathcal{P}$ is completely continuous. \end{lemma} \begin{proof} For any $x\in\mathcal{P}$, Lemmas \ref{lemma13}, \ref{lemma15} and Lemma \ref{lemma23} imply that $(Tx)(t)\ge\theta \|Tx\|$ on $[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$, so that $T(\mathcal{P})\subseteq\mathcal{P}$. By a standard application of the Arzela-Ascoli Theorem, $T$ is completely continuous. \end{proof} To establish an existence result we will employ the following fixed point theorem due to Guo and Krasnoselskii \cite{gu}, and seek a fixed point of $T$ in $\mathcal{P}$. \begin{theorem}\label{fixedpt} Let $E$ be a Banach space, $P\subseteq E$ be a cone, and suppose that $\mathcal{S}_1$, $\mathcal{S}_2$ are bounded open balls of $E$ centered at the origin with $\overline{\mathcal{S}}_1\subset\mathcal{S}_2$. Suppose further that $L:P\cap(\overline{\mathcal{S}}_2\setminus\mathcal{S}_1)\to P$ is a completely continuous operator such that either \begin{itemize} \item[$(i)$] $\|Ly\| \le \|y\|$, $y\in P\cap\partial\mathcal{S}_1$ and $\|Ly\| \ge \|y\|$, $y\in P\cap\partial\mathcal{S}_2$, or \item[$(ii)$] $\|Ly\| \ge \|y\|$, $y\in P\cap\partial\mathcal{S}_1$ and $\|Ly\| \le \|y\|$, $y\in P\cap\partial\mathcal{S}_2$ \end{itemize} holds. Then $L$ has a fixed point in $P\cap(\overline{\mathcal{S}}_2\setminus\mathcal{S}_1)$. \end{theorem} \begin{theorem}\label{ethm1} Assume {\rm (H1)} through {\rm (H5)}. Then there exists $\lambda^*>0$ such that the third-order nonlocal time scale boundary value problem \eqref{bvp}, \eqref{bvpbc} has at least one positive solution in $\mathcal{P}$ for any $\lambda\in(0,\lambda^*)$. \end{theorem} \begin{proof} By Lemma \ref{lemma31}, $T:\mathcal{P}\to\mathcal{P}$ given by \eqref{T} is completely continuous. Take $\mathcal{S}_1:=\{x\in\mathcal{B}:\|x\|<\Theta\}$ for $\Theta$ given in \eqref{Theta}, and let $$\lambda^*:=\min\Bigg\{1,\; \frac{\int_{\rho(t_1)}^{t_3}y(s)\nabla s}{\theta \Big(\max\limits_{0\le\tau\le\Theta}h(\tau)+1\Big) \int_{\rho(t_1)}^{t_3}(z(s)+y(s))\nabla s}\Bigg\}.$$ Then for any $x\in\mathcal{P}\cap\partial\mathcal{S}_1$, $$0 \le x^\dagger(s) \le x(s) \le \|x\| = \Theta, \quad s\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T},$$ and, for $\bar{A}$ as in the statement of Lemma \ref{lemma23}, \begin{align*} (Tx)(t) &\le \lambda\Big(\int_{\rho(t_1)}^{t_3} G(t_2,s)f^*(s,x(s))\nabla s +A(f^*)u_2(t_2) \Big) \\ &\le \lambda \Big(\max_{0\le\tau\le\|x\|}h(\tau)+1\Big) \int_{\rho(t_1)}^{t_3}G(t_2,s)\left(z(s)+y(s)\right)\nabla s \\ &\quad + \lambda \bar{A}\Big(\max_{0\le\tau\le\|x\|}h(\tau)+1\Big) \Big(\int_{\rho(t_1)}^{t_3}\big(z(s)+y(s)\big)\nabla s\Big)u_2(t_2) \\ &\le \lambda^* u_2(t_2)\left(1+\bar{A}\right) \Big(\max_{0\le\tau\le\|x\|}h(\tau)+1\Big)\int_{\rho(t_1)}^{t_3} \left(z(s)+y(s)\right)\nabla s \\ &\le \Theta=\|x\|. \end{align*} Hence $\|Tx\|\le\|x\|$ for $x\in\mathcal{P}\cap\partial\mathcal{S}_1$. Pick $\Upsilon\in\mathbb{R}$ such that $\Upsilon>0$ and $$1 \le \frac{\lambda \Upsilon \theta}{\Theta+1}\int_{\xi_1}^{t_2}G(\xi_1,s) \nabla s.$$ By (H4), for any $t\in[\xi_1,t_2]_\mathbb{T}$, there exists a constant $K>0$ such that $f(t,y)>\Upsilon y$ for $y>K$. Pick $Q:=\max\big\{\lambda(\Theta+1),\Theta+1,\frac{K(\Theta+1)}{\theta}\big\}$. If \$\mathcal{S}_2:=\{y\in\mathcal{B}:\|y\|