\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 18, pp. 1--20.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/18\hfil The spreading of charged micro-droplets] {The spreading of charged micro-droplets} \author[Joseph Iaia \hfil EJDE-2015/18\hfilneg] {Joseph Iaia} \address{Joseph Iaia \newline Department of Mathematics, University of North Texas, 1155 Union Circle \#311430, Denton, TX 76203-5017, USA} \email{iaia@unt.edu} \thanks{Submitted August 7, 2014. Published January 20, 2015.} \subjclass[2000]{35B07, 35B09} \keywords{Spreading fluids; circular symmetry; charged droplet} \begin{abstract} This article considers the analysis of the Betelu-Fontelos model of the spreading of a charged microdroplet on a flat dielectric surface whose spreading is driven by surface tension and electrostatic repulsion. This model assumes the droplets are circular and spread according to a power law. This leads to a third-order nonlinear ordinary differential equation on $[0,1]$ that gives the evolution of the height profile. We examine existence of solutions for this equation. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} In this article we perform the analysis of the Betelu-Fontelos (BF) model \cite{BF} of the spreading of a charged microdroplet. With this model a charged microdroplet spreads over a flat surface. In the absence of charge, it has been shown experimentally \cite{D} that the radius of a circular drop, $a(t)$, spreads according to the law $a(t) = A t^{1/10}$ for some constant $A$. The mathematical analysis of the uncharged case is difficult, the main result \cite{B} being that in the absence of molecular forces, the ``paradox of the contact line'' arises and the drop does \emph{not} spread. In \cite{BF}, electric charges were included into the model, and it was shown that the drop \emph{does} spread resolving the paradox of the contact line. Also surprisingly, the presence of the charges does \emph{not} alter the similarity exponent for the spreading of the drop and it, too, spreads according to a $ t^{1/10}$ law. This is also surprising since if we add gravity instead then the exponent \emph{does} change \cite{D2}. This model has practical implications in physical processes on which electrically charged droplets spread on surfaces such as electro-painting. This model uses the \emph{lubrication} approximation which assumes that the fluid spreads over a solid surface and that the droplet is thin so that the horizontal component of the velocity is much larger than the vertical component and that the stresses are mostly due to gradients of the velocity in the direction perpendicular to the surface. Using this approximation it is shown in \cite{BF} that the height profile $h(r,t)$ of a circular drop satisfies \begin{equation} h_t + \frac{1}{r} \frac{\partial}{\partial r} \Big[ \frac{r}{3 \mu} h^3 \frac{\partial}{\partial r} \Big( \frac{Q^2}{2 \epsilon_0 (4\pi a(t))^2 } \frac{1}{a^2(t) - r^2} + \gamma (h_{rr} + \frac{h_{r}}{r}) \Big) \Big] = 0 \label{PDE} \end{equation} and \begin{equation} \int_0^{a(t)} 2\pi r h(r,t) \, dr = V_0 = \text{constant} \label{volume} \end{equation} where $a(t)$ is the radius of the drop and the boundary conditions are \begin{equation} h_{r}(0,t) = h_{rrr}(0,t) = 0 \label{PDE1} \end{equation} (due to the circular symmetry), and \begin{equation} h(a(t),t) = 0. \label{PDE2} \end{equation} Here $\gamma$ is the free surface tension coefficient, $\epsilon_0$ is the permittivity of the gas above the drop, $\mu$ is the viscosity, and $Q$ is the total charge. Equation \eqref{volume} states that the volume of the drop remains constant throughout this process. We seek a self-similar solution such that the radius of the drop $a(t)$ satisfies a power law, i.e. $a(t) = At^{\beta}$. The height profile will then, by conservation of mass, be of the form $$ h(r,t) = \frac{B}{t^{2\beta}} H\big( \frac{r}{a(t)}\big) = \frac{B}{t^{2\beta}} H\big( \frac{r}{At^{\beta}}\big) $$ where $\rho = \frac{r}{a(t)}$ and $0 \leq \rho \leq 1$. In terms of $V_0$ this gives: $$ V_0 = \int_0^{a(t)} 2\pi r h(r,t) \, dr = A^{2}B\int_0^{1} 2 \pi \rho H(\rho) \, d\rho $$ where \begin{equation} Y\equiv\int_0^{1} 2 \pi \rho H(\rho) \, d\rho \label{Y equation} \end{equation} denotes the dimensionless ``shape factor'' of the drop. Remarkably, with $\beta = \frac{1}{10}$ equation \eqref{PDE} becomes $$ \Big[ \rho H^3 \Big( H_{\rho\rho} + \frac{H_{\rho}}{\rho} + \frac{XY}{1-\rho^2} \Big)_{\rho} \Big]_{\rho} = Z(\rho^2 H_{\rho} + 2 \rho H) $$ where $$ X = \frac{Q^2}{32\pi^2\epsilon_0 \gamma V_0},\quad Z = \frac{3\mu A^4}{10 \gamma B^3 }. $$ Integrating once, using \eqref{PDE1}, and rewriting yields \begin{gather} H''' + \frac{H''}{\rho} - \frac{H'}{\rho^2} = \big( H'' + \frac{H'}{\rho} \big)' = \frac{Z\rho}{H^2} - \frac{2XY\rho}{(1-\rho^2)^2} \quad \text{for } 0 < \rho < 1, \label{DE} \\ H'(0) = 0, \label{A} \\ H(1) = 0. \label{B} \end{gather} Note that $X,Y,Z$ are all positive constants. Throughout this paper we will also assume: \begin{equation} H(0)=1. \label{C} \end{equation} Note that \begin{equation} H(\rho) = 1 - \rho^2 \label{solution} \end{equation} is a solution of \eqref{DE}-\eqref{C} when $Z=2XY$. A natural question is whether there are other solutions of \eqref{DE}-\eqref{C}. In an earlier paper \cite{IA} we showed that \eqref{solution} is the only solution of \eqref{DE}-\eqref{C} which is differentiable on all of $[0,1]$. If we weaken the assumption and only look for solutions $H\in C^{3}[0,1) \cap C[0,1]$, then in an earlier paper \cite{I} we showed that if $0< Z< 2XY$ then there is a solution of \eqref{DE}-\eqref{C} with $H\in C^{3}[0,1) \cap C[0,1]$ such that $\lim_{\rho \to 1^{-}}H'(\rho) = -\infty$. More specifically we showed that: if $H\in C^{3}[0,1) \cap C[0,1]$ is a solution of \eqref{DE}-\eqref{C} which is not differentiable at $x=1$ then $$ \lim_{\rho \to 1^{-}} \frac{H'(\rho)}{\ln(1-\rho^2)} = \frac{XY}{4} = \lim_{\rho \to 1^{-}} -\frac{1}{4}(1-\rho^2)H''(\rho) = \lim_{\rho \to 1^{-}} -\frac{1}{8}(1-\rho^2)^2 H'''(\rho). $$ In this paper we examine the case when $ Z > 2XY$. In attempting to solve \eqref{DE}-\eqref{C}, we first thought of using the \emph{shooting} method. That is, we would solve \eqref{DE} with \begin{gather} H(0)=1, \label{initial position} \\ H'(0) = 0, \label{initial velocity} \\ H''(0)=k \label{initial acceleration} \end{gather} where $k$ is arbitrary and show that if $k$ is sufficiently large then $H>0$ on $[0,1)$ and if $k$ is sufficiently small then $H$ must have a zero on $[0,1)$. Then making an appropriate choice for $k$ we could show that $H(1)=0$. Therefore we conjectured that there would be at least one value of $k$ such that $H$ is a solution. However, we discovered that this method does not quite work for this problem. In fact, in \cite{I} we proved the following result. \begin{theorem} \label{thm1} Let $H\in C^{3}(\rho_0,1) \cap C[\rho_0,1) $ be a solution of \eqref{DE} such that $0\leq\rho_0<1$ and $H(\rho_0)>0$. Then $H> 0 $ on $(\rho_0,1)$. \end{theorem} We were able to eventually show that if we look at a slightly different differential equation then it is possible to solve this new problem by the shooting method. The key turned out to be to look at the function: \begin{equation} W = H- \sqrt{\frac{Z}{2XY}}(1 - \rho^2). \label{W equation} \end{equation} Using \eqref{DE} it is straightforward to see that \begin{equation} \Big( W'' + \frac{W'}{\rho} \Big)' = \Big( H'' + \frac{H'}{\rho} \Big)' = \frac{-2XY \rho}{ (1 - \rho^2)^2} \frac{W \big( H+ \sqrt{\frac{Z}{2XY}}(1 - \rho^2)\big)}{ H^2 } \label{DE3} \end{equation} for $0 < \rho < 1$. The initial conditions for $W$ are related to \eqref{initial position}-\eqref{initial acceleration} by \eqref{W equation}: \begin{gather} W(0)=1-\sqrt{\frac{Z}{2XY}}, \label{initial position W} \\ W'(0) = 0, \label{initial velocity W} \\ W''(0)=k +\sqrt{\frac{2Z}{XY}}. \label{initial acceleration W} \end{gather} In \cite{I} we proved the following theorem. \begin{theorem} \label{thm2} For each $ 0< Z < 2XY $ there is a positive $C^{3}(0,1)\cap C^{1}[0,1) \cap C[0,1]$ solution of \eqref{DE3} with $W'(0)=0$, $W(1) = 0$, and $W'(1) = -\infty$. (And thus $H$ solves \eqref{DE} with $H'(0)=0$, $H(1) = 0$, and $H'(1) = -\infty$). If $Z=2XY$ then $W\equiv 0$ is a solution of \eqref{DE3}. (And thus $H= \sqrt{\frac{Z}{2XY}}(1 - \rho^2)$ solves \eqref{DE} with $H'(0)=0$, $H(1) = 0$, and $H'(1) =-\sqrt{\frac{2Z}{XY}} $). Thus we see that there is a solution of \eqref{DE}-\eqref{C} for $0 2XY$. Then there exist real numbers $k^{-}$ and $k^{+}$ with $k^{-} \leq k^{+} $such that there are no $C^{3}(0,1) \cap C^{1}[0,1) \cap C[0,1]$ solutions of \eqref{DE3} with \eqref{initial position W}-\eqref{initial acceleration W} and $W(1)=0$ if $k> k^{+}$ or if $k< k^{-}$. (Thus there are no $C^{3}(0,1) \cap C^{1}[0,1) \cap C[0,1]$ solutions of \eqref{DE} with $H(0)=1$, $H'(0)=0$, $H''(0) =k$, and $H(1) = 0$ if $k> k^{+}$ or if $k< k^{-}$). \end{theorem} In the proof of the Main Theorem (Theorem \ref{main-thm}) we show that $H(1,k) \to \infty $ as $k\to \infty$ and $k \to -\infty$. We also know from Theorem \ref{thm1} that $H(1,k) \geq 0$ for all $k \in \mathbb{R}$. We then define $k_0$ to be a value of $k$ so that $H(1, k_0) \leq H(1,k)$ for all $k \in \mathbb{R}$. We attempted to prove that either $H(1, k_0) =0$ or $H(1,k_0)>0$ but we were not able to prove either of these. However, numerics in \cite{BF} strongly suggest that $H(1, k_0) >0$ and so we conjecture that there are no solutions of \eqref{DE} with $H(0)=1$, $H'(0)=0$, and $H(1)=0$ for $Z> 2XY$. \section{Preliminaries} Rewriting \eqref{DE3} we see that \begin{equation} W''' + \frac{W''}{\rho} - \frac{W'}{\rho^2} + \frac{2XY \rho}{(1 - \rho^2)^2 } \frac{W \big( H+ \sqrt{\frac{Z}{2XY}}(1 - \rho^2)\big)}{ H^2} = 0. \label{DE4} \end{equation} Now we note the following: \begin{lemma} \label{lem2.1} $W'$ does not have a positive local maximum on the set where $W\leq 0$ and $0<\rho <1$. \end{lemma} \begin{proof} If there were such a point, $p$, then at this point we would have $W(p)\leq 0$, $W'(p)>0$, and since $W'$ has a local maximum at $p$, then from calculus it follows that $W''(p)=0$ and $W'''(p)\leq 0$. This however contradicts \eqref{DE4} and Theorem \ref{thm1}. \end{proof} \begin{lemma} \label{lem2.2} $W'$ does not have a negative local minimum on the set where $W\geq 0$ and $0<\rho < 1$. \end{lemma} \begin{proof} If there were such a point, $p$, then at this point we would have $W(p)\geq 0$, $W'(p)<0$, and since $W'$ has a local minimum at $p$, then from calculus it follows that $W''(p)=0$ and $W'''(p)\geq 0$. This again contradicts \eqref{DE4} and Theorem \ref{thm1}. \end{proof} \begin{lemma} \label{lem2.3} Suppose $W \in C^{3}(0,1) \cap C^{1}[0,1) \cap C[0,1]$ satisfies \eqref{DE4} with $W(0)<0$ and $W'(0)=0$. Then $W$ must get positive somewhere on $(0,1)$. \end{lemma} \begin{proof} Suppose by way of contradiction that there is a solution, $W$, with $W\leq 0$ on $(0,1)$. Integrating \eqref{DE4} on $(\rho_0, \rho)$ where $0< \rho_0 < \rho < 1$ gives for some constant $C_0$: $$ W'' + \frac{W'}{\rho} = C_0 - \int_{\rho_0}^{\rho} \frac{2XY t W \big( H+ \sqrt{\frac{Z}{2XY}}(1 - t^2)\big)}{ (1 - t^2)^2 H^2} \, dt. $$ Multiplying by $ \rho$ and integrating on $(\rho_0, \rho)$ gives \begin{equation} \rho W' = \rho_0 W'(\rho_0) + \frac{C_0}{2}(\rho^2 - \rho_0^2) + \int_{\rho_0}^{\rho} t \int_{\rho_0}^{t} \frac{(-2XY) s W \big( H+ \sqrt{\frac{Z}{2XY}}(1 - s^2)\big)}{ (1 - s^2)^2 H^2} \, ds \, dt. \label{ID1} \end{equation} Assuming $W\leq 0$ on $(0,1)$ and Theorem \ref{thm1} then the integrand on the right-hand side of \eqref{ID1} is nonnegative, and so the integral term is an increasing function. Thus it follows that \begin{equation} \lim_{\rho \to 1^{-}} W'(\rho) \text{ exists (and is possibly} +\infty). \label{ID2} \end{equation} Since we are also assuming $W\leq 0$ on $(0,1)$, it follows by continuity that $W(1)\leq 0$. On the other hand, we also know from \eqref{W equation} and Theorem \ref{thm1} that $W(1)=H(1)\geq 0$. Thus $W(1)=0$. Now we know from section 3 in \cite{I} that if the limit in \eqref{ID2} is finite then in fact the limit must be zero. Thus, it follows from \eqref{ID2} that either: $ \lim_{\rho \to 1^{-}} W'(\rho) = 0$ or $ \lim_{\rho \to 1^{-}} W'(\rho) = +\infty$. Suppose first that $ \lim_{\rho \to 1^{-}} W'(\rho) = 0$. We also know that since $W\leq 0$ and $W(0)<0$ then $W$ has a local and absolute minimum, $m$, with $0 \leq m < 1$ such that $W(m) < 0$ and $W'(m) = 0$. By the mean value theorem, $0< W(1)-W(m) = W'(c)(1-m)$ for some $c\in(m,1)$ and so we see that $W'$ must get positive somewhere on $(m,1)$. Since $W'(m) = 0 = \lim_{\rho \to 1^{-}} W'(\rho)= W'(1)$ we see that $W'$ has a positive local maximum on $(m,1)$ with $W\leq 0$ contradicting Lemma \ref{lem2.1}. Therefore, the assumption that $\lim_{\rho \to 1^{-}} W'(\rho) = 0$ must be false. Thus it must be the case that $ \lim_{\rho \to 1^{-}} W'(\rho) = \infty$. Since $W(1)=0$ then it follows from L'Hopital's rule that \begin{equation} \lim_{\rho \to 1^{-}} \frac{1-\rho}{W(\rho)} = 0. \label{ID3} \end{equation} Now rewriting \eqref{DE3} we see that \begin{equation} -\Big( W'' + \frac{W'}{\rho} \Big)' = \frac{2XY \rho}{(1 - \rho^2)^2} \bigg( \frac{1+ 2\sqrt{\frac{Z}{2XY}}\frac{1 - \rho^2}{W}} {(1+ \sqrt{\frac{Z}{2XY}}\frac{1 - \rho^2}{W})^2} \bigg). \label{ID4} \end{equation} By \eqref{ID3} the term in parentheses on the right-hand side of \eqref{ID4} goes to $1$ as $\rho \to 1^{-}$ and hence is larger than $1/2$ for $\rho$ close to $1$. Thus, for $\rho$ close to $1$ with $\rho < 1$ we have: $$ -\Big( W'' + \frac{W'}{\rho} \Big)' \geq \frac{XY \rho}{(1 - \rho^2)^2}. $$ Integrating this on $(\rho_0, \rho)$ gives for some constant $C_{1}$, $$ -\big( W'' + \frac{W'}{\rho} \big) \geq C_{1} + \frac{XY}{1 - \rho^2}. $$ Multiplying by $\rho$ and integrating on $(\rho_0, \rho)$ gives for some constant $C_{2}$, \begin{equation} \rho W' \leq C_{2} - \frac{C_{1}}{2}\rho^2 + \frac{XY}{2}\ln(1 - \rho^2). \label{yaz} \end{equation} This implies $W' \to -\infty$ as $\rho \to 1^{-}$ contradicting that $W' \to \infty$. This completes the proof. \end{proof} We next show that if $k$ is chosen sufficiently large and $Z> 2XY$ then $W$ has a first positive zero, $z$, and $W>0$ on $(z,1]$. This then proves the existence of the number $k^{+}$ referred to in Theorem \ref{main-thm}. We begin by integrating \eqref{DE} on $(0, \rho)$ and using \eqref{initial velocity}-\eqref{initial acceleration} gives $$ H'' + \frac{H'}{\rho} + \frac{XY}{1-\rho^2} = 2k + XY + \int_0^{\rho} \frac{Zt}{H^2} \, dt. $$ Multiplying by $\rho$, integrating on $(0,\rho)$, and simplifying gives $$ H' = \frac{XY}{2}\frac{\ln(1-\rho^2)}{\rho} + \big( k+ \frac{XY}{2} \big) \rho + \frac{1}{\rho} \int_0^{\rho} s \int_0^{s} \frac{Zt}{H^2} \, dt \, ds. $$ Thus by \eqref{W equation}, we obtain \begin{equation} W' = \frac{XY}{2}\frac{\ln(1-\rho^2)}{\rho} + \Big( k+ \frac{XY}{2} +\sqrt{\frac{2Z}{XY}} \Big) \rho + \frac{1}{\rho} \int_0^{\rho} s \int_0^{s} \frac{Zt}{H^2} \, dt \, ds. \label{billy} \end{equation} Thus for $k$ sufficiently large we see by \eqref{initial acceleration W} and \eqref{billy} that $W$ is increasing on say $[0, 1-\epsilon]$ and since $W$ is bounded below by $-\sqrt{\frac{Z}{2XY}}$ (from Theorem \ref{thm1}) we see that there is a $z$ with $00$ Integrating \eqref{billy} again on $(0, \rho)$ and using \eqref{initial position} gives \begin{equation} H = 1 + \frac{XY}{2} \int_0^{\rho} \frac{\ln(1-t^2)}{t} \, dt + \big( \frac{k}{2}+ \frac{XY}{4} \big) \rho^2 + \int_0^{\rho}\frac{1}{x} \int_0^{x} t \int_0^{t} \frac{Zs}{H^2} \, ds \, dt \, dx. \label{seeger} \end{equation} Thus by \eqref{W equation} we see that \begin{equation} W \geq (1 - \sqrt{\frac{Z}{2XY}}) + \frac{XY}{2} \int_0^{\rho} \frac{\ln(1-t^2)}{t} \, dt + \Big( \frac{k}{2}+ \frac{XY}{4} +\sqrt{\frac{Z}{2XY}} \Big) \rho^2. \label{bragg} \end{equation} We note by L'Hopital's rule that $$ \lim_{\rho\to 0^{+}}\frac{ \frac{XY}{2} \int_0^{\rho} \frac{\ln(1-t^2)}{t} \, dt + (\frac{k}{2}+ \frac{XY}{4} +\sqrt{\frac{Z}{2XY}}){\rho^2}} {\rho^2} = \frac{k}{2} + \sqrt{\frac{Z}{2XY}}. $$ Also, $\frac{\ln(1-t^{2})}{t}$ is integrable at $t=1$ so we see that it follows from \eqref{bragg} that $W$ remains positive on all of $[z,1]$ provided $k$ is chosen large enough. Thus we see that if $k$ is sufficiently large then $W$ has exactly one zero on $[0,1]$. Therefore there exists $k^{+}>0$ such that if $k> k^{+}$ then there are no $C^{3}(0,1) \cap C^{1}[0,1) \cap C[0,1]$ solutions of \eqref{DE3} with \eqref{initial position W}-\eqref{initial acceleration W} and $W(1)=0$. The rest of the paper is devoted to proving the existence of $k^{-}$. \section{Proofs} We now write $W_k$ instead of $W$ to emphasize the dependence of $W$ on $k$. Throughout this section we assume: \begin{gather} W_k \in C^{3}(0,1) \cap C^{1}[0,1) \cap C[0,1] \text{ solves } \eqref{DE3}, \label{assume} \\ -\sqrt{\frac{Z}{2XY}} < W_k(0)<0, \quad\text{and}\quad W_k'(0)=0. \label{assume2} \end{gather} \begin{lemma} \label{lem3.1} Suppose $W_k$ satisfies \eqref{assume}-\eqref{assume2} and $k < -\sqrt{\frac{2Z}{XY}}$. Then there exist points $p_{1,k}$, $m_k$, $z_k$, $p_{2,k}$, $M_k$ with $0< p_{1,k}< m_k< z_k < p_{2,k} < M_k<1$ such that $W_k$ has a local minimum at $m_k$, a zero at $z_k$, a local maximum at $M_k$, and inflection points at $p_{1,k}$ and $p_{2,k}$. Furthermore, $W_k$ has no other local extrema on $(0,M_k)$ and $W_k$ has no inflection points on $(m_k, z_k)$. \end{lemma} \begin{proof} By assumption $W_k(0)<0$ and $W_k$ is continuous so $W_k$ has a zero on $(0,1)$ by Lemma \ref{lem2.3}. Thus there exists a $z_k$ with $0 0$ for if $W_k''(m_k) = 0$ we would have $W_k'(m_k)=W_k''(m_k)=0$ and from \eqref{DE3} it follows that $W_k'''(m_k)>0$. Thus $W_k''$ would be increasing in a neighborhood of $m_k$ and since $W_k''(m_k)=0$, then $W_k''<0$ on $(m_k-\delta, m_k)$ for some $\delta >0$ which implies $W_k'$ is decreasing on $(m_k-\delta, m_k)$, and since $W_k'(m_k)=0$, it follows that $W_k'>0$ on $(m_k-\delta, m_k)$ contradicting that $W_k'<0$ on $(0, m_k)$. Thus $W_k''(m_k)>0$ and therefore $m_k$ is a local minimum of $W_k$. Also since $W_k''(0)<0$ and $W_k''(m_k)>0$ it follows that there must be an inflection point, $p_{1,k}$, with $0< p_{1,k} < m_k$ such that $W_k''< 0$ on $(0, p_{1,k})$ and $W_k''\geq 0$ on $(p_{1,k}, p_{1,k}+\delta_{2})$ for some $\delta_{2} > 0$. Next, we observe that $W_k''>0$ on $(m_k, z_k)$ for if there were an $r_k$ with $m_k < r_k < z_k$ with $W_k''> 0$ on $(m_k,r_k)$ and $W_k''(r_k) = 0$ then from \eqref{DE4} we see that since $W_k(r_k)<0$ and $W_k'(r_k)>0$ then $W_k'''(r_k) >0$. Thus, $W_k''$ is increasing in a neighborhood of $r_k$ and since $W_k''(r_k)=0$ then this would imply $W_k''<0 $ on $(r_k - \delta_{3}, r_k)$ for some $\delta_{3} > 0$ which contradicts that $W_k''> 0$ on $(m_k,r_k)$. Thus $W_k''>0$ on $(m_k, z_k)$ and since $W_k'(m_k)=0$ it follows that $W_k'(z_k)>0$. Thus there is a $\delta_4>0$ such that $W_k>0$ on $(z_k, z_k + \delta_4)$. Now either $W_k$ has a second zero, $z_{2,k}$, on $(z_k,1)$ or $W_k>0$ on $(z_k,1)$. \textbf{Case 1:} If $W_k$ has another zero, $z_{2,k}$, on $(z_k, 1)$ then there exists $M_k$ with $W_k'>0$ on $[z_k, M_k)$, $W_k'(M_k)=0$, and $W_k(M_k) > 0$. This implies $W_k''(M_k)\leq 0$. Now since $W_k'(m_k) =0$, $W_k''(m_k)>0$ (shown earlier in this proof), and $W_k'(M_k)=0$ it follows that $W_k'$ has a positive local maximum, $p_{2,k}$, on $(m_k, M_k)$ with $W_k''>0$ on $(m_k, p_{2,k})$ and $m_k < p_{2,k} < M_k$. From Lemma \ref{lem2.1} it follows that $W_k(p_{2,k})>0$ and so $z_k 0$ on $(z_k, 1]$ then there is a constant $c_0>0$ such that $$ \Big(W_k'' + \frac{W_k'}{\rho}\Big)' + \frac{c_0\rho}{(1-\rho^2)^2} \leq 0 $$ near $\rho =1$. Integrating this on $(\rho_0, \rho)$ gives \begin{equation} W_k'' + \frac{W_k'}{\rho} + \frac{c_0}{2(1-\rho^2)} \leq W_k''(\rho_0) + \frac{W_k'(\rho_0)}{\rho_0} + \frac{c_0}{2(1-\rho_0^2)} \equiv b_k \label{IKE} \end{equation} and thus we see that $ W_k'' + \frac{W_k'}{\rho}$ must get negative as $\rho \to 1^{-}$ since the right-hand side of \ref{IKE} is fixed. Multiplying \eqref{IKE} by $\rho$ and integrating on $(\rho_0, \rho)$ gives \begin{equation} \rho W_k' \leq \rho_0W_k'(\rho_0) + b_k(\rho - \rho_0) + \frac{c_0}{4}\ln\Big( \frac{1-\rho^2}{1-\rho_0^2} \Big). \label{FDR} \end{equation} Thus we see that $W_k'$ becomes negative as $\rho \to 1^{-}$. Therefore, we see that there is an $M_k$ with $m_k < z_k < M_k<1$ such that $W_k'>0$ on $(m_k,M_k)$, $W_k(M_k)>0$, and $W_k'(M_k)=0$. As in the proof of Case 1, it is possible to show that $M_k$ is a local maximum, $W_k''(M_k)<0$, and there is an inflection point $p_{2,k}$ with $m_k < z_k< p_{2,k} < M_k$ with $W_k(p_{2,k})>0$. This completes the proof of the lemma for Case 2. \end{proof} \begin{lemma} \label{lem3.2} Suppose $W_k$ satisfies \eqref{assume}-\eqref{assume2}. Then $H_k(m_k) \to 0$ as $k \to -\infty$ (for some subsequence of $k$'s). \end{lemma} \begin{proof} Suppose on the contrary that there exists an $l_0>0$ such that $H_k(m_k) \geq l_0>0$ for all $k$ sufficiently negative. Then on $[0,m_k]$ we have $0\geq W_k' = H_k' + \sqrt{\frac{2Z}{XY}} \rho \geq H_k'$. Therefore, $H_k$ is decreasing on $[0,m_k]$ and hence $H_k \geq H_k(m_k) \geq l_0$ on $[0,m_k]$. Then $\frac{1}{H_k^2} \leq \frac{1}{l_0^2}$ on $[0, m_k]\subset[0,1]$ so we see by integrating \eqref{DE3} on $(0, \rho)$, using \eqref{initial velocity W}-\eqref{initial acceleration W}, and assuming $k$ is sufficiently negative \begin{equation} \begin{aligned} W_k'' + \frac{W_k'}{\rho } + \frac{XY}{1-\rho^{2}} &= 2\Big(k + \sqrt{\frac{2Z}{XY}} \Big) + XY + \int_0^{\rho}\frac{Zt}{H_k^2} \, dt \\ &\leq 2\Big(k + \sqrt{\frac{2Z}{XY}}\Big) + XY + \frac{Z}{2l_0^2} \rho^2 < 0. \end{aligned} \label{galveston} \end{equation} Evaluating \eqref{galveston} at $m_k$ gives $$ 0 < W_k''(m_k) +\frac{XY}{1-m_k^{2}} \leq 2\Big(k + \sqrt{\frac{2Z}{XY}}\Big) + XY + \frac{Z}{2l_0^2}m_k^2 < 0 $$ which is a contradiction to \eqref{galveston}. Thus, the lemma holds. \end{proof} \begin{lemma} \label{lem3.3} Suppose $W_k$ satisfies \eqref{assume}-\eqref{assume2}. Then $p_{1,k} \to 0$ as $k \to -\infty$ (for some subsequence of $k$'s). \end{lemma} \begin{proof} Since $W_k'' = H_k'' + \sqrt{\frac{2Z}{XY}}$ we see that $H_k''\leq 0$ when $W_k''\leq 0$. Also, by Lemma \ref{lem3.1} we note that $H_k$ is concave down on $[0, p_{1,k}]$ and so on this interval the graph of $H_k$ lies above the line through $(0,1)$ and $(p_{1,k}, H_k(p_{1,k}))$. That is $$ H_k(\rho) \geq 1 - \frac{1-H_k(p_{1,k})}{p_{1,k}} \rho \text{ on } [0, p_{1,k}). $$ Thus $$ H_k(\rho) \geq 1 - A_k \rho >0 \quad \text{on } [0, p_{1,k}), $$ where \begin{equation} A_k = \frac{1-H_k(p_{1,k})}{p_{1,k}}. \label{abe} \end{equation} Thus \begin{equation} \frac{1}{H_k^2} \leq \frac{1}{(1-A_k\rho)^2} \quad \text{on } [0, p_{1,k}) . \label{ruth} \end{equation} And so integrating \eqref{DE} on $(0, \rho) \subset (0,1)$ and using \eqref{ruth} we see \begin{align*} H_k'' + \frac{H_k'}{\rho} + \frac{XY}{1-\rho^2} &= 2k + XY + \int_0^{\rho} \frac{Zt}{H_k^2} \, dt\\ &\leq 2k + XY + \int_0^{\rho} \frac{Zt}{ (1-A_kt)^2} \, dt \\ &\leq 2k + XY + \frac{\rho Z}{(1-A_k\rho)}. \end{align*} Since $\frac{XY}{1-\rho^2} \geq XY$ on $[0,1]$ this reduces to $$ H_k'' + \frac{H_k'}{\rho} \leq 2k + \frac{\rho Z}{d(d-A_k\rho)}\quad \text{on } [0, p_{1,k}). $$ Multiplying by $\rho$, integrating on $(0, \rho)$, and simplifying gives $$ H_k' \leq k \rho - \frac{Z}{A_k \rho}\ln(1 - A_k \rho). $$ Integrating on $(0, p_{1,k})$ we obtain \begin{equation} 0\leq H_k(p_{1,k})\leq 1 +\frac{k}{2}p_{1,k}^2 -\frac{Z}{A_k}\int_0^{p_{1,k}} \frac{\ln(1 - A_kt)}{t} \, dt. \label{WW} \end{equation} Making the change of variables $u= A_kt$ and using \eqref{abe}-\eqref{WW} we obtain \begin{equation} \frac{|k|}{2}p_{1,k}^2 \leq 1 + \frac{Zp_{1,k}}{d^2} \frac{1}{\big(1-H_k(p_{1,k})\big)} \int_0^{(1-H_k(p_{1,k}))} \Big(-\frac{\ln(1-u)}{u}\Big) \, du. \label{carter} \end{equation} Now let $$ g(x) = \frac{1}{x}\int_0^{x}\Big(-\frac{\ln(1-u)}{u}\Big) \, du. $$ It follows by using the power series for $\ln(1-u)$ that \[ g(x)= \frac{1}{x}\int_0^{x}\Big(-\frac{\ln(1-u)}{u}\Big) \, du = 1 + \frac{x}{2^2} + \frac{x^2}{3^2} + \frac{x^3}{4^2} + \cdots \] which converges for $0\leq x \leq 1$. In addition, for these $x$ we have $1 \leq g(x)\leq g(1) = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6} < 2$. Now notice that since $p_{1,k} \leq 1$ then \eqref{carter} can be rewritten as $$ \frac{|k|}{2}p_{1,k}^2 \leq 1 + Zp_{1,k}g\left(1- H_k(p_{1,k})\right) \leq 1 + Z g\left(1- H_k(p_{1,k}) \right). $$ Thus $$ \frac{|k|}{2}p_{1,k}^2 \leq 1 + 2Z $$ and so we see that $p_{1,k} \to 0$ as $k \to -\infty$. \end{proof} \begin{lemma} \label{lem3.4} Suppose $W_k$ satisfies \eqref{assume}-\eqref{assume2}. Then $H_k(p_{1,k}) \to 0$ as $k \to -\infty$ (for some subsequence of $k$'s). \end{lemma} \begin{proof} Suppose not and so assume there is an $s_0>0$ such that $H_k(p_{1,k}) \geq s_0$. Now recall that $W_k'$ has a local minimum at $p_{1,k}$. Therefore from calculus we have $W_k''(p_{1,k})=0$ and $W_k'''(p_{1,k})\geq 0$. Using \eqref{DE} and \eqref{DE3} we see that $$ 0\leq - \frac{W_k'(p_{1,k})}{p_{1,k}^2} \leq \frac{Z p_{1,k}}{H_k^{2}(p_{1,k})}. $$ Thus, by Lemma \ref{lem3.3}, \begin{equation} 0 \leq - \frac{W_k'(p_{1,k})}{p_{1,k}} \leq \frac{Z p_{1,k}^2}{H_k^{2}(p_{1,k})} \leq \frac{Z p_{1,k}^2}{s_0^2} \to 0 \quad \text{as } k \to -\infty. \label{george} \end{equation} On the other hand, since $W_k' = H_k' + \sqrt\frac{2Z}{XY}\rho $ and $W_k$ is decreasing on $[0, p_{1,k}]$, it follows that $H_k$ is also decreasing on $[0, p_{1,k}]$ and therefore $H \geq s_0$ on $[0, p_{1,k})$. Using this fact and integrating \eqref{DE3} on $(0, \rho)$ gives $$ W_k'' + \frac{W_k'}{\rho} + \frac{XY}{1-\rho^2} = 2k + XY + \int_0^{\rho} \frac{Zt}{H_k^2} \, dt \leq 2k + XY + \frac{Z}{2s_0^2}\rho^2 \text{ on } [0, p_{1,k}]. $$ Further since $\frac{XY}{1-\rho^2} \geq XY$ on $[0,1]$ we obtain $$ W_k'' + \frac{W_k'}{\rho} \leq 2k + \frac{Z}{2s_0^2}\rho^2 \quad\text{on } [0, p_{1,k}]. $$ Multiplying by $\rho$ and integrating on $(0, \rho)$ gives: \[ W_k' \leq k \rho + \frac{Z}{8s_0^2}\rho^4 \] and so $$ \frac{W_k'(p_{1,k})}{p_{1,k}} \leq k + \frac{Z}{8s_0^2}p_{1,k}^3 \leq k + \frac{Z}{8s_0^2} \to -\infty \quad\text{as } k \to -\infty $$ contradicting \eqref{george}. Thus $H(p_{1,k}) \to 0$ as $k \to -\infty$. \end{proof} \begin{lemma} \label{lem3.5} Suppose $W_k$ satisfies \eqref{assume}--\eqref{assume2}. Then $m_k \to 0$ as $k \to -\infty$ (for some subsequence of $k$'s). \end{lemma} \begin{proof} Suppose not. So then there is a $t_0$ such that $m_k \geq t_0 > 0$. Using Theorem \ref{thm1} and the fact that $W_k$ is decreasing on $[0, m_k]$ we obtain \begin{equation} \begin{aligned} H_k(p_{1,k}) - \sqrt{\frac{Z}{2Y}}(1-p_{1,k}^2) &= W_k(p_{1,k}) \geq W_k(m_k) \\ &= H_k(m_k) - \sqrt{\frac{Z}{2XY}}(1-m_k^2) \\ &\geq - \sqrt{\frac{Z}{2XY}}(1-t_0^2). \end{aligned}\label{gehrig} \end{equation} By Lemmas \ref{lem3.3} and \ref{lem3.4} we see that the left-hand side of \eqref{gehrig} goes to $- \sqrt{\frac{Z}{2XY}}$ as $k \to -\infty$ and so this implies $t_0=0$ which is a contradiction. \end{proof} Now let $0<\epsilon \leq 1$ and observe that $ \frac{W_k'}{\rho^{1 - \epsilon}}$ is zero at $m_k$, $M_k$, and is positive on $(m_k,M_k)$. Therefore $ \frac{W_k'}{\rho^{1 - \epsilon}}$ has an absolute and local maximum on $(m_k,M_k)$. \begin{lemma} \label{lem3.6} Let $0<\epsilon \leq 1$. Suppose $W_k$ satisfies \eqref{assume}-\eqref{assume2}. At an absolute maximum, $q_k$, of $ \frac{W_k'}{\rho^{1 - \epsilon}}$ on $(m_k, M_k)$ we have $q_k\leq p_{2,k}$ and $$ \frac{W_k'(q_k)}{q_k^{1 - \epsilon}} \leq \frac{2XY q_k^{2+\epsilon}}{\epsilon(2-\epsilon)(1-q_k^2)^2}. $$ \end{lemma} \begin{proof} At $q_k$ we have $$ \Big(\frac{W_k'}{\rho^{1 - \epsilon}}\Big)' = 0 \quad\text{and}\quad \Big(\frac{W_k'}{\rho^{1 - \epsilon}}\Big)'' \leq 0. $$ Thus, $q_kW''(q_k) - (1- \epsilon) W_k'(q_k)=0$ and $q_kW'''(q_k) + \epsilon W_k''(q_k)\leq 0$. Therefore by \eqref{DE}, $$ \frac{W_k'(q_k)}{q_k^{1 - \epsilon}} \leq \frac{2XY q_k^{2+\epsilon}}{\epsilon(2-\epsilon)(1-q_k^2)^2}. $$ In addition, since $q_k$ is the maximum of $\frac{W_k'}{\rho^{1-\epsilon}}$ and $m_k < p_{2,k} < M_k$ then $$ \frac{W_k'(p_{2,k})}{p_{2,k}^{1 - \epsilon}} \leq \frac{W_k'(q_k)}{q_k^{1 - \epsilon}}. $$ Also, we know that $p_{2,k}$ is the maximum of $W_k'$ on $(m_k, M_k)$ so: $$ \frac{W_k'(q_k)}{q_k^{1 - \epsilon}} \leq \frac{W_k'(p_{2,k})}{q_k^{1 - \epsilon}}. $$ Since $W_k'>0 $ at $p_{2,k}$ and $q_k$, and also that $0<\epsilon \leq 1$, it then follows from the two previous inequalities that $$ q_k \leq p_{2,k}. $$ This completes the proof. \end{proof} \begin{lemma} \label{lem3.7} Suppose $W_k$ satisfies \eqref{assume}-\eqref{assume2}. Then $p_{2,k} \to 1$ as $k \to -\infty$ (for some subsequence of $k$'s). (And hence $M_k \to 1$ since $p_{2,k} \leq M_k$ (for some subsequence of $k$'s)). \end{lemma} \begin{proof} We first show that $p_{2,k} \not \to 0$. If $p_{2,k} \to 0$ then using Lemma \ref{lem3.6} with $\epsilon =1$ we see that \begin{equation} 0 \leq W_k'(p_{2,k}) \leq \frac{2XY p_{2,k}^3}{(1-p_{2,k}^2)^2} \to 0 \quad \text{as } k \to -\infty. \label{wpa} \end{equation} On the other hand, by the mean value theorem we have \begin{equation} -W_k(m_k) = W_k(z_k) - W_k(m_k) = W_k'(c_k)(z_k - m_k) \label{woody} \end{equation} for some $c_k$ with $m_k< c_k < z_k$. Also, since $p_{2,k}$ is a local maximum for $W_k'$ on $(m_k, M_k)$, it follows that $0\leq W_k'(c_k) \leq W_k'(p_{2,k})$. Substituting this into \eqref{woody}, using \eqref{wpa}, and that $0 \leq m_k \leq z_k \leq 1$ gives \begin{equation} \begin{aligned} -W_k(m_k) &= W_k(z_k) - W_k(m_k) \\ &= W_k'(c_k)(z_k - m_k)\\ &\leq W_k'(p_{2,k})(z_k-m_k) \\ &\leq \frac{2XY p_{2,k}^3}{(1-p_{2,k}^2)^2}. \end{aligned} \label{leadbelly} \end{equation} Next by Lemma \ref{lem3.2} and \eqref{W equation} it follows that $W_k(m_k) \to -\sqrt{\frac{Z}{2XY}}$ as $k \to -\infty$. Therefore, the left-hand side of \eqref{leadbelly} goes to $\sqrt{\frac{Z}{2XY}}>0$ as $k \to -\infty$ but the right-hand side goes to zero as $k \to -\infty$ by \eqref{wpa} which is a contradiction. Thus we see that $p_{2,k} \not \to 0$. So now suppose that the lemma is not true and that there is a $u_0$ and $v_0$ with $00$, $W_k'> 0$ on $(m_k, M_k)$, as well as $H_k'(m_k) + \sqrt{\frac{2Z}{XY}} m_k = W_k'(m_k) = 0$, $H_k''(m_k) + \sqrt{\frac{2Z}{XY}} = W_k''(m_k) >0$, and $H_kH_k'^{2} \geq 0$ gives \begin{align*} H_k^2 \left(H_k'' + \frac{H_k'}{\rho}\right) &\geq -2 \Big(\sqrt{\frac{2Z}{XY}}H_k^2(m_k) + \frac{Z}{XY} m_k^2 H_k(m_k) \Big)\\ &\quad + \frac{Z}{2}(\rho^{2}-m_k^2) - \int_{m_k}^{\rho}\frac{2XYt H_k^2}{(1- t^2)^2} \, dt - \int_{m_k}^{\rho} {H_k'}^{3} \, dt . \end{align*} Multiplying by $\rho$ and integrating by parts on $(m_k,\rho)$ gives \begin{align*} &\rho H_k^2 H_k' - \int_{m_k}^{\rho} 2 t H_kH_k'^2 \, dt \\ &\geq -\Big( \sqrt{\frac{2Z}{XY}}H_k^2(m_k) + \frac{Z}{XY} m_k^2 H_k(m_k)\Big) (\rho^2-m_k^2)+ \int_{m_k}^{\rho}\frac{Z}{2}t(t^{2}-m_k^2) \, dt \\ &\quad - \int_{m_k}^{\rho} t \int_{m_k}^{t}\frac{2XYs H_k^2}{(1- s^2)^2}\, ds \, dt -\int_{m_k}^{\rho} t \int_{m_k}^{t} H_k'^3 \, ds \, dt. \end{align*} Thus, since $H_k>0$ and $H_kH_k'^2\geq 0$, \begin{equation} \begin{aligned} & \rho H_k^2 H_k' + \int_{m_k}^{\rho} t \int_{m_k}^{t} \frac{2XYs H_k^2}{(1- s^2)^2} \, ds \, dt +\int_{m_k}^{\rho} t \int_{m_k}^{t} H_k'^3 \, ds \, dt \\ &\geq -\Big( \sqrt{\frac{2Z}{XY}}H_k^2(m_k) + \frac{Z}{XY} m_k^2 H_k(m_k)\Big)(\rho^2-m_k^2) \\ &\quad + \frac{Z}{2} \Big( \frac{(\rho^4 -m_k^4)}{4} - \frac{m_k^2(\rho^2-m_k^2)}{2} \Big). \end{aligned} \label{ID5} \end{equation} Since $p_{2,k}\leq v_0<1$, it follows from Lemma \ref{lem3.6} and \eqref{W equation} that for fixed $\epsilon$ with $0< \epsilon < 1$, \begin{equation} H_k' \leq W_k' \leq C_4\rho^{1-\epsilon} \quad\text{on } [m_k,p_{2,k}] \text{ where } C_4= \frac{2XYv_0^{2 + \epsilon}}{\epsilon(2-\epsilon)(1-v_0^2)^2}. \label{ID7} \end{equation} Substituting \eqref{ID7} into \eqref{ID5} gives \begin{equation} \begin{aligned} &C_4\rho^{1-\epsilon} H_k^2 + \int_{m_k}^{\rho} t \int_{m_k}^{t}\frac{2XYs H_k^2}{(1- s^2)^2} \, ds \, dt + C_4^3\int_{m_k}^{\rho} t \int_{m_k}^{t} s^{3-3 \epsilon} \, ds \, dt \\ &\geq -\Big( \sqrt{\frac{2Z}{XY}}H_k^2(m_k) + \frac{Z}{XY} m_k^2 H_k(m_k)\Big)(\rho^2-m_k^2)\\ &\quad +\frac{Z}{2} \Big(\frac{(\rho^4 -m_k^4)}{4} - \frac{m_k^2(\rho^2-m_k^2)}{2}\Big). \end{aligned}\label{ID6} \end{equation} Also, integrating \eqref{ID7} on $(m_k,\rho)$ and using \eqref{W equation} as well as Theorem \ref{thm1} gives \begin{equation} -\sqrt{\frac{Z}{2XY}}(1-\rho^2) \leq W_k(\rho) \leq W_k(m_k) + \frac{C_4 }{2-\epsilon}(\rho^{2-\epsilon} - m_k^{2-\epsilon}) \leq \frac{C_4 }{2-\epsilon}(\rho^{2-\epsilon} - m_k^{2-\epsilon}). \label{ID8} \end{equation} Now we know from Lemma \ref{lem3.1} that $W_k'\geq 0$ on $[m_k,M_k]$ and so it follows from \eqref{ID7}-\eqref{ID8} and Lemma \ref{lem3.6} that $ W_k$ and $ W_k'$ are uniformly bounded on $[m_k,M_k]$. So since $m_k$ and $p_{2,k}$ are bounded there exists a subsequence (still labeled $k$) such that $m_k \to m=0$ (by Lemma \ref{lem3.5}) and $p_{2,k} \to p_{2}$ with $00$ such that $W_k(M_k) \leq C$ for all $k$. Now we know from Lemma \ref{lem3.1} that on $(m_k, p_{2,k})$ we have $W_k'\geq 0$ and $W_k'' \geq 0$, and so it follows from \eqref{ID16} and \eqref{DE3} that $W_k, W_k'$, and $W_k''$ are uniformly bounded on compact subsets of $(m_k, p_{2,k})$. It then follows by the Arzela-Ascoli theorem that there is a subsequence (still labeled $k$) such that $W_k \to W$, $W_k' \to W'$, and $W_k'' \to W''$ uniformly on compact subsets of $(0, 1)$ since $m_k\to 0$ by Lemma \ref{lem3.5} and $p_{2,k} \to 1$ by Lemma \ref{lem3.7}. In addition, $W_k' \geq 0$ on $(m_k, p_{2,k})$ and $W_k'' \geq 0 $ on $(m_k, p_{2,k})$. Since $m_k \to 0$ by Lemma \ref{lem3.5} and $p_{2,k} \to 1$ by Lemma \ref{lem3.7} it follows that $W$ and $W'$ is increasing on $(0,1)$. Thus we may define $W(1)\equiv\lim_{\rho \to 1^{-}} W(\rho) $, $W'(1)\equiv\lim_{\rho \to 1^{-}} W'(\rho) $ and $W(0)\equiv\lim_{\rho \to 0^{+}} W(\rho) $, $W'(0)\equiv\lim_{\rho \to 0^{+}} W'(\rho)$. And so there is a corresponding function $H$ such that $H_k\to H$, $H_k'\to H'$, and $H_k'' \to H''$ uniformly on compact subsets of $(0,1)$. We have similar definitions for $H(0)$, $H'(0)$, $H(1)$, and $H'(1)$. It now follows from \eqref{DE3} that $W_k'''$ is uniformly bounded in a neighborhood of any $\rho_0$ with $0 < \rho_0 <1 $ where $H(\rho_0) >0$. Along with the boundedness of $W_k$, $W_k'$, and $W_k''$ in this neighborhood, it follows that $W$ satisfies \eqref{DE3} at any $0< \rho_0<1$ for which $H(\rho_0)>0$. Suppose now that there exists $\rho_0$ with $0 < \rho_0 < 1$ such that $H(\rho_0)=0$ and $H(\rho)>0$ for $\rho_0< \rho < 1$. This would contradict Theorem \ref{thm1} and so it must be the case that either $H>0$ on $(0,1)$ or $H \equiv 0$ on $(0,1)$. \textbf{Case 1:} Suppose first that $H>0$ on $(0,1)$. Next we note that after multiplying \eqref{ID16} by $\rho$ and using the fact that $W_k(M_k) \leq C$ we see that $W_k'$ and hence $H_k'$ is uniformly bounded, say by $T$, on $(m_k, M_k)$. Thus, integrating the inequality $H_k' \leq T$ on $(m_k,\rho)$ and using Theorem \ref{thm1} gives $0< H_k(\rho) \leq H_k(m_k) + T(\rho - m_k)$. We know $H_k(m_k) \to 0$ by Lemma \ref{lem3.2} and $m_k \to 0$ by Lemma \ref{lem3.5}, so taking limits as $k \to -\infty$ we see that $0 < H(\rho) \leq T\rho$ on $(0,1)$. Thus, we see $ H(0) \equiv \lim_{\rho \to 0^{+}} H(\rho) =0$. Now integrating \eqref{DE3} on $(\rho, \rho_{1})$ where $0< \rho < \rho_{1} < 1$ we see that $$ W''(\rho_{1}) + \frac{W'(\rho_{1})}{\rho_{1}} + \frac{XY}{1-\rho_{1}^2} = W''(\rho) + \frac{W'(\rho)}{\rho} + \frac{XY}{1-\rho^2} + \int_{\rho}^{\rho_{1}} \frac{Z t}{H^{2}(t)} \, dt. $$ Therefore, \begin{equation} \lim_{\rho \to 0^{+}} \Big( W''(\rho) + \frac{W'(\rho)}{\rho} + \int_{\rho}^{\rho_{1}} \frac{Z t}{H^{2}(t)} \, dt \Big) \text{ exists and is finite. } \label{USG} \end{equation} Now since $0 < H(\rho) \leq T\rho$ it follows that $$ \int_{\rho}^{\rho_{1}} \frac{Z t}{H^{2}(t)} \, dt \geq \int_{\rho}^{\rho_{1}} \frac{Z t}{T^2 t^2} \, dt = \frac{Z}{T^2} \ln\big(\frac{\rho_{1}}{\rho}\big) \to \infty \quad\text{as } \rho\to 0^{+} $$ which along with the fact that $W'\geq 0$ and $W'' \geq 0$ contradicts \eqref{USG}. Thus the assumption that $H>0$ on $(0,1)$ must be false. \textbf{Case 2:} Next suppose that $H \equiv 0$ on $(0,1)$. We first show that $H_k$ cannot be decreasing on $(m_k,1)$. Integrating \eqref{DE3} on $(m_k, \rho)$ gives \begin{equation} H_k'' + \frac{H_k'}{\rho} + \frac{XY}{1-\rho^2} = H_k''(m_k) + \frac{H_k'(m_k)}{m_k} + \frac{XY}{1-m_k^2} + \int_{m_k}^{\rho} \frac{Z t}{H_k^2} \, dt. \label{scooter} \end{equation} Substituting $ H_k'(m_k) + \sqrt{\frac{2Z}{XY}}m_k = W_k'(m_k)=0$, $ H_k''(m_k) + \sqrt{\frac{2Z}{XY}} = W_k''(m_k)\geq 0$, and assuming $H_k$ is decreasing on $(m_k,1)$ gives \begin{equation} H_k'' + \frac{H_k'}{\rho} + \frac{XY}{1-\rho^2} \geq -2\sqrt{\frac{2Z}{XY}} + \frac{Z}{2 H_k^2(m_k)}(\rho^2 - m_k^2). \end{equation} Multiplying by $\rho$ and integrating again on $(m_k,\rho)$ gives \begin{equation} \rho H_k' \geq - \sqrt{\frac{2Z}{XY}}\rho^2 +\frac{XY}{2} \ln\left(\frac{1-\rho^2}{1-m_k^2}\right) + \frac{Z}{2 H_k^2(m_k)} \int_{m_k}^{\rho}t(t^2 - m_k^2) \, dt. \label{big man} \end{equation} Dividing by $\rho$ and integrating again on $(m_k, \rho)$ gives \begin{equation} \begin{aligned} H_k &\geq H_k(m_k) - \frac{1}{2}\sqrt{\frac{2Z}{XY}}(\rho^2-m_k^2) + \int_{m_k}^{\rho}\frac{XY}{2t} \ln\left(\frac{1-t^2}{1-m_k^2}\right) \, dt \\ &\quad + \frac{Z}{2 H_k^2(m_k)} \int_{m_k}^{\rho} \frac{1}{s} \int_{m_k}^{s} t(t^2 - m_k^2) \, dt \, ds. \end{aligned} \label{patton} \end{equation} Next, making the substitution $u = \frac{1-t}{1-m_k}$ we observe that \begin{equation} \begin{aligned} \int_{m_k}^{1} \frac{1}{t} |\ln\Big(\frac{1-t^2}{1-m_k^2}\Big)| \, dt &\leq \int_{m_k}^{1} \frac{1}{t} |\ln [2 \Big(\frac{1-t}{1-m_k}\Big)]| \, dt\\ & \leq \int_0^{1} \frac{|\ln(2u)|}{1-u} \, du < \infty. \end{aligned} \label{ringo} \end{equation} Thus we see that $\frac{1}{t} \ln\big(\frac{1-t^2}{1-m_k^2}\big)$ is uniformly integrable on $(m_k,1)$. Combining this with the fact that $H_k(m_k) \to 0$ by Lemma \ref{lem3.2} and $m_k \to 0$ by Lemma \ref{lem3.5} we see that for fixed $\rho>0$ and $k$ sufficiently negative that the right-hand side of \eqref{patton} goes to $+\infty$. However, the left-hand side of \eqref{patton} is bounded (because by assumption \[ 0< H_k = W_k + \sqrt{\frac{Z}{2XY}}(1-\rho^2) \leq W_k(M_k) + \sqrt{\frac{Z}{2XY}} \leq C + \sqrt{\frac{Z}{2XY}}. \] This is a contradiction and so the assumption that $H_k$ is decreasing on $(m_k,1)$ must be false. Thus $H_k$ has a minimum, $n_k$, with $m_k < n_k<1$ and $H_k'\leq 0 $ on $(0, n_k)$. Also, by Lemma \ref{lem3.2} we have \begin{equation} 0< H_k(n_k) \leq H_k(m_k) \to 0. \label{vz} \end{equation} We next claim that $n_k\to 1$ as $k \to -\infty$. Repeating the above argument with $m_k$ being replaced by $n_k$ in \eqref{scooter} and using that $H_k'(n_k) =0$ and $H_k''(n_k) \geq 0$ gives: $$ H_k'' + \frac{H_k'}{\rho} + \frac{XY}{1-\rho^2} \geq \int_{n_k}^{\rho} \frac{Zt}{H_k^2} \, dt.$$ Multiplying by $\rho$ and integrating on $(n_k, \rho)$ gives: $$ \rho H_k' \geq \frac{XY}{2} \ln\Big(\frac{1-\rho^2}{1-n_k^2}\Big) + \int_{n_k}^{\rho} s \int_{n_k}^{s} \frac{Zt}{H_k^2} \, dt \, ds. $$ Dividing by $\rho$ and integrating again gives \begin{equation} H_k(\rho) \geq H_k(n_k) + \int_{n_k}^{\rho} \frac{XY}{2\rho} \ln\Big(\frac{1-\rho^2}{1-n_k^2}\Big) + \int_{n_k}^{\rho} \frac{1}{t}\int_{n_k}^{t} s \int_{n_k}^{s} \frac{Zx}{H_k^2} \, dx \, ds \, dt. \label{rhizuto} \end{equation} We know $H_k(n_k) \to 0$ by \eqref{vz}. Suppose now that $n_k \to n$ where $0\leq n < 1$. Then for fixed $\rho$ with $0<\rho < 1$ we have $H_k(\rho) \to H(\rho)\equiv 0$ and similarly as in \eqref{ringo} the term $\frac{1}{\rho}\ln\left(\frac{1-\rho^2}{1-n_k^2}\right)$ is uniformly integrable on $(n_k,1)$ but the last term on the right in \eqref{rhizuto} goes to infinity since $H_k \to H \equiv 0 $ uniformly on compact subsets of $(n,\rho)$ yielding a contradiction. Thus it must be the case that $n_k \to 1$. Next, taking limits in \eqref{patton} for fixed $\rho$ and using that $m_k \to 0$ also yields a contradiction because $\frac{1}{\rho}\ln\big(\frac{1-\rho^2}{1-m_k^2}\big)$ is uniformly integrable on $(m_k,1)$ and so the right-hand side goes to infinity while the left-hand side is bounded. Thus the assumption that $H \equiv 0$ must also be false and therefore it must be that $W_k(M_k) \to \infty$. This completes the proof. \end{proof} \begin{lemma} \label{lem3.9} Suppose $W_k$ satisfies \eqref{assume}-\eqref{assume2}. Then either $z_k \to 0$ as $k \to -\infty$ (for some subsequence of $k$'s) or $z_k \to 1$ as $k \to -\infty$ (for some subsequence of $k$'s). \end{lemma} \begin{proof} Suppose not. Then there is a $u_0$ and a $v_0$ with $0< u_0 \leq v_0 <1$ such that $0< u_0\leq z_k \leq v_0<1$ for all $k$. From Lemma \ref{lem3.8} we know that $W_k(M_k) \to \infty$ as $k \to -\infty$. We now define $$ Q_k(\rho) = \frac{W_k(\rho)}{W_k(M_k)}. $$ Note that $Q_k \leq 1$ on $[0, M_k]$ and since $W_k(\rho)\geq -\sqrt\frac{Z}{2XY}(1-\rho^2) \geq -\sqrt\frac{Z}{2XY} $ (by Theorem \ref{thm1}) it follows that $Q_k$ is bounded from below independent of $k$. Thus the $Q_k$ are uniformly bounded on $[0, M_k]$. It follows then from \eqref{ID16} that on $[\rho ,M_k]$ we have \begin{equation} \begin{aligned} &(1 - \rho)^2 \Big( Q_k'' + \frac{Q_k'}{\rho} \Big) + 2 (1-\rho)Q_k' + \frac{2}{\rho}Q_k \\ & \leq \frac{2}{M_k} + \frac{1}{W_k(M_k)}\int_{\rho}^{M_k} \frac{2XY t}{(1+t)^2} \, dt + \int_{\rho}^{M_k} \frac{2}{t^2}Q_k \, dt. \end{aligned}\label{LBJ} \end{equation} In addition, since $Q_k' \geq 0$ and $Q_k''\geq 0$ on $(m_k, p_{2,k})$ and $p_{2,k} \to 1$ by Lemma \ref{lem3.7}, it follows from \eqref{LBJ} that $Q_k$, $Q_k'$, and $Q_k''$ are uniformly bounded on compact subsets of $(m_k, v_0]$. Since $W_k \geq 0$ on $[z_k,M_k]$ we see from \eqref{DE3} that $(W_k'' + \frac{W_k'}{\rho})' \leq 0$ on $[z_k,M_k]$. In particular, since $z_k \leq v_0$ it follows that $$ W_k'' + \frac{W_k'}{\rho} \leq W_k''(v_0) + \frac{W_k'(v_0) }{v_0} \quad\text{ for } v_0 \leq \rho \leq M_k. $$ Therefore \begin{equation} Q_k'' + \frac{Q_k'}{\rho} \leq Q_k''(v_0) + \frac{Q_k'(v_0)}{v_0} \quad \text{for } v_0 \leq \rho \leq M_k. \label{JFK} \end{equation} Since $0\leq Q_k \leq 1$ on $[z_k, M_k]$ and also $[v_0, M_k] \subset [z_k, M_k] $ it follows from \eqref{LBJ} and since $v_0<1$ that $Q_k''(v_0) + \frac{Q_k'(v_0)}{v_0}$ is bounded independent of $k$. Since $Q_k' \geq 0$ and $Q_k'' \geq 0$ on $(m_k, p_{2,k})$ then it follows from \eqref{JFK} that $Q_k$, $Q_k'$, and $Q_k''$ are also uniformly bounded on $[v_0,p_{2,k}]$. In addition, earlier in this proof we showed that $Q_k$, $Q_k'$, and $Q_k''$ are uniformly bounded on compact subsets of $(m_k, v_0]$. Combining these results we see that $Q_k$, $Q_k'$, and $Q_k''$ are uniformly bounded on compact subsets of $(m_k, p_{2,k}]$. From Lemmas \ref{lem3.5} and \ref{lem3.7} we know that $m_k\to 0$ and $p_{2,k} \to 1$ as $k \to -\infty$ and so by the Arzela-Ascoli theorem there is a further subsequence (again labeled $k$) such that $Q_k \to Q$ and $Q_k' \to Q'$ uniformly on compact subsets of $(0, 1)$. In particular, $Q$ and $Q'$ are continuous on $(0, 1)$. Further, since $Q_k\geq 0$ and $Q_k'\geq 0$ on $(m_k, p_{2,k})$ and $m_k\to 0$ and $p_{2,k} \to 1$, it follows that $Q\geq 0$ and $Q' \geq 0$ on $(0,1)$. Now let $\rho$ and $\rho_0$ satisfy $z_k< \rho < \rho_0< 1$. Then on $[z_k, \rho_0]$ we know that $W_k\geq 0$ so $H_k \geq \sqrt\frac{Z}{2XY}(1- \rho_0^2) > 0$ on this set and so we see that $\frac{1}{H_k^2}$ is bounded on $[z_k, \rho_0]$. Also since $u_0 \leq z_k \leq v_0$ there is a subsequence (still labeled $k$) so that $z_k \to z$ with $00$ then $Q'<0$ on $(z-\delta,z)$ for some $\delta >0$ contradicting $Q'\geq 0$. Thus it must be that $Q''(z)=0$. Solving \eqref{gorka} along with the conditions $Q(z)=Q'(z)=Q''(z) = 0$ implies $Q\equiv 0$. Next recall from earlier in the proof that $Q_k'$ is uniformly bounded, say by $L$, on $(m_k, M_k)$. Thus $$ 0 \leq 1-Q_k(\rho) = Q_k(M_k)- Q_k(\rho) = \int_{\rho}^{M_k} Q_k'(t) \, dt \leq L(M_k -\rho). $$ So taking limits as $ k \to -\infty$ gives $$ 0\leq 1- Q(\rho) \leq L(1-\rho). $$ Now taking the limit as $\rho \to 1^{-}$, it follows that $Q(1)=1$. But we showed earlier that $Q \equiv 0$ so we obtain a contradiction. Therefore, it must be the case that either $z_k\to 0$ or $z_k \to 1$. This completes the proof. \end{proof} \begin{lemma} \label{lem3.10} Suppose $W_k$ satisfies \eqref{assume}-\eqref{assume2}. Then $z_k \to 0$ as $k \to -\infty$ (for some subsequence of $k$'s). \end{lemma} \begin{proof} From Lemma \ref{lem3.9}, we know that $z_k\to 0$ or $z_k \to 1$ as $ k \to -\infty$. Let us assume that $z_k \to 1$ as $k \to -\infty$. First, since $m_k \to 0$ by Lemma \ref{lem3.5} and $z_k \to 1$ by assumption we may choose a $\delta>0$ such that $m_k \leq 1-2\delta$ and $1-z_k \leq \frac{\delta}{2}$ for $k$ sufficiently negative. Then by the mean value theorem $ - W_k(1-\delta) = W_k(z_k) - W_k(1-\delta) = W_k'(c_k)( z_k-(1-\delta))$ for some $c_k$ with $ 1-\delta< c_k < z_k$. Thus, for $k$ sufficiently negative, $$ 0\leq W_k'(c_k) = \frac{- W_k(1-\delta)}{ z_k-(1-\delta)} \leq \frac{\sqrt{\frac{2Z}{XY}}}{\delta} $$ since $ 0\leq -W_k(1-\delta) \leq \sqrt{\frac{Z}{2Y}}$ and $\frac{1}{z_k - (1-\delta)} \leq \frac{2}{\delta}$. Also, since $W_k'' \geq 0$ on $(m_k, z_k)$ we see that for $\rho$ with $m_k \leq \rho \leq 1-\delta$ we have \begin{equation} 0 \leq W_k' \leq W_k'(c_k) \leq \frac{\sqrt{\frac{2Z}{XY}}}{\delta}\quad \text{on } (m_k, 1-\delta).\label{wade} \end{equation} Then by the mean value theorem there is an $x_k$ with for $1- 2\delta < x_k < 1 -\delta$ such that $$ \frac{\sqrt{\frac{2Z}{XY}}}{\delta} \geq W_k'(1-\delta) - W_k'(1-2\delta) = W_k''(x_k)\delta. $$ Thus $$ 0\leq W_k''(x_k) \leq \frac{\sqrt{\frac{2Z}{XY}}}{\delta^2}. $$ In addition, by \eqref{wade} we have \begin{gather} 0 \leq W_k'(x_k) \leq \frac{\sqrt{\frac{2Z}{XY}}}{\delta}, \label{mantle} \\ 0 \leq \frac{W_k'(x_k)}{x_k} \leq \frac{\sqrt{\frac{2Z}{XY}}}{\delta(1-2\delta)}. \label{maris} \end{gather} It follows then from \eqref{ID10} that for $m_k < x_k< \rho 0$. Along with the boundedness of $W_k$, $W_k'$, and $W_k''$ in this neighborhood it follows that $W$ solves \eqref{DE3} in this neighborhood. Suppose now that there exists $\rho_0$ with $0 < \rho_0 < 1$ such that $H(\rho_0)=0$ and $H(\rho)>0$ for $\rho_0< \rho < 1$. This would contradict Theorem \ref{thm1} and so it must be the case that either $H>0$ on $(0,1)$ or $H \equiv 0$ on $(0,1)$. If $H>0$ on $(0,1)$ then it actually follows that $W'(0)=0$. To see this, we first observe that since $W' \geq 0$ and $W''\geq 0$ on $(0,1)$ it follows that $\lim_{\rho \to 0^{+}} W'(\rho) = A \geq 0$. We would like to show $A=0$ so we will assume $A>0$. Multiplying \eqref{DE3} by $\rho^2$ and taking the limits as $\rho \to 0^{+}$ gives $\lim_{\rho \to 0^{+}} (\rho^2 W''' + \rho W') = A$. Thus for small positive $\rho$ we have $(\rho W'')' \geq \frac{A}{2\rho}$. Integrating on $(\rho, \rho_0)$ gives $ \rho W'' \leq \frac{A}{2} \ln(\rho) + C_0$ for some constant $C_0$. Dividing by $\rho$ and integrating again on $(\rho, \rho_0)$ gives $W \geq \frac{A}{4} \frac{\ln^{2}(\rho)}{\rho} + C_0 \ln(\rho) + C_{1}$ for some constant $C_{1}$. This implies $W\to \infty$ as $\rho \to 0^{+}$ which contradicts that $W(\rho)$ is bounded. Thus we see that $ W'(0) \equiv \lim_{\rho \to 0^{+}} W'(\rho) = 0$. Then by Lemma \ref{lem2.3} we see that this implies that $W$ must get positive on $(0,1)$ but this contradicts that $W\leq 0$ on $(0,1)$. On the other hand, if $H \equiv 0$ on $(0,1)$ then using a nearly identical argument as in Case 2 of Lemma \ref{lem3.7}, we can arrive at equation \eqref{patton} and from this it follows that $H_k$ must get large and hence a minimum, $n_k$, with $m_k < n_k$ must exist. The rest of the argument is the same and so again this leads to a contradiction. Thus we see that $z_k \not \to 1$ as $k \to -\infty$ and consequently by Lemma \ref{lem3.9} we see that $z_k \to 0$ as $k \to -\infty$. This completes the proof. \end{proof} \begin{proof}[Proof Theorem \ref{main-thm}] If there is such a solution then we know from Lemma \ref{lem3.10} that there is a subsequence (again labeled $k$) such that $z_k \to 0$. Integrating \eqref{DE3} on $(m_k, \rho)$ gives $$ W_k'' + \frac{W_k'}{\rho} + \frac{XY}{1-\rho^2} = W_k''(m_k) + \frac{XY}{1-m_k^2} + \int_{m_k}^{\rho} \frac{Zt}{H_k^2} \, dt \geq 0. $$ Multiplying by $\rho$, integrating on $(m_k, \rho)$, and using that $m_k$ is a local minimum so that $W_k''(m_k) \geq 0$ gives $$ 0 \geq \rho W_k' \geq \ln\Big( \frac{ 1-\rho^2}{1-m_k^2} \Big). $$ Suppose now that $W_k(z_{2,k}) = 0$ with $W_k>0$ on $(z_k, z_{2,k})$. Then we know that $W_k' \leq 0$ on $(M_k, z_{2,k})$ by Lemma \ref{lem2.2}. Thus, on $(M_k, z_{2,k})$ we have $$ W_k'^2 \leq \frac{\ln^2\big( \frac{ 1-\rho^2}{1-m_k^2} \big)}{\rho^2}. $$ Recall that $m_k \to 0$ by Lemma \ref{lem3.5} and $M_k \to 1$ by Lemma \ref{lem3.7}. Thus for some positive constant $A$ independent of $k$ we have: $$ \int_{M_k}^{z_{2,k}} W_k'^2 \, dt \leq \int_{M_k}^{z_{2,k}} \frac{\ln^2\big( \frac{ 1-\rho^2}{1-m_k^2} \big)}{\rho^2} \, d \rho \leq \int_{M_k}^{1} \frac{\ln^2\big( \frac{ 1-\rho^2}{1-m_k^2} \big)}{\rho^2} \, d \rho \leq A < \infty. $$ Thus, \begin{equation} \int_{M_k}^{z_{2,k}} Q_k'^2 \, dt = \frac{1}{W_k^2(M_k)} \int_{M_k}^{z_{2,k}} W_k'^2 \, dt \to 0 \quad \text{as } k \to -\infty. \label{dirk} \end{equation} Finally by Holder's inequality and \eqref{dirk} we have \begin{align*} 1 &=|0-1| = |Q_k(z_{2,k}) - Q_k(M_k)| = |\int_{M_k}^{z_{2,k}} Q_k'(t) \, dt| \\ &\leq \sqrt{z_{2,k}-M_k}\sqrt{\int_{M_k}^{z_{2,k}} Q_k'^2 \, dt } \to 0 \quad \text{as } k \to -\infty \end{align*} which is impossible. This completes the proof. \end{proof} \subsection*{Acknowledgments} We would like to thank Santiago Betelu for calling our attention to this project and for many useful discussions. \begin{thebibliography}{99} \bibitem{B} F. Bernis, R. Ferreira; Source type solutions to thin-film equations in higher dimensions, \emph{European Journal of Applied Mathematics}, Vol. 0, Issue 5, (1997) 507-524 . \bibitem{BF} S. Betelu, M. Fontelos; Spreading of a charged microdroplet, \emph{Physica D} 209 (2005) 28-35. \bibitem{D2} S. Betelu, J. A. Diez, R. Gratton, B. Marino, L. P. Thomas; Quasi-self-similarity for wetting drops, \emph{Physical Review E} 53, (1996) 3563-3572. \bibitem{BR} J. Birkhoff, G. C. Rota; \emph{Ordinary Differential Equations}, Ginn and Company, 1959. \bibitem{D} J. A. Diez, R. Gratton, B. Marino, L.P. Thomas; Laplace pressure driven drop spreading: quasi-self-similar solution, \emph{Journal of Colloid and Interface Science}, Vol. 168, Issue 1, (1994) 15-20. \bibitem{I} J. Iaia; Spreading of Charged Microdroplets, \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 202, 1-22. \bibitem{IA} J. Iaia; Spreading Charged Microdroplets, submitted to \emph{Differential and Integral Equations}, 2014. \end{thebibliography} \end{document}