% include figures \documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{graphics} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 214, pp. 1--13.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/214\hfil Direction of bifurcation] {Direction of bifurcation for some non-autonomous problems} \author[P. Korman \hfil EJDE-2012/214\hfilneg] {Philip Korman} % in alphabetical order \address{Philip Korman \newline Department of Mathematical Sciences \\ University of Cincinnati \\ Cincinnati, Ohio 45221-0025, USA} \email{kormanp@math.uc.edu} \thanks{Submitted June 28, 2012. Published November 27, 2012.} \subjclass[2000]{34B15, 65L10, 80A25} \keywords{Global solution curves; direction of bifurcation; \hfill\break\indent continuation in a global parameter} \begin{abstract} We study the exact multiplicity of positive solutions, and the global solution structure for several classes of non-autonomous two-point problems. We present two situations where the direction of turn can be computed rather directly. As an application, we consider a problem from combustion theory with a sign-changing potential. We illustrate our results by numerical computations, using a novel method. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} In recent years bifurcation theory methods were applied to study the exact multiplicity of positive solutions, and the global solution structure of non-autonomous two-point problems \label{i1} u''+\lambda f(x,u)=0, \quad a0$($I<0$) the direction of the turn is to the left (right) in the$(\lambda ,u)$plane". If one can show that a turn to the left occurs at any critical point, then there is at most one critical point. Usually, there is exactly one critical point, which provides us with the exact shape of solution curve, and the exact multiplicity count for solutions. \smallskip In the present paper we present two situations in which the sign of$I$can be computed in a rather direct way, differently from the previous works. As an application, we obtain three new exact multiplicity results for non-autonomous equations, including one for sign-changing equations of combustion theory. Sign-changing equations present several new challenges, which we overcome in case of symmetric potentials. We also present some improvements of earlier results. In the last section we develop an algorithm for the numerical computation of global solution curves for non-autonomous equations. This is accomplished by continuation in a global parameter. \section{The direction of bifurcation } We consider positive solutions of a two point non-autonomous boundary value problem \label{1} u''+\lambda f(x,u)=0, \quad a0$ on $(a,b)$. If we have, for some $c>0$, $$\label{3} u^2f_{uu}(x,u) \geq c \left( uf_{u}(x,u)-f(x,u) \right), \quad \text{for all u>0, and x \in (a,b)}$$ then $$\label{3.1} \int_a^b f_{uu}(x,u)w^3 \, dx>0 \,.$$ If, on the other hand, for some $c>0$, $$\label{4} u^2f_{uu}(x,u) \leq -c \left( uf_{u}(x,u)-f(x,u) \right), \quad \text{for all u>0, and x \in (a,b)}$$ then $$\label{4.1} \int_a^b f_{uu}(x,u)w^3 \, dx<0 \,.$$ \end{lemma} \begin{proof} We multiply the equation \eqref{2} by $\frac{w^2}{u}$, and subtract from that the equation \eqref{1} multiplied by $\frac{w^3}{u^2}$, then integrate $\lambda \int_a^b \Big[\frac{f_u(x,u)}{u}-\frac{f(x,u)}{u^2} \Big] w^3 \, dx =\int_a^b \Big(\frac{w^3}{u^2} u''-\frac{w^2}{u} w'' \Big) \, dx \,.$ In the last integral we integrate by parts. The boundary terms vanish, since $u'(a)$ and $u'(b)$ are not zero by Hopf's boundary lemma, and hence $u(x)$ is asymptotically linear near the end points. We have \begin{align*} \int_a^b \left(\frac{w^3}{u^2} u''-\frac{w^2}{u} w'' \right) \, dx &=\int_a^b \frac{2 w^3 u{u'}^2-4w^2w'u^2u'+2w{w'}^2u^2}{u^4} \, dx \\ & =\int_a^b \frac{2w (wu'-uw')^2}{u^3} \, dx >0 \,. \end{align*} If condition \eqref{3} holds, then $\int_a^b f_{uu}(x,u)w^3 \, dx \geq c \int_a^b \Big[\frac{f_u(x,u)}{u}-\frac{f(x,u)}{u^2} \Big] w^3 \, dx > 0 \,.$ Similarly, the condition \eqref{4} implies \begin{equation*} -\int_a^b f_{uu}(x,u)w^3 \, dx \geq c \int_a^b \Big[\frac{f_u(x,u)}{u}-\frac{f(x,u)}{u^2} \Big] w^3 \, dx > 0 \,. \qedhere \end{equation*} \end{proof} \noindent \textbf{Remarks} \begin{enumerate} \item After this paper was written, we became aware that a similar result was proved in Shi \cite{S}. \item The condition \eqref{3}, when $c=1$, is equivalent to $\Big[u \Big( \frac{f(u)}{u} \Big)' \Big]' = \Big(f'(u)-\frac{f(u)}{u} \Big)'>0 \,.$ In Ouyang and Shi \cite{OS} it has been pointed out that the turning direction is sometimes related to monotonicity of the function $f(u)/u$. This form of \eqref{3} again shows a connection to the function $f(u)/u$. \end{enumerate} \medskip \noindent \textbf{Example} $f(u)=a+u^p+u^q$, with a constant $a \geq 0$. One computes, with $c=1$, $u^2f''(u)-uf'(u)+f(u)=(p-1)^2 u^p+(q-1)^2 u^q+a>0 \quad \text{for all u>0} \,.$ The case when $00$. Similarly, the condition \eqref{3} holds for $f(x,u)=\Sigma _{i=1}^m a_i(x) u^{p_i}$, with $a_i(x)>0$ for all $x$, and any positive $p_i$. While in the case of constant $a_i(x)$, the direction of bifurcation was known before (see the Theorem \ref{thm:1+} below), the non-autonomous case is new. \section{A class of symmetric nonlinearities} We study positive solutions of a class of symmetric problems of the type \label{s1} u'' + \lambda f(x,u)=0 \quad \text{for $-10$}, \\ \label{s3} f_x(x,u) \leq 0 \quad \text{for all $00$}. \end{gather} Under the above conditions the following facts, similar to those for autonomous problems, have been established. \smallskip \noindent \textbf{1.} Any positive solution of \eqref{s1} is an even function, with $u'(x)<0$ for all $x \in (0,1]$, so that $x=0$ is a point of global maximum. This follows from B. Gidas, W.-M. Ni and L. Nirenberg \cite{GNN}. \smallskip \noindent \textbf{2.} Assume, additionally, that $f(x,u)>0$. Then the maximum value of solution, $\alpha=u(0)$, uniquely identifies the solution pair $(\lambda, u(x))$, as proved in Korman and Shi \cite{KS}. i.e., $\alpha=u(0)$ gives a global parameter on any solution curve. We shall generalize this result below, dropping the condition that $f(x,u)>0$. \smallskip \noindent \textbf{3.} Any non-trivial solution of the corresponding linearized problem \label{s4} w'' + \lambda f_u(x,u)w=0 \quad \text{for $-10, \quad a_i(-x)=a_i(x) \quad \text{for$x \in (-1,1)$}, \quad i=1,2\,,\\ \label{s7} a'_i(x)<0 \quad \text{for$x \in (0,1)$}, \quad i=1,2 \,. \end{gather} Then there is a critical$\lambda_0 > 0$, such that for$\lambda > \lambda_0$the problem \eqref{s5} has no positive solutions, it has exactly one positive solution for$\lambda = \lambda_0$, and exactly two positive solutions for$0<\lambda < \lambda_0$. Moreover, all positive solutions lie on a single smooth solution curve$u(x,\lambda)$, which for$0<\lambda < \lambda_0$has two branches denoted by$0 < u^-(x,\lambda) < u^+(x,\lambda)$, with$u^-(x,0)=0$,$u^-(x,\lambda)$strictly monotone increasing in$\lambda$for all$x \in (-1,1)$, and$\lim_{\lambda\to 0} u^+(0,\lambda) = \infty$. The maximal value of solution,$u(0,\lambda)$, serves as a global parameter on this solution curve. \end{theorem} \begin{proof} Conditions \eqref{s6} and \eqref{s7} imply that the above mentioned results on symmetric problems apply, and in particular any non-trivial solution of the linearized problem, corresponding to \eqref{s5} is positive on$(-1,1)$. Then by Lemma \ref{lma:1} only turns to the left are possible on the solution curve. The rest of the proof is similar to that for similar results in Korman and Ouyang \cite{KO1}, or Korman and Shi \cite{KS}, so we just sketch it. By the Implicit Function Theorem, there is curve of positive solutions of \eqref{s5} starting at$(\lambda=0,u=0)$. By Sturm's comparison theorem, this curve cannot be continued indefinitely in$\lambda$, so that it will have to reach a critical point$(\lambda _0,u_0)$at which the Crandall-Rabinowitz Theorem \cite{CR} applies. By Lemma \ref{lma:1}, a turn to the left occurs at$(\lambda _0,u_0)$, and at any other critical point. Hence, there are no other turning points, and the solution curve continues for all decreasing$\lambda >0$, tending to infinity as$\lambda \to 0+$. \end{proof} \noindent \textbf{Remark} This theorem also holds for more general$f(u)=\Sigma _{i=1}^m a_i(x) u^{p_i}$, with$a_i(x)$satisfying \eqref{s6} and \eqref{s7}, and$p_i \geq 0$, with at least one of$p_i$less than one, and at least one of$p_i$greater than one. \section{Non-symmetric nonlinearities} Without the symmetry assumptions on$f(x,u)$, the problem is much harder. We restrict to a subclass of such problems; i.e., we now consider positive solutions of the boundary value problem \label{g1} u'' + \lambda \alpha (x) f(u)=0 \quad \text{for$a 0 \quad \text{for $u>0$}, \quad \alpha (x)>0 \quad \text{for $x \in [a,b]$}. As before, it is crucial for bifurcation analysis to prove positivity for the corresponding linearized problem \label{g4} w'' + \lambda \alpha (x) f'(u)w=0 \quad \text{for $a0$ on $(a,b)$. \end{lemma} Using Lemma \ref{lma:1}, we have the following exact multiplicity result, whose proof is similar to that of Theorem \ref{thm:1}. \begin{theorem}\label{thm:2} Consider the problem \label{g6} \quad u'' + \lambda \alpha (x) \Sigma _{i=1}^m a_i u^{p_i}=0 \quad \text{for $a 0$, such that for $\lambda > \lambda_0$ the problem \eqref{g6} has no positive solutions, it has exactly one positive solution for $\lambda = \lambda_0$, and exactly two positive solutions for $0<\lambda < \lambda_0$. Moreover, all positive solutions lie on a single smooth solution curve $u(x,\lambda)$, which for $0<\lambda < \lambda_0$ has two branches denoted by $0 < u^-(x,\lambda) 0$; i.e., $u(0)$ is the global maximum of solution $u(x)$. \begin{theorem}\label{thm:100} Consider the problem \eqref{s1}, with $f(x,u)$ satisfying \eqref{s2} and \eqref{s3}, with the inequality \eqref{s3} being strict for almost all $x \in (-1,1)$ and $u>0$. Then the set of positive solutions of \eqref{s1} can be globally parameterized by the maximum values $u(0)$. (I.e., the value of $u(0)$ uniquely determines the solution pair $(\lambda,u(x))$.) \end{theorem} \begin{proof} Assume, on the contrary, that $v(x)$ is another solution of \eqref{s1}, with $v(0)=u(0)$, and $v'(0)=u'(0)=0$. We may assume that $\mu >\lambda$. Setting $x=\frac{1}{\sqrt{\lambda}} t$, we see that $u=u(t)$ satisfies $$\label{s9} u''+f(\frac{1}{\sqrt{\lambda}} t,u)=0, \quad u'(0)=u(\sqrt{\lambda})=0 \,.$$ Similarly, letting $x=\frac{1}{\sqrt{\mu}} z$, and then renaming $z$ by $t$, we see that $v=v(t)$ satisfies $v''+f(\frac{1}{\sqrt{\mu}} t,v)=0, \quad v'(0)=v(\sqrt{\mu})=0 \,,$ and in view of \eqref{s3}, $$\label{s10} v''+f(\frac{1}{\sqrt{\lambda}} t,v) < 0 \,;$$ i.e., $v(t)$ is a supersolution of \eqref{s9}. We may assume that \label{s11} v(t)0$and small} \,. Indeed, the opposite inequality is impossible by the strong maximum principle (a supersolution cannot touch a solution from above, and the possibility of infinitely many points of intersection of$u(t)$and$v(t)$near$t=0$is ruled out by the Sturm comparison theorem, applied to$w=u-v$). Since$v(t)$is positive on$(0,\sqrt{\lambda})$, we can find a point$\xi \in (0,\sqrt{\lambda})$so that$u(\xi)=v(\xi)$,$|u'(\xi)| \geq |v'(\xi)|$, and \eqref{s11} holding on$(0,\xi)$. We now multiply the equation \eqref{s9} by$u'(t)$, and integrate over$(0,\xi)$. Since the function$u(t)$is decreasing, its inverse function exists. Denoting by$t=t_2(u)$, the inverse function of$u(t)$on$(0,\xi)$, we have $$\label{s12} \frac{1}{2} {u'}^2(\xi) +\int_{u(0)}^{u(\xi)} f(\frac{1}{\sqrt{\lambda}} t_2(u),u) \, du=0.$$ Similarly denoting by$t=t_1(u)$the inverse function of$v(t)$on$(0,\xi)$, we have multiplying \eqref{s10} by$v'(t)<0$, and integrating $$\label{s14} \frac{1}{2} {v'}^2(\xi) +\int_{u(0)}^{u(\xi)} f(\frac{1}{\sqrt{\lambda}} t_1(u),u) \, du > 0.$$ Subtracting \eqref{s14} from \eqref{s12}, we have $\frac{1}{2}\Big[ {u'}^2(\xi)-{v'}^2(\xi)\Big] +\int_{u(\xi)}^{u(0)} \Big[f(\frac{1}{\sqrt{\lambda}} t_1(u),u) -f(\frac{1}{\sqrt{\lambda}} t_2(u),u) \Big] \, du < 0.$ Notice that$t_2(u)>t_1(u)$for all$u \in (u(\xi), u(0))$. Using the condition \eqref{s3}, both terms on the left are non-negative and the integral is positive, and we obtain a contradiction. \end{proof} We now consider a boundary value problem arising in combustion theory, see e.g., Bebernes and Eberly \cite{B}, Wang \cite{W30}, and Wang and Lee \cite{W2} \label{c1} u''+\lambda \alpha (x) e^u=0, \quad -10$ (the result of this section is known if $\alpha (x)$ is positive on $(-1,1)$). The linearized problem corresponding to \eqref{c1} is \label{c2} w''+\lambda \alpha (x)e^uw=0, \quad -10$on$(-1,1)$, and$\alpha (x) \in C[-1,1]$. Then $$\label{c3} \int_{-1}^1 \alpha (x)e^uw^3 \, dx>0, \quad \text{and} \quad \int_{-1}^1 \alpha (x)e^uw \, dx>0 \,.$$ \end{lemma} \begin{proof} Multiplying equation \eqref{c2} by$w^2$, and integrating $\int_{-1}^1 \alpha (x)e^uw^3 \, dx=2 \int_{-1}^1 w{w'}^2 \, dx>0 \,.$ Integrating \eqref{c2} $\int_{-1}^1 \alpha (x)e^uw \, dx=w'(-1)-w'(1)>0 \,,$ as claimed. \end{proof} We need the following simple lemma. \begin{lemma}\label{lma:6} Let$u(x)$be a solution of the problem \label{c4} -u''=\alpha (x), \quad \text{$00 \quad \text{for $x \in [0,1)$} \,, \\ \label{c6} \int _0^1 \alpha (\xi) \, d \xi >0 \,. \end{gather} Then $u(x)>0$ on $[0,1)$, and $u'(1)<0$. \end{lemma} \begin{proof} The formulas \eqref{c5} and \eqref{c6} give $u(x)$ and $-u'(1)$ respectively. \end{proof} We shall assume that $\alpha (x) \in C^1[-1,1]$ satisfies \begin{gather} \label{c6a} \alpha (-x)=\alpha (x) \quad \text{for all $x \in [0,1]$} \,, \\ \label{c6b} \alpha '(x)<0 \quad \text{for almost all $x \in [0,1]$} \,. \end{gather} As before, we know that under these conditions any non-trivial solution of \eqref{c2} is positive on $(-1,1)$. Observe also that under these conditions a sign-changing $\alpha (x)$ changes sign exactly once on $(0,1)$ (and on $(-1,0)$). \begin{theorem}\label{thm:3} Consider the problem \eqref{c1}, and assume that the function $\alpha (x)$ satisfies \eqref{c6a} and \eqref{c6b}. We also assume that $\alpha (x)$ is sign-changing, and it satisfies \eqref{c5} and \eqref{c6}. Then there is a critical $\lambda_0 > 0$, such that for $\lambda > \lambda_0$ the problem \eqref{c1} has no positive solutions, it has exactly one positive solution for $\lambda = \lambda_0$, and exactly two positive solutions for $0<\lambda < \lambda_0$. Moreover, all positive solutions lie on a single smooth solution curve $u(x,\lambda)$, which for $0<\lambda < \lambda_0$ has two branches denoted by $0 < u^-(x,\lambda) < u^+(x,\lambda)$, with $u^-(x,0)=0$, and $\lim_{\lambda\to\ 0} u^+(0,\lambda) = \infty$. The maximal value of solution, $u(0,\lambda)$, serves as a global parameter on this solution curve. \end{theorem} \begin{proof} We begin with the solution $(\lambda=0,u=0)$. This solution is non-singular (the corresponding linearized problem \eqref{c2} has only the trivial solution), so that by the Implicit Function Theorem we have a curve of solutions $u(x,\lambda)$ passing through the point $(\lambda=0,u=0)$. We claim that these solutions are positive for small $\lambda$. Indeed, $u_{\lambda}(x,0) \equiv u_{\lambda}$ satisfies $-u''_{\lambda}=\alpha (x), \quad \text{-10 on [0,\xi) and \alpha (x)<0 on (\xi,1). Denote u(\xi)=u_0. Then from the equation \eqref{c1} \[ -u''> \lambda e^{u_0} \alpha (x), \quad \text{0\lambda e^{u_0} A(x) for all x \in (0,1), and then u'(1) \leq \lambda e^{u_0}A'(1)<0. We claim next that the curve of positive solutions cannot be continued in \lambda beyond a certain point. With \xi denoting the root of \alpha (x), as above, fix any \eta \in (0,\xi), and denote a_0=\alpha (\eta)>0. If we denote \varphi (x) = \cos \frac{\pi}{2 \eta} x and \lambda _1 =\frac{\pi ^2}{4 \eta ^2}, then \[ \varphi ''+ \lambda _1 \varphi=0, \quad \text{on (0,\eta)}, \quad \varphi'(0)= \varphi (\eta)=0 \,.$ We have $\int_0^{\eta} u'' \varphi \, dx=\int_0^{\eta} u \varphi '' \, dx-u(\eta) \varphi ' (\eta) > \int_0^{\eta} u \varphi '' \, dx=-\lambda _1 \int_0^{\eta} u \varphi \, dx \,.$ Then multiplying the equation \eqref{c1} by $\varphi$ and integrating, we have $\lambda \int_0^{\eta} \alpha (x) e^u \varphi \, dx< \lambda _1 \int_0^{\eta} u \varphi \, dx \,.$ Also $\int_0^{\eta} \alpha (x) e^u \varphi \, dx > a_0 \int_0^{\eta} u \varphi \, dx \,.$ We conclude that $\lambda <\frac{\lambda _1}{a_0} \,.$ The next step is to show that solutions of \eqref{c1} remain bounded, if $\lambda$ is bounded away from zero. Indeed, for large $u$, $\lambda \alpha (x)e^u>Mu$, with arbitrarily large constant $M$, when $x$ belongs to any sub-interval of $(-\xi, \xi)$. By Sturm's comparison theorem, the length of the interval on which $u(x)$ becomes large, must tend to zero. But that is impossible, because $u(x)$ is concave on $(-\xi, \xi)$. We now return to the curve of positive solutions, emanating from $(\lambda=0,u=0)$. Solutions on this curve are bounded, while the curve cannot be continued indefinitely in $\lambda$. Hence, a critical point $(\lambda _0,u_0)$ must be reached on this curve; i.e., at $(\lambda _0,u_0)$ the corresponding linearized problem \eqref{c2} has a non-trivial solution. By the results on symmetric problems, reviewed above, any non-trivial solution of the corresponding linearized problem \eqref{c2} is of one sign; i.e., we may assume that $w(x)>0$ on $(-1,1)$. By Lemma \ref{lma:5}, the Crandall-Rabinowitz theorem applies at any critical point, and a turn to the left occurs. Hence, after the turn at $(\lambda _0,u_0)$, the curve continues for decreasing $\lambda$, without any more turns. By the Theorem \ref{thm:100}, $u(0,\lambda)$ is a global parameter on the solution curve. Along the solution curve, the global parameter $u(0,\lambda)$ is increasing and tending to infinity. By above, this may happen only as $\lambda \to 0$. \end{proof} \noindent \textbf{Example} The function $\alpha (x)=1-c x^2$ for $10 \,, \] and $A(x)=\frac{cx^4}{12}-\frac{x^2}{2}-\frac{c}{12}+\frac{1}{2}>0, \quad \text{for 00 on (0,u_0), and f''(u)<0 on (u_0, \infty) for some u_0>0, and the definition of concave-convex functions is similar. \begin{theorem}[\cite{KLO,OS}]\label{thm:1+} \textbf{(i)} Assume f(0) \leq 0, and f(u) is convex-concave. Then at any critical point, with u_0(0)>u_0 and u'_0(1)<0, we have I<0, and hence a turn to the right occurs. \\ \textbf{(ii)} Assume f(0) \geq 0, and f(u) is concave-convex. Then at any critical point, with u_0(0)>u_0, we have I>0, and hence a turn to the left occurs. \end{theorem} \begin{proof} In case f(0) \geq 0, we have u'_0(1)<0 by Hopf's boundary lemma. We shall write (\lambda,u) instead of (\lambda _0,u_0). It is known that any non-trivial solution of the linearized problem \eqref{a3} is of one sign, see e.g., \cite{KLO} or \cite{K}, and so we may assume that w(x)>0 on (-1,1), which implies that $$\label{a8} w'(1)<0 \,.$$ From the equations \eqref{a2} and \eqref{a3} it is straightforward to verify the following identities \begin{gather} \label{a9} u'(x)w'(x)-u''(x)w(x)=constant=u'(1)w'(1) \,; \\ \label{a10} \left( u''w'-u'w'' \right)'=\lambda f''(u){u'}^2w \,. \end{gather} Assume that the first set of conditions hold. Integrating \eqref{a10}, $$\label{a11} \lambda \int _0^1 f''(u){u'}^2w \, dx=u''(1)w'(1)=-\lambda f(0)w'(1) \leq 0 \,.$$ Consider the function p(x) \equiv \frac{w(x)}{-u'(x)}. Since p(1)=0, and by \eqref{a9} \[ p'(x)=-\frac{u'(1)w'(1)}{{u'(x)}^2}<0 \,,$ the function$p(x)$is positive and decreasing on$(0,1)$. The same is true for$p^2(x)=\frac{w^2(x)}{{u'}^2(x)}$. Let$x_0$be the point where$f''(u(x))$changes sign (i.e.,$f''(u(x))<0$on$(0,x_0)$, and$f''(u(x))>0$on$(x_0,1)$). By scaling$w(x)$, we may achieve that$w^2(x_0)={u'}^2(x_0)$. Then$w^2(x)>{u'}^2(x)$on$(0,x_0)$, and the inequality is reversed on$(x_0,1)$. Using \eqref{a11}, we have $$\label{a12} \int _0^1 f''(u(x)) w^3 \, dx < \int _0^1 f''(u(x)) {u'}^2w \, dx \leq 0 \,.$$ Integrating \eqref{a9}, we have $$\label{a14} \int _0^1 f(u)w \, dx=\frac{1}{2 \lambda} u'(1)w'(1)>0 \,.$$ The formulas \eqref{a12} and \eqref{a14} imply that$I<0$. The second part of the theorem is proved similarly. \end{proof} Observe that, in case$f=f(u)$, this theorem and the Lemma \ref{lma:1} have intersecting domains of applicability, but neither one is more general than the other. \section{Numerical computation of the solution curves } In this section we present computations of the global curves of positive solutions for the problem \label{n1} u'' + \lambda f(x,u)=0 \quad \text{for$-1