\documentclass[reqno]{amsart} \usepackage{graphicx,amssymb,hyperref} % Note dpflatex torres.tex did not rotate the fiures. I have to rotate and save them \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 130, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/130\hfil Method of straight lines] {Method of straight lines for a Bingham problem as a model for the flow of waxy crude oils} \author[G. A. Torres, C. Turner\hfil EJDE-2005/130\hfilneg] {Germ\'an Ariel Torres, Cristina Turner} \address{ Germ\'an Ariel Torres \hfill\break Universidad Nacional de C\'ordoba \\ CIEM - CONICET, C\'ordoba, Argentina} \email{torres@mate.uncor.edu \quad http://www.famaf.unc.edu.ar/torres} \address{Cristina Turner \hfill\break Universidad Nacional de C\'ordoba \\ CIEM - CONICET, C\'ordoba, Argentina} \email{turner@mate.uncor.edu \quad http://www.famaf.unc.edu.ar/turner} \date{} \thanks{Submitted April 15, 2005. Published November 24, 2005.} \subjclass[2000]{35A40, 35B40, 35R35, 65M20, 65N40} \keywords{Bingham fluid; straight lines; non-newtonian fluids} \begin{abstract} In this work, we develop a method of straight lines for solving a Bingham problem that models the flow of waxy crude oils. The model describes the flow of mineral oils with a high content of paraffin at temperatures below the cloud point (i.e. the crystallization temperature of paraffin) and more specifically below the pour point at which the crystals aggregate themselves and the oil takes a jell-like structure. From the rheological point of view such a system can be modelled as a Bingham fluid whose parameters evolve according to the volume fractions of crystallized paraffin and the aggregation degree of crystals. We prove that the method is well defined for all times, a monotone property, qualitative behaviour of the solution, and a convergence theorem. The results are compared with numerical experiments at the end of this article. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \section{Introduction} As a justification for using this model, we quote statements made by Farina and Fasano in \cite{farina-fasano}: \begin{quote} Crude oils in many reservoirs throughout the world contain significant quantities of wax which can crystallize during production, transportation, and storage \cite{tuttle}. This can cause severe difficulties in pipelining and storage. At sufficiently high temperatures, the waxy crude oils (i.e., oils which contain a great deal of wax), although chemically very complex, are simple Newtonian fluids. As the temperature is reduced, the flow properties of these crudes can radically change from the simple Newtonian flow to a very complex behavior due to the crystallization of waxes \cite{denis-durand}. The waxes basically consist of n-alkanes, usually ranging from C$_{18}$H$_{38}$ to C$_{40}$H$_{82}$, which crystallize (as soon as the equilibrium temperature and pressure is reached), forming an interlocking structure of plate, needle, or malformed crystals \cite{ferris-cowles}. When the oil is cooled to a temperature lower than the crystallization point (generally called pour point), the crystals, growing and agglomerating, entrap the oil into a jell-like structure. Consequently, the flow properties of the oil become distinctly non-Newtonian. A yield-stress (the minimum stress required to start the flow) can be detected. Moreover, the flow properties are complicated by their critical dependence upon their mechanical and thermal history'. The viscosity of the waxy crudes can be greatly reduced by a continued shear. This fact seems to indicate a kind of thixotropy. The disintegration of large wax agglomerates appears to be the primary cause of the lower viscosity \cite{wardhaugh-boger}. \end{quote} This paper deals with the model proposed by Farina and Fasano in \cite{farina-fasano}. In that work they study the low temperature behavior of a waxy crude oil in a laboratory experimental loop. In the second section we describe the physical model for this class of fluids, and the study of the related mathematical problem. In the third section we describe the mathematical problem, while in the fourth we remember some results present in \cite{farina-tesilaurea}. In the fifth section a method of straight lines for a Bingham problem as a model for the flow of a waxy crude oils is developed. We prove that the method is well defined for all times, a monotone property, qualitative behaviour of the solution, and asymptotic convergence for large times. In the sixth section the results are compared with numerical experiments. \section{The physical problem} First, we state the physical assumptions. \begin{itemize} \item[(F1)] \label{fis-1} Low, uniform, and constant temperature. A first simplifying assumption is that the temperature is uniform, constant, and below the so-called pour point'', so that the density of the crystallized wax is constant in space and time. Therefore, the non-Newtonian behaviour of the fluid has to be only atributed to the aglomeration of wax crystals. If the temperature field is denoted by $T(\vec{x},t)$, in a domain $V$ in $\mathbb{R}^2$: $$\label{TleqTpp} \mbox{constant} = T(\vec{x},t) \leq T_{pp}, \quad \vec{x} \in V, \quad t \geq 0,$$ where $T_{pp}$ is the pour point''. \item[(F2)] Incompressible Fluid. A very reasonable assumption, consistent with the previous one, is to take, $$\label{rhoconst} \rho = \mbox{constant},$$ where $\rho$ is the oil density. If $\vec{v}(\vec{x},t)$ is the velocity field of the fluid, using (\ref{rhoconst}) and the continuity equation, we conclude that $\nabla \cdot \vec{v} = 0$ in $V$. \item[(F3)] Laminar Flow. This assumption is justified by the fact that, for low temperatures, the Reynolds number (evaluated for typical pipelines values) is less than the threshold of turbulent flow. \end{itemize} Now let us pass to define a rheological model which takes into account the experimental data. Waxy crude oils show, at low temperature, the presence of a yield-stress \cite{douglas-gallie}. According to this experimental evidence, we describe them as Bingham fluids. Roughly speaking, a Bingham fluid is a non-newtonian fluid which behaves like a rigid body when the shear stress $\tau$ is less than a threshold value $\tau_0$, while it behaves like a viscous fluid when the stress exceeds $\tau_0$, and for which the relationship between the stress $\tau$ and the shear strain $\gamma$ is linear, that is $$\label{tautau0etagamma} \tau = \tau_0 + \eta \gamma$$ where $\eta$ is the viscosity. To consider the evolution of the sheared system exhibiting a kind of thixotropy'' we introduce a time-dependent parameter $\alpha$, defined as the ratio between the aggregated solid paraffin mass by volume unit, and the total mass of paraffin present in the fluid. Therefore, $\alpha$ is a quantity ranging in the interval $[0,1]$. We assume that the yield-stress $\tau$ is influenced only by the agglomeration factor, that is $$\label{dependenciatau0alfa} \tau_0 = \tau_0(\alpha),$$ with \begin{itemize} \item[(Y1)] $\tau_0 : [0,1] \to [\tau_m, \tau_M]$, where $0 \leq \tau_m < \tau_M < +\infty$. \item[(Y2)] $\tau_0 \in C^1([0,1])$. \item[(Y3)] $\tau_0$ is a non-decreasing function. \end{itemize} A natural way of writing down an evolution equation for $\alpha$ accounting both for the spontaneous aggregation of paraffin crystals and agglomerates fragmentation (explaining `thixotropy'') is the following $$\label{evoluciodealfa} \begin{gathered} \alpha'(t) = K_1 (1-\alpha (t)) - K_2\alpha (t)|\overline{W}(t)|,\\ \alpha(0) = \alpha_0. \end{gathered}$$ where $\overline{W}(t) = \frac{1}{V} \int_V W(\vec{x},t) d\vec{x}$ is the power dissipated in the flow by the viscous force, and $K_1$, $K_2$ are constants (they can be obtained experimentally). \section{The mathematical problem} In this section we will consider the Bingham problem in plane geometry. We consider a fluid between two parallel plates. Using the Navier-Stokes equation for the viscous region and the Newton's law for the rigid zone, we model the behavior of the system. The boundary that separates the two regions is an unknown that evolves in time. It is one of the most important unknown quantities of the problem. We assume that the fluid is incompressible, laminar, and with constant density $\rho$. Fixing the $x$ coordinate along the direction of motion, $y$ the perpendicular coordinate to the plates, and $z$ the remaining coordinate, we make the following assumptions: \begin{enumerate} \item The pressure gradient, $\nabla p$, is applied in only one direction, that is, $\frac {\partial p}{\partial y} = \frac {\partial p}{\partial z} = 0$. \item The fluid is laminar, that is, the velocities $v$ and $w$ satisfy $v = w = 0$. \item The non-zero component of the velocity $u$ depends only on time, $t$, and on the perpendicular position, $y$, that is, $\frac {\partial u}{\partial z} = \frac {\partial u}{\partial x} = 0$. \item There is no transport of fluid through the free boundary, $y=s(t)$. This is a condition of no deformation, that is, $u_y (s(t),t) = 0$, for all $t > 0$. \item The velocity of the fluid $u$ at the walls of the plates is zero. This is an adherence condition. \end{enumerate} Using the above hypotheses, we obtain a system of partial differential equations, which we call problem (P), $$\label{problemP} \begin{gathered} \rho u_t - \eta u_{yy} = f(t), \quad s(t) < y < L, \; t > 0, \\ u(L,t) = 0, \quad t > 0, \\ u(y,0) = u_0 (y), \quad s(0) = s_0, \; 0 < s_0 < y < L, \\ u_y (s(t),t) = 0, \quad t > 0, \\ u_t (s(t),t) = \frac{1}{\rho} \big( f(t) - \frac {\tau_0}{s(t)} \big), \quad t > 0. \end{gathered}$$ where $f(t)$ represents $- \partial p / \partial x$. We add to the problem above the equation (\ref{evoluciodealfa}) for the evolution of $\alpha$. In this case the dissipated power is $$\label{Wgeometriaplana} W(t) = \frac{\eta}{L} \int_{s(t)}^{L} u_y^2(y,t) dy + \frac{\tau_0(\alpha (t))}{L} u(s(t),t).$$ Then the evolution equation for $\alpha$ is $$\label{ecuacionevolucionalfageometriaplana} \alpha'(t) = K_1 (1-\alpha (t)) - \alpha (t)\frac{K_2}{L} \Big| \eta \int_{s(t)}^{L} u_y^2(y,t) dy + \tau_0(\alpha (t)) u(s(t),t) \Big|,$$ \noindent with the initial condition $$\label{condicioninicialevolucionalfa} \alpha(0) = \alpha_0.$$ Now we have the following model for the problem \eqref{problemP}, called $(P^{\alpha})$, \begin{gather} \rho u_t - \eta u_{yy} = f(t), \quad s(t) < y < L, t > 0, \label{problemaPtau0var-1} \\ u(L,t) = 0, \quad t > 0, \label{problemaPtau0var-2} \\ u_y(s(t),t) = 0, \quad t > 0, \label{problemaPtau0var-3} \\ u_t(s(t),t) = \frac{1}{\rho} \big( f(t) - \frac{\tau_0(\alpha)}{s(t)} \big), \quad t > 0, \label{problemaPtau0var-4} \\ s(0) = s_0, \label{problemaPtau0var-8} \\ u(y,0) = u_0(y), \quad s_0 \leq y \leq L, \label{problemaPtau0var-5} \\ \alpha'(t) = K_1(1-\alpha(t)) - \frac{\alpha(t)K_2}{L} \big| \eta \int_{s(t)}^{L} u_y^2 dy - \tau_0(\alpha(t)) \int_{s(t)}^{L} u_y dy \big|, \quad t > 0, \label{problemaPtau0var-6} \\ \alpha(0) = \alpha_0. \label{problemaPtau0var-7} \end{gather} To get the equation (\ref{problemaPtau0var-6}) we use the equation (\ref{ecuacionevolucionalfageometriaplana}) and the fact that $u(y,t) = - \int _{y}^{L} u_y(\xi,t) d\xi$. The following problems will be useful for obtaining the discrete solution. We transform the problem $(P^{\alpha})$ using the function $w=u_y$ to obtain a new problem, denoted by $(P_y^{\alpha})$, \begin{gather} \rho w_t - \eta w_{yy} = 0, \quad s(t) < y < L, \quad t > 0, \label{problemaPytau0var-1} \\ w_y(L,t) = -f(t)/\eta, \quad t > 0, \label{problemaPytau0var-2} \\ w(s(t),t) = 0, \quad t > 0, \label{problemaPytau0var-3} \\ w_y(s(t),t) = - \tau_0(\alpha) / \eta s(t), \quad t > 0, \label{problemaPytau0var-4} \\ s(0) = s_0, \label{problemaPytau0var-8} \\ w(y,0) = u_0'(y), \quad 0 < s_0 \leq y \leq L, \label{problemaPytau0var-5} \\ \alpha'(t) = K_1(1-\alpha(t)) - \frac{\alpha(t)K_2}{L} \Big| \eta \int_{s(t)}^{L} w^2 dy - \tau_0(\alpha(t)) \int_{s(t)}^{L} wdy \Big|, \quad t > 0, \label{problemaPytau0var-6} \\ \alpha(0) = \alpha_0. \label{problemaPytau0var-7} \end{gather} If $z = u_t$ then the function $z$ satisfies the problem, $(P_t^{\alpha})$, \begin{gather} \rho z_t - \eta z_{yy} = f'(t), \quad s(t) < y < L, \quad t > 0, \label{problemaPztau0var-1} \\ z(L,t) = 0, \quad t > 0, \label{problemaPztau0var-2} \\ z(s(t),t) = \frac{1}{\rho} \big( f(t) - \frac{\tau_0(\alpha)}{s(t)} \big), \quad t > 0, \label{problemaPztau0var-3} \\ z_y(s(t),t) = \frac{\tau_0(\alpha) s'(t)}{\eta s(t)}, \quad t > 0, \label{problemaPztau0var-4} \\ s(0) = s_0, \label{problemaPztau0var-5} \\ z(y,0) = \frac{\eta}{\rho}u_0''(y) + \frac{f(0)}{\rho}, \quad 0 < s_0 \leq y \leq L, \label{problemaPztau0var-6} \\ \alpha'(t) = K_1(1-\alpha(t)) - \frac{\alpha(t)K_2}{L} \Big| \eta \int_{s(t)}^{L} w^2 dy - \tau_0(\alpha(t)) \int_{s(t)}^{L} wdy \Big|, \quad t > 0, \label{problemaPztau0var-7} \\ \alpha(0) = \alpha_0. \label{problemaPztau0var-8} \end{gather} In \cite{farina-tesilaurea} a theorem of local existence and uniqueness of the problem $(P_t^{\alpha})$ is proved. \section{Theoretical results} Now, we recall some hypotheses and we will add a few more. \\ {\bf Y-Hypotheses on $\tau_0$:} \begin{itemize} \item[(Y1)] $\tau_0 : [0,1] \to [\tau_m,\tau_M]$, with $0 < \tau_m < \tau_M < \infty$. \label{datosY1} \item[(Y2)] $\tau_0 \in C^1([0,1])$. \label{datosY2} \item[(Y3)] $\tau_0$ is monotone non-decreasing. \label{datosY3} \item[(Y4)] $\tau_m \leq \tau_0(\alpha_0) \leq \tau_M$. \label{datosY4} \item[(Y5)] $\tau_0$ is Lipschitz with constant $N$; that is, $$\label{datosY5} | \tau_0(\alpha_1) - \tau_0(\alpha_2) | \leq N | \alpha_1 - \alpha_2 |.$$ \end{itemize} {\bf P-Hypotheses on $f$:} \begin{itemize} \item[(P1)] $0 < f_m < f(t) < f_M$ for all $t > 0$. \label{datosP1} \item[(P2)] Operability condition: $f_m > \tau_M/L$. \label{datosP2} \end{itemize} This implies that the pressure gradient is bounded, but is big enough to allow that the fluid can circulate inside the pipe. \noindent{\bf A-Hypotheses on $\alpha_0$ and $s_0$:} \begin{itemize} \item[(A1)] $s_m < s_0 < s_M$ with $s_m = \tau_m/f_M$ y $s_M = \tau_M/f_m$. \label{datosA1} \item[(A2)] $0 \leq \alpha_0 \leq 1$. \label{datosA2} \end{itemize} The aggregation factor is in the interval $[0,1]$ by definition. \noindent {\bf U-Hypotheses on $u_0(y)$:} \begin{itemize} \item[(U1)] $u_0(y) \in C^3([s_0,L])$. \label{datosU1} \item[(U2)] $u_0(y) \geq 0$ for $s_0 \leq y \leq L$; $u_0(L) = 0$. \label{datosU2} \item[(U3)] $u_0'(y) \leq 0$ for $s_0 \leq y \leq L$; $u_0'(s_0) = 0$. \label{datosU3} $u_m'(y) \leq u_0'(y) \leq u_M'(y)$ in $s_0 \leq y \leq L$ where \begin{gather}\label{umprima} u_m'(y) = \begin{cases} 0, & 0 \leq y \leq s_m, \\ -\frac{f_M}{\eta}(y-s_m), & s_m \leq y \leq L, \end{cases} \\ \label{uMprima} u_M'(y) = \begin{cases} 0, & 0 \leq y \leq s_M, \\ -\frac{f_m}{\eta} (y - s_M), & s_M \leq y \leq L. \end{cases} \end{gather} \item[(U4)] $u_0''(y) < 0$ in $s_0 < y < L$; $u_0''(s_0) = - \tau_0(\alpha_0)/s_0$. \label{datosU4} These are conditions of smoothness on $u_0$. \end{itemize} As a consequence of the Maximum Principle, and the Hopf's Lemma, we have \begin{gather} u(y,t) \geq 0, \quad s(t) < y < L, \quad 0 < t < T_0, \label{uleq0tau0var} \\ u_y(y,t) \leq 0, \quad s(t) < y < L, \quad 0 < t < T_0, \label{uyleq0tau0var} \\ u_{yy}(y,t) \leq 0, \quad s(t) < y < L, \quad 0 < t < T_0. \label{uyyleq0tau0var} \end{gather} where $T_0$ is the maximum time of existence. Besides that, properties of the free boundary problem and the relationship of the velocity field with respect to the initial conditions can be proved. \begin{gather} s_m < s(t) < s_M < L, \label{sm0$and$g'(s) > 0$for$s \in [0,L]$. \section{Method of the Straight Lines.}\label{Method} Some examples of this method can be found in \cite{bachelis-melamed-shlyaifer}-\nocite{davis-kapadia} \nocite{meyer}\nocite{meyer1}\nocite{vasilev} \nocite{torres-turner}\cite{torres-turner1}. In this case, the idea is to discretize the time, and get an ordinary differential equation system. The difficulty is that we do not know if the numerical solution exists, because the domain is also an unknown. Choosing a fixed time step$\Delta t >0$, we define: \begin{gather*} t_n = (n-1) \Delta t, \quad n \in \mathbb{N}, \\ s_n = s(t_n), \quad n \in \mathbb{N}, \\ f_n = f(t_n), \quad n \in \mathbb{N}, \\ w_n(r) = w(r,t_n), \quad n \in \mathbb{N}, \\ q = \sqrt{\frac{1}{\Delta t}}. \end{gather*} We approximate time derivatives with the incremental quotient, and the evolution equation for$\alpha$is discretized with the Euler's method. In this way, the$(P_y^{\alpha})$system is transformed into the$(P_{y,d}^{\alpha})$system. For all$n$in$\mathbb{N}$we have: \begin{gather} w_{n+1}'' - \frac{\rho}{\eta}q^2 w_{n+1} = - \frac{\rho}{\eta}q^2w_n, \quad s_{n+1} < y < L, \quad n \in \mathbb{N}, \label{lineasdiscretizadotau0var-1} \\ w_{n+1}(s_{n+1}) = 0, \quad n \in \mathbb{N}, \label{lineasdiscretizadotau0var-3} \\ w_{n+1}'(s_{n+1}) = - \tau_0(\alpha_{n+1})/\eta s_{n+1}, \quad n \in \mathbb{N}, \label{lineasdiscretizadotau0var-4} \\ w_{n+1}'(L) = - f_{n+1}/\eta, \quad n \in \mathbb{N}, \label{lineasdiscretizadotau0var-2} \\ s_1 = s_0, \label{lineasdiscretizadotau0var-5} \\ w_1(y) = u_0'(y), \quad s_0 \leq y \leq L, \label{lineasdiscretizadotau0var-6} \\ \alpha_{n+1} = \alpha_n + \frac{1}{q^2} \Big( K_1 (1-\alpha_n) - \alpha_n \frac{K_2}{L} \Big| \eta \int_{s_n}^{L} w_n^2 dy - \tau_0(\alpha_n) \int_{s_n}^{L} w_n dy \Big| \Big), \quad n \in \mathbb{N}. \label{lineasdiscretizadotau0var-7} \\ \alpha_1 = \alpha_0. \label{lineasdiscretizadotau0var-8} \end{gather} Now we will try to prove existence and uniqueness of the solution of the problem$(P_{y,d}^{\alpha})$, but before we will prove a technical lemma. \begin{lemma}\label{lemadespejewn+1tau0var} Consider the system of equations $$\label{ecuaciondespejewn+1tau0var} \begin{gathered} w'' - \frac{\rho}{\eta}q^2 w = g, \quad s < y < L,\\ w(s) = 0, \\ w'(s) = - T/\eta s, \end{gathered}$$ where$\rho$,$\eta$,$q$,$s$,$L$y$T$are positive numbers and$gis a continuous function. Then \begin{gather} w(y) = - \frac{T}{\sqrt{\rho \eta} sq} \sinh \big( \sqrt{\frac{\rho}{\eta}}q(y-s) \big) + \int_{s}^{y} \frac{g(\xi)}{\sqrt{\rho/\eta} q} \sinh \big( \sqrt{\frac{\rho}{\eta}} q(y-\xi) \big) d\xi, \label{ecuaciondespejewwn+1tau0var} \\ w'(y) = - \frac{T}{\eta s} \cosh \big( \sqrt{\frac{\rho}{\eta}} q(y-s) \big) + \int_{s}^{y} g(\xi) \cosh \big( \sqrt{\frac{\rho}{\eta}} q(y -\xi) \big) d\xi. \label{ecuaciondespejewprimawn+1tau0var} \end{gather} \end{lemma} \begin{proof} The proof of this lemma can be found in \cite{torres-turner}. The only thing to do is to convert the second order differential equation into a differential equation system of first order. The system can be uncoupled and the problem is reduced to the resolution of two first order ordinary differential equations. \end{proof} We have just deduced that the solution of the equations (\ref{lineasdiscretizadotau0var-1})-(\ref{lineasdiscretizadotau0var-4}) is \begin{gather} \begin{aligned} w_{n+1}(y) &= - \frac{\tau_0(\alpha_{n+1})}{\sqrt{\rho \eta}s_{n+1}q} \sinh \big( \sqrt{\frac{\rho}{\eta}} q(y-s_{n+1}) \big) \\ &\quad - \int_{s_{n+1}}^{y} \sqrt{\frac{\rho}{\eta}} qw_n(\xi) \sinh \big( \sqrt{\frac{\rho}{\eta}}q(y-\xi) \big) d\xi, \end{aligned}\label{wn+1tau0var} \\ \begin{aligned} w_{n+1}'(y) &= - \frac{\tau_0(\alpha_{n+1})}{\eta s_{n+1}} \cosh \big( \sqrt{\frac{\rho}{\eta}} q(y-s_{n+1}) \big) \\ &\quad - \int_{s_{n+1}}^{y} \frac{\rho}{\eta} q^2 w_n(\xi) \cosh \big( \sqrt{\frac{\rho}{\eta}} q(y-\xi) \big) d\xi. \end{aligned}\label{wn+1primatau0var} \end{gather} We need to know the value ofs_{n+1}(we do not know yet if it exists). Replacing our solution (\ref{wn+1primatau0var}) into the equation (\ref{lineasdiscretizadotau0var-2}), we deduce that \begin{align*} - \frac{f_{n+1}}{\eta} &= w_{n+1}'(L)\\ & = - \frac{\tau_0(\alpha_{n+1})}{\eta s_{n+1}} \cosh \big( \sqrt{\frac{\rho}{\eta}} q(L-s_{n+1}) \big) \\ &\quad - \int_{s_{n+1}}^{L} \frac{\rho}{\eta} q^2 w_n(\xi) \cosh \big( \sqrt{\frac{\rho}{\eta}} q(L-\xi) \big) d\xi. \end{align*} We define \label{Fn+1tau0var} \begin{aligned} F_{n+1}(s) &= \frac{f_{n+1}}{\eta} - \frac{\tau_0(\alpha_{n+1})}{\eta s} \cosh \big( \sqrt{\frac{\rho}{\eta}} q (L-s) \big)\\ &\quad - \int_{s}^{L} \frac{\rho}{\eta}q^2 w_n(\xi) \cosh \big( \sqrt{\frac{\rho}{\eta}} q(L-\xi) \big) d\xi. \end{aligned} Thens_{n+1}$must be a root of$F_{n+1}$in the interval$(0,L)$. \begin{theorem}\label{teoremaexistenciaunicidadtau0vardiscreto} If the hypotheses {\rm (Y1)-(Y5), (P1)-(P2), (A1)-(A2),(U1)-(U4)}, hold and we have the following condition, for the discretization parameter, $$\label{condicionqtau0var} \max \Big( K_1, \frac{K_2}{L} \eta \int_{s_m}^{L} u_m^{'2}dy - \frac{K_2}{L} \tau_M \int_{s_m}^{L} u_m'dy \Big) \leq q^2,$$ then there exists a unique solution of the problem$(P_{y,d}^{\alpha})$. \end{theorem} \begin{proof} We will prove the theorem by induction. For$n=1$, the quantities$\alpha_1$,$s_1$and$w_1$are defined by the initial conditions. Moreover, it is true that \begin{gather} 0 \leq \alpha_1 \leq 1, \label{condicionn=1tau0var-1} \\ w_1 \equiv 0, \quad \mbox{in }[0,s_1], \label{condicionn=1tau0var-2} \\ w_1 \leq 0, \label{condicionn=1tau0var-3} \\ w_1' \leq 0, \label{condicionn=1tau0var-4} \\ s_1 \in (0,L). \label{condicionn=1tau0var-5} \\ u_m' \leq w_1 \leq u_M' \label{condicionn=1tau0var-6} \\ s_m \leq s_1 \leq s_M \label{condicionn=1tau0var-7} \end{gather} We observe that (\ref{condicionn=1tau0var-2}) is satisfied since we can extend the function$u_0'$continuously by zero in the interval$[0,s_0]$. Then, suppose that exists a unique solution until the level$n$with the following properties:$0 \leq \alpha_n \leq 1$,$w_n \equiv 0$in$[0,s_n]$,$w_n \leq 0$,$w_n' \leq 0$and$s_n \in (0,L)$,$u_m' \leq w_n \leq u_M'$and$s_m \leq s_n \leq s_M$. Now we consider the level$n+1$. \noindent$\blacktriangleright0 \leq \alpha_{n+1} \leq 1$. Using the inductive hypothesis we have $$\label{0leqalfan+1leq1-1} \alpha_{n+1} \geq \frac{\alpha_n}{q^2} \Big[ q^2 - \frac{K_2}{L} \eta \int_{s_n}^{L} w_n^2 dy + \frac{K_2}{L} \tau_0(\alpha_n) \int_{s_n}^{L} w_n dy \Big].$$ Using the bounds and sign of$w_n$and the bounds of$s_n$, we obtain $$\label{0leqalfan+1leq1-2} \alpha_{n+1} \geq \frac{\alpha_n}{q^2} \Big[ q^2 - \frac{K_2}{L} \eta \int_{s_m}^{L} u_m^{'2}dy + \frac{K_2}{L} \tau_0(\alpha_n) \int_{s_m}^{L} u_m' dy \Big].$$ Using the bounds of$\tau_0$and the condition on$q$, $$\label{0leqalfan+1leq1-3} \alpha_{n+1} \geq \frac{\alpha_n}{q^2} \Big[ q^2 - \frac{K_2}{L} \eta \int_{s_m}^{L} u_m^{'2} dy + \frac{K_2}{L} \tau_M \int_{s_m}^{L} u_m' dy \Big] \geq 0.$$ Moreover using the signs of$w_n$, $$\label{0leqalfan+1leq1-4} \alpha_{n+1} \leq \frac{1}{q^2} \left( \alpha_n q^2 + K_1 (1-\alpha_n) \right).$$ Due to the condition on$q$, we obtain $$\label{0leqalfan+1leq1-5} \alpha_{n+1} \leq \frac{1}{q^2} \left( \alpha_n q^2 + q^2(1-\alpha_n) \right) = \alpha_n \leq 1.$$ Then$\tau_0(\alpha_{n+1})$is a well-defined number. \noindent$\blacktriangleright$There exists at least a root of$F_{n+1}$in$(0,L)$. The function$F_{n+1}$is continuous in the interval$(0,L]$. Moreover: $F_{n+1}(L) = \frac{1}{\eta} \big( f_{n+1} - \frac{\tau_M}{L} \big) > 0,$ due to the operability condition; $\lim_{s \to 0^{+}} F_{n+1}(s) = - \infty,$ because the integral is bounded. Then, by continuity, it exists a root of$F_{n+1}$in the interval$(0,L)$. \noindent$\blacktrianglerightF_{n+1}$may have more than one root in$(0,L). From (\ref{Fn+1tau0var}) we obtain \begin{aligned} F_{n+1}'(s) =& \frac{\tau_0(\alpha_{n+1})}{\eta s^2} \cosh \Big(\sqrt{\frac{\rho}{\eta}} q(L-s) \Big) + \frac{\tau_0(\alpha_{n+1})}{\eta s} \sqrt{\frac{\rho}{\eta}} q \sinh \Big( \sqrt{\frac{\rho}{\eta}} q (L-s) \Big) \\ &+ \frac{\rho}{\eta} q^2 w_n(s) \cosh \Big( \sqrt{\frac{\rho}{\eta}} q (L-s) \Big). \end{aligned}\label{Fn+1primatau0var} IfF_{n+1}$has no critical points the proof is concluded. Otherwise, suppose that there exists at least a$s_{*}$such that$F_{n+1}'(s_{*}) = 0; that is, \begin{aligned} 0 =& \frac{\tau_0(\alpha_{n+1})}{\eta s_*^2} \cosh \Big( \sqrt{\frac{\rho}{\eta}} q(L-s_*) \Big) + \frac{\tau_0(\alpha_{n+1})}{\eta s_*} \sqrt{\frac{\rho}{\eta}} q \sinh \Big( \sqrt{\frac{\rho}{\eta}} q (L-s_*) \Big) \\ &+ \frac{\rho}{\eta} q^2 w_n(s_*) \cosh \Big( \sqrt{\frac{\rho}{\eta}} q (L-s_*) \Big). \end{aligned}\label{Fn+1primas_*tau0var} Clearlys_{*} \neq 0$since$w_n \equiv 0$in$[0,s_n]$. Multiplying (\ref{Fn+1primas_*tau0var}) by the inverse of the first member of the sum, we have that$s_{*}$is a zero of the function $$\label{ecuacionparas*tau0var} B(s) = h(s) + \frac{\rho q^2 s^2 w_n(s)}{\tau_0(\alpha_{n+1})},$$ where $$\label{htau0var} h(s) = 1 + s \sqrt{\frac{\rho}{\eta}} q \tanh \Big( \sqrt{\frac{\rho}{\eta}} q (L-s) \Big).$$ It can be shown that the function$h$is concave and positive in the interval$(0,L)$. See \cite{torres-turner}. The function$B$is a sum of a positive concave function$h$(where$h(0) = h(L) = 1$), and a negative decreasing function. Now, the second term of the right hand side of the equation (\ref{ecuacionparas*tau0var}) could eventually equal the function$h$in several points in the interval where$h$is increasing. Therefore the statement is concluded. \noindent$\blacktrianglerights_{n+1}$can be chosen as the minimum root of$F_{n+1}$. We already know that there exists at least one root of$F_{n+1}$in the interval$(0,L)$, and that$\lim_{s \to 0^{+}} F_{n+1}(s) = - \infty$(that is,$s=0$can not be a root of$F_{n+1}$). Now, let's suppose that we have a set$R = \{ r_i \}$(finite or infinite) of roots of$F_{n+1}$. If$R$is finite, we define$s_{n+1}$as the minimum of the roots of$F_{n+1}$. If$R$is infinite, then we take$s_{n+1}$as the infimum of$R$. Then there exists a subsequence of roots that converges to$s_{n+1}$. As$F_{n+1}$is a continuous function,$s_{n+1}$is also a root and therefore the minimum of$R$. \smallskip Now we can solve$w_{n+1}$using (\ref{wn+1tau0var}) and (\ref{wn+1primatau0var}). Until now we have solved the level$n+1$, but we need the properties of$w_{n+1}$and$s_{n+1}$in order to continue the inductive step. \noindent$\blacktrianglerightw_{n+1} \equiv 0$in$[0,s_{n+1}]$. We extend$w_{n+1}$by zero in$[0,s_{n+1}]$and due to (\ref{lineasdiscretizadotau0var-3}) this extension is continuous, but the derivative does not exits at$s_{n+1}$because of (\ref{lineasdiscretizadotau0var-4}). \noindent$\blacktrianglerightw_{n+1} \leq 0$in$[0,L]$. We know that$w_{n+1} = 0$in$[0,s_{n+1}]$. If we denote$w_{n+1}$by$A$,$s_{n+1}$by$s$, and$\rho q^2 /\eta$by$\gamma^2$, it is clear that$w_{n+1}$satisfies the system $$\label{ecuacionlemaparawnegativa} \begin{gathered} A'' - \gamma^2A \geq 0, \quad s < y < L, \\ A(s) \leq 0, \\ A'(L) \leq 0. \end{gathered}$$ Here we have used (\ref{lineasdiscretizadotau0var-1}), (\ref{lineasdiscretizadotau0var-3}) and (\ref{lineasdiscretizadotau0var-2}), and the inductive hypothesis. In \cite{torres-turner} it is proved that$A \leq 0$in the interval$[0,L]$, that is equivalent to$w_{n+1} \leq 0$in$[s_{n+1},L]$. That means the statement is finished. \noindent$\blacktrianglerightw_{n+1}' \leq 0$in$[0,L]$. In$[0,s_{n+1})$is clear that$w_{n+1}' = 0$. In$[s_{n+1},L]$, it holds $$\label{sistemawn+1'''parawn+1primanegativatau0var} \begin{gathered} w_{n+1}''' - \frac{\rho}{\eta}q^2 w_{n+1}' = - \frac{\rho}{\eta}q^2 w_n', \quad s_{n+1} < y < L, \\ w_{n+1}'(s_{n+1}) = - \tau_0(\alpha_{n+1})/\eta s_{n+1}, \\ w_{n+1}'(L) = - f_{n+1}/\eta. \end{gathered}$$ We denote$w_{n+1}'$by$A$,$s_{n+1}$by$s$and$\rho q^2/\eta$by$\gamma^2$, then$w_{n+1}'$satisfies the system $$\label{ecuacionlemaparawprimanegativa} \begin{gathered} A'' - \gamma^2 A \geq 0, \quad s < y < L, \\ A(s) < 0, \\ A(L) < 0. \end{gathered}$$ In \cite{torres-turner} it is proved that$A \leq 0$in the interval$[s,L]$, that means that$w_{n+1}' \leq 0$in$[s_{n+1},L]$. The statement is proved. \noindent$\blacktrianglerights_m \leq s_{n+1} \leq s_M$. Let$s$be such that$0 < s < s_m$. Note that the function $$\label{funcioncreciente} \frac{1}{s}\cosh\Big(\sqrt{\frac{\rho}{\eta}}q(L-s)\Big)$$ is decreasing in the interval$(0,L]$. Taking into account that$w_n = 0$in$[s,s_m]$(because$s_m \leq s_n$using inductive hypothesis), we can deduce that $$\label{cotafn+1<=} F_{n+1}(s) \leq F_{n+1}(s_m)$$ Now, using that$\tau_0 \geq \tau_m$and$w_n \geq u_m'we can obtain \begin{aligned} F_{n+1}(s_m) \leq& \frac{f_{n+1}}{\eta} - \frac{\tau_m}{\eta s_m} \cosh \Big( \sqrt{\frac{\rho}{\eta}} q (L-s_m) \Big) \\ &+ \int_{s_m}^{L} \frac{\rho}{\eta} q^2 \frac{f_M}{\eta} (\xi - s_m) \cosh \Big( \sqrt{\frac{\rho}{\eta}} q(L-\xi) \Big) d\xi. \end{aligned}\label{sm1} Integrating by parts in the second member of the inequality in the integral (\ref{Fn+1>1}), and using that\tau_0 \leq \tau_M$we obtain \label{sm 0. Then$F_{n+1}(s) > 0$for all$s$in$[s_M,L]$; therefore,$s_{n+1} \leq s_M$, so we conclude the statement. \noindent$\blacktrianglerightw_{n+1} \leq u_M'$. Let$B_{n+1} = w_{n+1} - u_M'$be. If$y \in [0,s_{n+1}]$then$B_{n+1}(y) = 0$. If$y \in [s_{n+1},s_M]$then$B_{n+1}(y) = w_{n+1}(y) \leq 0$. If$y \in [s_M,L]\$ it satisfies \label{umprima