\documentclass[twoside]{article} \usepackage{amsfonts, amsmath} \pagestyle{myheadings} \markboth{\hfil Asymptotic analysis of the Carrier-Pearson problem \hfil EJDE/Conf/10} {EJDE/Conf/10 \hfil Chunqing Lu \hfil} \begin{document} \setcounter{page}{227} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent Fifth Mississippi State Conference on Differential Equations and Computational Simulations, \newline Electronic Journal of Differential Equations, Conference 10, 2003, pp 227--240. \newline http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Asymptotic analysis of the Carrier-Pearson problem % \thanks{ {\em Mathematics Subject Classifications:} 34B15, 34E05, 34E20. \hfil\break\indent {\em Key words:} Boundary layers, interior layer, transcendentally small terms. \hfil\break\indent \copyright 2003 Southwest Texas State University. \hfil\break\indent Published February 28, 2003.} } \date{} \author{Chunqing Lu} \maketitle \begin{abstract} This paper provides a rigorous analysis of the asymptotic behavior of the solution for the boundary-value problem \begin{gather*} \epsilon ^2u''+u^2-1 =0,\quad -10$, the initial value problem (\ref{eq:cp})-(\ref{ini:0}) has a unique solution. If$\alpha =-1$, the initial-value problem has the constant solution$u(x)\equiv -1$. If$\alpha <-1$, then$u''<0$initially, and hence,$u'<0$,$u''<0$as long as$u<-1$which cannot provide the solution with a spike. Thus, we only consider$\alpha \in (-1,0)$. Define$f(u)=3u-u^{3}$. From (\ref{eq:u'2}), $$u'=\frac{1}{\epsilon }\sqrt{(2/3}\sqrt{f(u)-f(\alpha )} \label{eq:u'}$$ where$\alpha \in (-1,0)$. Let$u'(x_{+})=0$for an$x_{+}>0$. Then,$u(x_{+})=\beta $is the greatest zero point of the equation$ f(u)-f(\alpha )=0$, and$\alpha $the mid-zero point of the equation, and therefore,$ f(u)-f(\alpha )=(u-\mu )(u-\alpha )(\beta -u) $with$\mu \leq \alpha <0<\beta $. Hence, \begin{eqnarray} f(u)-f(\alpha ) &=&-(u-\alpha )(u^2+\alpha u+\alpha ^2-3) \nonumber\\ &=&(\beta -u)(u^2+\beta u+\beta ^2-3) \end{eqnarray} with $$\beta =\frac{-\alpha +\sqrt{12-3\alpha ^2}}{2} \label{eq:beta}$$ It is also seen that$\beta >\sqrt{3}$($\alpha =0$implies$\beta =\sqrt{3}$and$\mu =-\beta )$and$\beta $is a decreasing function of$\alpha $because$\frac{d\beta }{d\alpha }=-\frac{1}{2}-\frac{3\alpha }{\sqrt{ 12-3\alpha ^2}}<0$for$\alpha \in (-1,0)$. We now define a function $$I\left( \alpha \right) =\int_{\alpha }^{\beta }\frac{du}{\sqrt{f(u)-f(\alpha )}} \label{ex:ia}$$ for$\alpha \in (-1,0)$and$\beta =\frac{-\alpha +\sqrt{12-3\alpha ^2}}{2} \in (\sqrt{3},2)$. Notice that this function is independent of$\epsilon . $From (\ref{eq:u'}), the value of$\alpha $required by the boundary-value problem can be determined by the equation $$I(\alpha )=\frac{\sqrt{2}}{\sqrt{3}\epsilon }. \label{eq:i(a)=1}$$ Thus, the existence and uniqueness of the one-spike solution of the boundary-value problem is equivalent to the existence and uniqueness of the value of$\alpha $, the solution of (\ref{eq:i(a)=1}) for given$ \epsilon >0$. Since the function$f(u)-f(\alpha )$has three distinct real zero points and the two limits of the improper integral$I(\alpha )$are regular, the improper integral$I(\alpha )$converges and$I(\alpha )$is well defined. \begin{lemma} \label{lm3} As$\epsilon \to 0$,$\alpha \to -1^{+}$. \end{lemma} \paragraph{Proof} Since \begin{eqnarray*} I(\alpha )&=&\Big( \int_{\alpha }^{0}+\int_{0}^{\beta }\Big) \frac{du }{\sqrt{f(u)-f(\alpha )}}\\ &=&\int_{\alpha }^{0}\frac{du}{\sqrt{3u-u^{3}-3\alpha +\alpha ^{3}}}+\int_{0}^{\beta }\frac{du}{\sqrt{3u-u^{3}-3\beta +\beta ^{3}}} \\ &\leq &\int_{\alpha }^{0}\frac{du}{\sqrt{3(u-\alpha )(1-\alpha ^2)}}+(\int_{0}^{c}+\int_{c}^{\beta }) \frac{du}{\sqrt{\beta u(\beta -u)}}\\ &\leq& \frac{2\sqrt{-\alpha }}{\sqrt{3-3\alpha ^2}}+ \frac{2\sqrt{c}}{\sqrt{\beta (\beta -c)}}+\frac{2\sqrt{\beta -c}}{\sqrt{ \beta c}}, \end{eqnarray*} for some constant$c\in (0,\beta )$. It follows that$\sqrt{2}/(\sqrt{3}\epsilon )\leq 2\sqrt{-\alpha }/\sqrt{3-3\alpha ^2}+O(1)$. This proves either$\alpha \to -1$as$\epsilon \to 0$. \hfill$\Box$\smallskip In the remainder of the paper, we use$\alpha \to -1$to denote the side limit,$\alpha \to -1^{+}$. \begin{lemma} \label{lm4} The function$I(\alpha )$is continuous for$\alpha \in (-1,0)$and decreasing for sufficiently small$(\alpha +1)>0$. \end{lemma} For the proof of this lemma, see \cite{Lu2}. More delicate work may prove that the function$I(\alpha )$is a monotone function in the open interval$(-1, 0)$. Because we only focus on asymptotic analysis as$\epsilon \to 0$in the paper, we omit that proof here. As stated above, it follows from Lemma \ref{lm4} that the single-spike solution of (\ref{eq:cp})-(\ref{bd:+-1}) exists and is unique. \section{Length of the interior layer} The asymptotic analysis begins in this section. From (\ref{eq:u'}), we see that $$\int_{\alpha }^{x}\frac{du}{\sqrt{f(u)-f(\alpha )}}=\sqrt{\frac{2}{3}}\frac{x }{\epsilon }. \label{eq:intg}$$ Through out the paper, we will study the asymptotic behavior of the integral function on the left side of (\ref{eq:intg}). \begin{lemma} \label{lm5} Let$x_{0}$be the turning point. Then, as$\epsilon \to 0$, $$1-x_{0}\sim \lbrack \sqrt{2}\ln (\sqrt{3}+\sqrt{2})]\epsilon.$$ \end{lemma} \paragraph{Proof} It is seen from (\ref{eq:beta}) that$\beta \to 2^{-}$as$ \alpha \to -1$. We will only consider the case that$(2-\beta \dot{)} $is positive and sufficiently small. From (\ref{eq:u'}), we see that $$\int_{0}^{u}\frac{du}{\sqrt{f(u)-f(\alpha )}}=\sqrt{\frac{2}{3}}\frac{x-x_{0} }{\epsilon }. \label{eq:int+}$$ Define$K(\beta )=\int_{0}^{\beta }\frac{du}{\sqrt{f(u)-f(\beta )}}, M(\beta )=\int_{0}^{\beta }\frac{du}{\sqrt{f(u)-f(2)}}$and$s(\beta )=K(\beta )-M(\beta )$. It follows from (\ref{eq:int+}) that$K(\beta )= \sqrt{(2/3)}\frac{x-x_{0}}{\epsilon }$. Since \begin{eqnarray} K(\beta ) &=&\Big( \int_{0}^{\sqrt{3}}+\int_{\sqrt{3}}^{\beta }\Big) \frac{du}{\sqrt{f(u)-f(\beta )}}\nonumber \\ &=&\int_{0}^{\sqrt{3}}\frac{du}{\sqrt{f(u)-f(\beta )}} +2\int_{\sqrt{3}}^{\beta }\frac{d( \sqrt{f(u)-f(\beta )}) }{f'(u)} \nonumber\\ &=&\int_{0}^{\sqrt{3}}\frac{du}{\sqrt{f(u)-f(\beta )}} -\frac{2\sqrt{f(\sqrt{3})-f(\beta )}}{f'(\sqrt{3})}\\ &&+\int_{\sqrt{3}}^{\beta }2\sqrt{f(u)-f(\beta )} \frac{f''(u)}{[ f'(u)] ^2}du. \nonumber \end{eqnarray} and \begin{eqnarray} M(\beta ) &=&\Big( \int_{0}^{\sqrt{3}}+\int_{\sqrt{3}}^{\beta }\Big) \frac{du}{\sqrt{f(u)-f(2)}}\nonumber \\ &=&\int_{0}^{\sqrt{3}}\frac{du}{\sqrt{f(u)-f(2)}} +2\int_{\sqrt{3}}^{\beta }\frac{d( \sqrt{f(u)-f(2)}) }{f'(u)} \nonumber\\ &=&\int_{0}^{\sqrt{3}}\frac{du}{\sqrt{f(u)-f(2)}}+\frac{2\sqrt{f(\beta )-f(2) }}{f'(\beta )}-\frac{2\sqrt{2}}{f'(\sqrt{3})} \\ &&+2\int_{\sqrt{3}}^{\beta }\sqrt{f(u)-f(2)}\frac{f''(u)}{ \left[ f'(u)\right] ^2}\,du, \nonumber \end{eqnarray} it follows that \begin{eqnarray} s(\beta ) &=&\int_{0}^{\sqrt{3}}\Big[ \frac{1}{\sqrt{f(u)-f(2)}}-\frac{1}{ \sqrt{f(u)-f(\beta )}}\Big] du \nonumber \\ &&-\frac{2\sqrt{-f(\beta )}}{f'(\sqrt{3})}-\Big[ \frac{2\sqrt{ f(\beta )-f(2)}}{f'(\beta )}-\frac{2\sqrt{2}}{f'(\sqrt{3})} \Big] \label{ex:s(b)} \\ &&+2\int_{\sqrt{3}}^{\beta }\left[ \sqrt{f(u)-f(\beta )}-\sqrt{f(u)-f(2)} \right] \frac{f''(u)}{[ f'(u)] ^2}\,du. \nonumber \end{eqnarray} The last integral in (\ref{ex:s(b)}) can be written as $$p(\beta )=\int_{\sqrt{3}}^{\beta }\frac{(2-\beta )(1+\beta )^2}{\big[ \sqrt{f(u)-f(\beta )}+\sqrt{f(u)-f(2)}\big] }\frac{(-6u)}{9(1-u^2)^2}\,du.$$ It then follows that$p(\beta )=O(\sqrt{2-\beta })$and$s(\beta )\to 0$as$\beta \to 2$. From$M(\beta )=-\frac{1}{\sqrt{3} }\big[ \ln \frac{\sqrt{3}+\sqrt{2-\beta }}{\sqrt{3}-\sqrt{2-\beta }}-\ln \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\big] $, we see that$\lim_{\beta \to 2}K(\beta ) =\lim_{\beta \to 2}M(\beta ) =\frac{2}{\sqrt{3}}\ln (\sqrt{3}+\sqrt{2})$. Since, as$\beta \to 2$, $\frac{\sqrt{2}(1-x_{0})}{\sqrt{3}\epsilon }=K(\beta )\to \frac{2}{ \sqrt{3}}\ln (\sqrt{3}+\sqrt{2}),$ or $\lim_{\epsilon \to 0}\Big\{ \frac{\sqrt{2}(1-x_{0})}{\sqrt{3}% \epsilon }-\frac{2}{\sqrt{3}}\ln (\sqrt{3}+\sqrt{2})\Big\} =0,$ it turns out that, as$\epsilon \to 0$, $1-x_{0}\sim \epsilon \sqrt{2}\ln (\sqrt{3}+\sqrt{2}),$ which is the conclusion of Lemma \ref{lm5}. \hfill$\Box$We are now ready to study the asymptotic behavior of the solution at the boundary layers.\qquad \section{Asymptotic analysis for$x\leq x_{0}$} Let$x0$for$u\in \lbrack \alpha ,u_{1}]$with some constant$u_{1}\in (\alpha ,0)$. Noting that$S(u)$is independent of$% \alpha$, we see that$u_{1}$is independent of$\alpha$. Thus,$S'(u)>0$for$u\in [\alpha ,u_{1}]$, and hence,$S(u)\geq S(\alpha )$. It follows that $$\ 0\leq \frac{J\breve{(}u,\alpha )}{L(u)}-1\leq \frac{\sqrt{3}|u|}{L(\alpha )}\to 0$$ uniformly for$u\in \lbrack \alpha ,u_{1}]$because$L(\alpha )\sim \frac{1}{% \sqrt{3}}\ln \frac{\sqrt{3}+\sqrt{2-\alpha }}{\sqrt{3}-\sqrt{2-\alpha }}$. For any fixed$u_{0}\in (u_{1},0)$, the convergence of$\frac{J(u,\alpha )}{% L(u)}\to 1$is uniform on$[u_{1},u_{0}]$. This can be obtained by the uniform convergence of$J(u,\alpha )\to L(u)$on the interval because$f(u)-f(\alpha )$uniformly converges to$f(u)-f(-1)$for$u\in [u_{1},u_{0}]$as$\alpha \to -1$. Next, we will prove that the convergence is also uniform on$[u_{0},0]$by choosing a$u_{0}$independent of$\alpha$. First, we see that$L(u)\sim - \frac{u}{\sqrt{2}}$because$-\frac{u}{\sqrt{2}}$is the linear part of$L(u)$for sufficiently small$|u|$. We then consider the function$ Q(u,\alpha )=\frac{J(u,\alpha )}{L(u)}$for$u<0$where$|u|$is sufficiently small, noting that$Q(0,\alpha )$is defined by an application of L'Hospital's rule. From $$\frac{\partial Q}{\partial \alpha }=\frac{\left( f(u)-f(\alpha )\right) ^{-3/2}f'(\alpha )}{L(u)},$$ we find$\frac{\partial Q}{\partial \alpha }>0$for$u<0$and for$|\alpha |<1$. This shows that$Q(u,\alpha )0$is sufficiently small and independent of$\alpha$, to prove the uniform convergence of$K(u,\alpha )\sim M(u)$where the function \begin{eqnarray} M(u) &=&\int_{0}^{u}\frac{du}{\sqrt{f(u)-f(2)}} \nonumber \\ &=&\frac{1}{\sqrt{3}}\ln \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{1% }{\sqrt{3}}\ln \frac{\sqrt{3}+\sqrt{2-u}}{\sqrt{3}-\sqrt{2-u}}. \end{eqnarray} Then for$u\in \lbrack u_{2},2-\delta ]$, where$\delta $is a sufficiently small positive number, the convergence$\int_{0}^{u}\frac{du}{\sqrt{f(u)-f(\alpha )}}\to M(u) $is also uniform because of the uniform convergence of$f(u)-f(\beta )$to$% f(u)-f(2)$. This means that$K(u,\beta )\sim M(u)$holds for$u\in \lbrack u_{2},2-\delta ]$. Finally, we prove that the uniform convergence holds on the interval$ [2-\delta ,\beta ]$. Following the computation in the proof of lemma \ref{lm5}, by integration by parts, we get \begin{eqnarray} K(u,\beta ) &=&\int_{0}^{\sqrt{3}}\frac{du}{\sqrt{f(u)-f(\beta )}}+\frac{2 \sqrt{f(u)-f(\beta )}}{f'(u)}-\frac{2\sqrt{f(\sqrt{3})-f(\beta )}}{ f'(\sqrt{3})} \nonumber \\ && +\int_{\sqrt{3}}^{u}2\sqrt{f(u)-f(\beta )}\frac{f''(u)}{% \left[ f'(u)\right] ^2}\,du. \end{eqnarray} and \begin{eqnarray} M(\beta ) &=&\newline \int_{0}^{\sqrt{3}}\frac{du}{\sqrt{f(u)-f(2)}}+\frac{2\sqrt{f(u)+2}}{ f'(u)}-\frac{2\sqrt{2}}{f'(\sqrt{3})} \nonumber \\ &&+2\int_{\sqrt{3}}^{u}\sqrt{f(u)-f(2)}\frac{f''(u)}{\left[ f'(u)\right] ^2}\,du. \end{eqnarray} Let$s(u)=K(u,\beta )-M(u)$. Then \begin{eqnarray} s(u) &=&\int_{0}^{\sqrt{3}}\Big[ \frac{1}{\sqrt{f(u)-f(2)}}-\frac{1}{\sqrt{% f(u)-f(\beta )}}\Big] du \nonumber \\ &&+\frac{2\sqrt{f(u)-f(\beta )}}{f'(u)}-\frac{2\sqrt{-f(\beta )}}{% f'(\sqrt{3})}-\Big[ \frac{2\sqrt{f(u)-f(2)}}{f'(u)}-\frac{% 2\sqrt{2}}{f'(\sqrt{3})}\Big] \nonumber \\ &&+2\int_{\sqrt{3}}^{u}\left[ \sqrt{f(u)-f(\beta )}-\sqrt{f(u)-f(2)}\right] \frac{f''(u)}{\left[ f'(u)\right] ^2}du. \label{ex:s(sb)} \end{eqnarray} The last integral in (\ref{ex:s(sb)}) can be written as $$p(u)=\int_{\sqrt{3}}^{u}\frac{(2-\beta )(1+\beta )^2}{\left[ \sqrt{f(u)-f(\beta )}+\sqrt{f(u)-f(2)}\right] }\frac{-6u}{9(1-u^2)^2}\,du.$$ It is observed that$p(u)=O(\sqrt{2-\beta })$. It follows that$% s(u)\to 0$uniformly for$u\in \lbrack 2-\delta ,b]$, as$\beta \to 2$. Therefore,$K(u,\alpha )\sim M(u)$uniformly for$u\in \lbrack 2-\delta ,\beta ]$. We now can conclude that $$\frac{1}{\sqrt{3}}\ln \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{1}{% \sqrt{3}}\ln \frac{\sqrt{3}+\sqrt{2-u}}{\sqrt{3}-\sqrt{2-u}}\sim \sqrt{\frac{% 2}{3}}\frac{x-x_{0}}{\epsilon }$$ which implies that $$u\sim -1+3 \mathop{\rm sech}{}^2\big( \frac{x_{0}-x}{\epsilon \sqrt{2}}+2\ln (\sqrt{3}+% \sqrt{2})\big) \label{ex:u:[0,1]}$$ uniformly for$x\in \lbrack x_{0},1]$Summing up, we have proved that, as$\epsilon \to 0$, the asymptotic formula $$\int_{\alpha }^{u}\frac{du}{\sqrt{f(u)-f(\alpha )}}\sim \int_{\alpha }^{u}% \frac{du}{\sqrt{f(u)-f(-1)}}$$ uniformly holds for all$u\in \lbrack \alpha ,\beta ]$and the asymptotic expression (\ref{ex:u:[0,1]}) holds uniformly for$x\in [0,1]$. One can then easily get the asymptotic solution for the homogeneous problem. Set$\Delta =1-x_{0}$. Applying the symmetry, by Lemma \ref{lm2}, to expand the solution, we can obtain the single spike solution defined on$[-1+\Delta ,1-\Delta ]$. Then, using the translation$x=y+\Delta $, one can get the solution of (\ref{eq:cp} )-(\ref{bd:+-1}) with the single spike. The interval$[-1+\Delta ,1-\Delta ]$is translated to$[-1,1]$, and$[0,1]$to$[-\Delta ,1-\Delta ]. $The uniform convergence proved above implies that the single spike solution$u(x,\epsilon )$of (\ref{eq:cp})-(\ref {bd:+-1}) satisfies $$u(x,\epsilon )\sim -1+3 \mathop{\rm sech}{}^2 \big( \frac{1-2\Delta -x}{\epsilon \sqrt{2}% }+2\ln (\sqrt{3}+\sqrt{2})\big) \label{ex:unif}$$ uniformly on$[-\Delta ,1-\Delta ]$. For$x\in \lbrack -1,-\Delta ]$, we use the symmetry about$x=-\Delta $, i.e.,$u(x)=u(-x-2\Delta )$, replacing$x$by$-x-2\Delta $in (\ref {ex:unif}) to get $$u\sim -1+3 \mathop{\rm sech}{}^2\big( \frac{1+x}{\epsilon \sqrt{2}}+2\ln (\sqrt{3}+% \sqrt{2})\big).$$ For$x\in \lbrack 1-\Delta ,1]$, we use the symmetry about$x=1-\Delta $,$% u(x)=u(1-\Delta -(x-1+\Delta ))=u(2-2\Delta -x)$, replacing$x=2-2\Delta -x$in (\ref{ex:unif}), to get $$u\sim -1+3 \mathop{\rm sech}{}^2\big( \frac{x-1}{\epsilon \sqrt{2}}+2\ln (\sqrt{3}+% \sqrt{2})\big),$$ Summing up, we have proved the main result of this paper: \begin{theorem} \label{thm1} The single spike solution$u(x,\epsilon )$of (\ref{eq:cp})-(\ref {bd:+-1}) satisfies $$u(x,\epsilon )\sim -1+3 \mathop{\rm sech}{}^2\big( \frac{1-2\Delta -x}{\epsilon \sqrt{2}% }+2\ln (\sqrt{3}+\sqrt{2})\big)$$ uniformly on$[-\Delta ,1-\Delta ]$, $$u\sim -1+3 \mathop{\rm sech}{}^2\big( \frac{x-1}{\epsilon \sqrt{2}}+2\ln (\sqrt{3}+ \sqrt{2})\big),$$ uniformly on$[1-\Delta, 1]$, and $$u\sim -1+3 \mathop{\rm sech}{}^2\big( \frac{1+x}{\epsilon \sqrt{2}}+2\ln (\sqrt{3}+% \sqrt{2})\big)$$ uniformly on$[-1, -\Delta]$, where$\Delta \sim \epsilon \sqrt{2}\ln (\sqrt{3}+\sqrt{2})$. \end{theorem} Applying this theorem, one can easily get the asymptotic solutions for the entire interval [-1,1]. This result supports the formal asymptotic work of \cite{Lange} and \cite{MacG}. \section{Nonhomogeneous problem} For the nonhomogeneous problem, equation (\ref{eq:cp}) with $$u(-1)=a,\quad u(1) =b \label{bd:nonhom}$$ where not both$a$and$b$are zero, and$-1