\documentclass[twoside]{article} \usepackage{texdraw} \pagestyle{myheadings} \markboth{ Instability and exact multiplicity } { Philip Korman \& Junping Shi } \begin{document} \setcounter{page}{311} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent Nonlinear Differential Equations, \newline Electron. J. Diff. Eqns., Conf. 05, 2000, pp. 311--322\newline http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Instability and exact multiplicity of solutions of semilinear equations % \thanks{ {\em Mathematics Subject Classifications:} 34B15. \hfil\break\indent {\em Key words:} Bifurcation of solutions, global solution curve. \hfil\break\indent \copyright 2000 Southwest Texas State University. \hfil\break\indent Published October 31, 2000. \hfil\break\indent Partially supported the Taft Faculty Grant at the University of Cincinnati } } \date{} \author{ Philip Korman \& Junping Shi \$12pt] {\em Dedicated to Alan Lazer} \\ {\em on his 60th birthday }} \maketitle \begin{abstract} For a class of two-point boundary-value problems we use bifurcation theory to show that a solution is unstable under a simple, geometric in nature, assumption on the non-linear term. As an application we obtain some new exact multiplicity results. \end{abstract} \newtheorem{thm}{Theorem} \newtheorem{lma}{Lemma} \section{Introduction} It is well known that for convex f(u), with f(0)>0, the set of all positive solutions for the boundary-value problem (depending on a parameter \lambda >0) \label{01} u'' + \lambda f(u)=0 \quad \mbox{for a0, with constants c_1, c_2>0, p>1, see e.g. \cite{KO1}. Both conditions restrict behavior of f(u) for large u>0. A condition not restricting behavior at infinity came up in the work on the S-shaped curves, see the references in \cite{KL}. Let F(u)=\int_0 ^u f(t) \, dt, h(u)=2F(u)-uf(u). The condition in question is that h(\alpha)<0 for some \alpha>0. One shows that the solution with maximal value equal to \alpha is unstable, from which one concludes that a turn must occur, since the solutions with small maximal value are stable. Previous proofs of this (in e.g. \cite{KL}) involved phase-plane analysis. In the present work we present a bifurcation theory proof of instability. One advantage of bifurcation theory is its flexibility. In Section 4 we extend our results to a class of non-autonomous problems, where phase-plane analysis is clearly not applicable. Inequality h(\alpha)<0 implies that the area under the graph of f(u) is smaller than that of the triangle with vertices (0,0), (0,u) and (u,f(u)). It means that f(u) is sufficiently convex" on (0,\alpha). As an application we obtain some new exact multiplicity results for both autonomous and non-autonomous problems. \section{Instability and multiplicity} Without loss of generality we may work on the interval (-1,1), and consider positive solutions of the boundary-value problem \label{1} u'' + \lambda f(u)=0 \quad \mbox{for -10 satisfies \label{2} w'' + \lambda f'(u)w+\mu w=0 \quad \mbox{for -10, and for some \alpha> \beta >0 we have: \begin{eqnarray} \label{4} & h'(u)\geq 0 \quad \mbox{for 0 0. It follows that h(u) has a unique critical point (a local maximum) on [0,\alpha], and it takes its positive maximum at u=\beta. (See Figure 1.) Define x_0\in (0,1) by u(x_0)=\beta. We then conclude \begin{eqnarray} \label{6} & f(u(x))-u(x)f'(u(x)) \leq 0 \quad \mbox{on (0,x_0)},& \\ & f(u(x))-u(x)f'(u(x)) \geq 0 \quad \mbox{on (x_0,1)}.& \nonumber \end{eqnarray} We also remark that by the condition (\ref{5}), $$\label{7} \int_0 ^1 \left[f(u)-uf'(u) \right]u'(x) \, dx=\int_0^1 \frac{d}{dx} h(u(x)) \, dx=-h(\alpha)\ge 0.$$ Assume now that u(x) is stable, i.e. \mu \geq 0 in (\ref{2}). Without loss of generality, we assume that w>0 in (-1,1). By the maximum principle, u'(1)<0, so near x=1 we have -u'(x)>w(x). Since -u'(0)=0, while w(0)>0, the functions w(x) and -u'(x) change their order at least once on (0,1). We claim that the functions w(x) and -u'(x) change their order exactly once on (0,1). % (We ignore the points where these functions merely touch".) Observe that -u'(x) satisfies $$\label{8} (-u')''+\lambda f'(u)(-u')=0 \quad \mbox{on (0,1)},$$ Let x_3 \in (0,1) be the largest point where w(x) and -u'(x) change the order. Assuming the claim to be false, let x_2, with 0-u' on (x_2,x_3), and the opposite inequality to the left of x_2. Since w(0)>-u'(0), there is another point x_10 and -u'(x)>0 on (0,1), we get a contradiction in (\ref{8.1}). Since the point of changing of order is unique, by scaling w(x) we can achieve \begin{eqnarray} \label{9} & -u'(x) \leq w(x) \quad \mbox{on (0,x_0)},& \\ & -u'(x) \geq w(x) \quad \mbox{on (x_0,1)}.& \nonumber \end{eqnarray} Using (\ref{6}), (\ref{9}), and also (\ref{7}), we have $$\label{10} \int_0 ^1 \left[f(u)-uf'(u) \right]w(x) \, dx <\int_0 ^1 \left[f(u)-uf'(u) \right](-u'(x)) \, dx\le 0.$$ On the other hand, multiplying the equation (\ref{2}) by u, the equation (\ref{1}) by w, subtracting and integrating over (0,1), we have \[ \int_0 ^1 \left[f(u)-uf'(u) \right]w(x) \, dx=\frac{\mu }{\lambda }\int_0 ^1 uw \, dx \geq 0,$ which contradicts (\ref{10}). So $\mu<0$. \hfill$\diamondsuit$ \paragraph{Remarks.} \begin{enumerate} \item Theorem \ref{thm:1} is stated in a way that we assume the existence of a solution with $u(0)=\alpha$. In fact, if $f(u)> 0$ for all $u\in [0,\alpha]$, then for any $d\in (0,\alpha]$, there exists a unique $\lambda(d)$ such that (\ref{1}) has a positive solution with $\lambda=\lambda(d)$ and $u(0)=d$, see e.g. \cite{KO1}. (See Figure 2.) \item It is easy to see that the condition (\ref{4}) holds if $$\label{10.1} f''(u)> 0 \quad \mbox{for 0 < u < \alpha},$$ and (\ref{5}) is also satisfied. So Theorem \ref{thm:1} is true if we replace (\ref{4}) by (\ref{10.1}). \item It is well-known that if for some $\beta>0$, $f(u)>0$ and $h'(u)\ge 0$ for $0\le u \le \beta$, then the solutions of (\ref{1}) with $u(0)=d$ and $0\alpha$, and $\lim_{u \to \infty} f(u)/u=0$. The proof used a bifurcation theoretic approach, except at one point, when the phase plane argument was used to show that when $u(0)=\alpha$ the solution curve travels to the left, i.e. $\lambda$ is decreasing when $u(0)$ is increasing (see formula (2.13) in \cite{KL}). Theorem 1 above provides an alternative proof of this fact. Indeed, the solution curve starts at $\lambda=0 ,u=0$, which is a stable solution (the principal eigenvalue of the corresponding linearized problem $=\pi^2/4$). As we increase $\lambda$, the solutions on the curve continue to be stable until a degenerate solution is reached. Since $f(u)$ is convex for $u<\alpha$, a standard bifurcation analysis shows that a turn to the left occurs at a degenerate solution, see e.g. \cite{KLO}. Hence the solution curve admits at most one turn for $u(0)<\alpha$, and since by Theorem 1 the solution at $u(0)=\alpha$ is unstable, this turn has already occurred, and so the solution curve travels to the left. The rest of the proof follows \cite{KL}. (See Figure 3.) \vspace{0.3in} \begin{texdraw} \arrowheadtype t:F \arrowheadsize l:0.08 w:0.04 \def\Rtext #1{\bsegment \textref h:L v:C \htext ( 0.08 0){#1} \esegment} \def\Ltext #1{\bsegment \textref h:R v:C \htext ( -0.08 0){#1} \esegment} \def\Ttext #1{\bsegment \textref h:C v:B \htext (0 0.06){#1} \esegment} \def\Btext #1{\bsegment \textref h:C v:T \htext (0 -0.06){#1} \esegment} \move (-1.3 0) \bsegment \move (0 0) \avec (1.8 0) \Rtext{$\lambda$} \move (0 0) \avec (0 1.1) \Ttext{$u(0)$} \move (0 0) \clvec (2.4 0.1)(-0.8 0.78)(1.8 1.0) \lpatt (0.033) \move (0 0.55) \Ltext{$\alpha$} \lvec (1.8 0.55) %\move (0 0.34) \Ltext{$\beta$}\lvec (1.8 0.34) \move (0.9 -0.1) \Btext{Figure 3: $S$-shaped curve} \esegment \move (1.2 0) \bsegment \move (0 0) \avec (1.8 0) \Rtext{$\lambda$} \move (0 0) \avec (0 1.1) \Ttext{$u(0)$} \move (0 0) \clvec (2 0.1)(0.3 0.3)(0.1 1.1) \lpatt (0.033) \move (0 0.55) \Ltext{$\alpha$} \lvec (1.8 0.55) %\move (0 0.34) \Ltext{$\beta$}\lvec (1.8 0.34) \move (0.9 -0.1) \Btext{Figure 4: $\supset$-shaped curve} \esegment \end{texdraw} \vspace{0.3in} We derive next several new exact multiplicity results, where the nonlinear term $f(u)$ does not have to be convex. \begin{thm}\label{thm:2} Assume $f \in C^1[0,\infty)$, $f(u)>0$ for all $u>0$, and assume that for some $\alpha >0$ we have (\ref{5}), (\ref{10.1}), and $$\label{11a} h'(u)=f(u)-uf'(u)<0 \quad \mbox{for all u>\alpha},$$ Then there exist two constants $0 \leq \bar \lambda <\lambda _0$ so that the problem (\ref{1}) has no solution for $\lambda >\lambda _0$, exactly two solutions for $\bar \lambda < \lambda < \lambda _0$, and in case $\bar \lambda >0$ it has exactly one solution for $0<\lambda <\bar \lambda$. Moreover, all solutions lie on a unique smooth solution curve. (See Figure 4.) If we moreover assume that $$\label{11c} \lim _{u \to \infty} \frac{f(u)}{u}=\infty$$ then $\bar \lambda =0$. \end{thm} \paragraph{Proof:} By the implicit function theorem there is a curve of positive solutions of (\ref{1}), starting at $\lambda =0$, $u=0$. This curve continues for increasing $\lambda$, with $u(0)$ increasing, see e.g. \cite{KLO}. From the Theorem 1 and the above remarks on \cite{KL}, we know that by the time the solution curve reaches $u(0)=\alpha$ it has made exactly one turn, and it travels to the left. If we can show that any solution with $u(0)>\alpha$ is also unstable (and in particular non-degenerate), the proof will follow, since the solution curve always travels to the left. To show that any solution with $u(0)>\alpha$ is unstable, we notice that from (\ref{11a}), $h'(u)\le 0$ for $u>\alpha$, then $h(u)\le h(\alpha)<0$ for all $u>\alpha$. Thus Theorem 1 can be applied to any $u>\alpha$ as well. Therefore we conclude that for all $u(0)>\alpha$ the solution $u(x)$ is unstable, and the solution curve always moves to the left. Finally, it is well known that the condition (\ref{11c}) prevents the solution curve from going to infinity at a positive $\bar \lambda$, see e.g. \cite{KO1}. \hfill$\diamondsuit$ \paragraph{Remark.} The result in Theorem \ref{2} is well-known under the conditions $f(u)>0$ and $f''(u)>0$ for all $u>0$. (See \cite{L}.) Here we only assume (\ref{11a}) for $u>\alpha$, so $f''$ can change sign for $u>\alpha$. We conjecture that this result is true without any convexity condition but just assuming $f(u)>0$ for all $u>0$, $h(\alpha)<0$ and \label{11.4} h'(u)\geq 0 \quad \mbox{for $0 0$ for all $u>0$, and assume that for some $\alpha >\beta>0$ we have (\ref{5}), (\ref{11a}) and \label{12} f''(u)<0 \quad \mbox{for $0 0 \quad \mbox{for$\beta\lambda _0$, exactly two solutions for$\bar \lambda < \lambda < \lambda _0$, and in case$\bar \lambda >0$it has exactly one solution for$0<\lambda <\bar \lambda$. Moreover, all solutions lie on a unique smooth solution curve. (See Figure 4.) If we moreover assume that (\ref{11c}) holds then$\bar \lambda =0$. \end{thm} \paragraph{Proof:} We follow the same scheme as in the Theorem \ref{thm:2}. Two things need to be checked: that when$u(0)<\alpha$a turn to the left occurs on any degenerate solution, while for$u(0) \geq \alpha$any solution of (\ref{1}) is unstable. \smallskip To see that only a turn to the left can occur when$u(0)<\alpha$, we follow the approach in P. Korman, Y. Li and T. Ouyang \cite{KLO} (see also T. Ouyang and J. Shi \cite{OS}). Assume$(\lambda ,u(x))$is a degenerate solution of (\ref{1}), i.e. the problem (\ref{2}) has a non-trivial solution$w(x)$at$\mu=0$. Differentiating the equation (\ref{1}), we have $$\label{13a} u_x^{''}+\lambda f'(u)u_x=0.$$ Similarly, differentiating the linearized equation for (\ref{1}) $$\label{13b} w_x^{''}+\lambda f'(u)w_x+\lambda f''(u)u_xw=0.$$ Multiply the equation (\ref{13b}) by$u_x$, the equation (\ref{13a}) by$w_x$, integrate over$(0,1)$and subtract. After expressing from the corresponding equations$w''(1)=-\lambda f'(u(1))w(1)=0$, and$u''(1)=-\lambda f(u(1))=-\lambda f(0)$, we obtain $$\label{13c} \int_0^1 f''(u)u_x^2w \, dx=-f(0)w'(1)>0.$$ Define$x_0 \in (0,1)$by$u(x_0)=\beta$. Arguing as in \cite{KLO} (see also the Theorem 1 of the present work) by scaling$w(x)$we can achieve the inequalities (\ref{9}) above. Using our condition (\ref{12}), and the inequality (\ref{13c}), we have $$\label{13d} \int_0^1 f''(u)w^3 \, dx >\int_0^1 f''(u)u_x^2w \, dx>0.$$ This implies that only turns to the left are possible, see \cite{KLO} for more details. We claim next for that$u(0) \geq \alpha$any solution of (\ref{1}) is unstable. We apply the Theorem 1 to any solution$u$with$u(0)\geq\alpha$. Since$h(0)=0$,$h'(0)=f(0)>0$, and$h''(u)=-uf''(u)>0$for$u \in (0,\beta)$, it follows that the function$h(u)$is positive and increasing on$(0,\beta)$. Since$h(u)$is concave for$u> \beta$, and eventually$h(u)$is non-positive (at$u=\alpha$), it follows that$h(u)$must have a unique point of maximum at some$u_1>\beta$, and then decrease for$u_1\alpha$, and so the Theorem 1 can be applied in case$u(0)>\alpha$, to prove the instability of the solution. \hfill$\diamondsuit$\paragraph{Example} Consider the problem \begin{eqnarray*} & u'' + \lambda (u^3-au^2+bu+c)=0 \quad \mbox{for$-1\beta$. So we only need to assume that$b>a^2/4$(to assure that$u^3-au^2+bu>0$(so$f(u)>0$) for all$u>0$) for the Theorem \ref{thm:3} to apply (with$\bar \lambda =0$). A similar result was previously obtained by S.-H. Wang and D.-M. Long \cite{W}, who required that$b>49a^2/160$, and$c$small enough. We remark though that the nice result of \cite{W} has some advantages over our Theorem \ref{thm:3}. Indeed, the result of \cite{W} does not involve the second derivative assumptions (they assume basically that the function$h(u)$has properties similar to ours, and some technical conditions), and they can allow some$f(u)$that change concavity more than once. \smallskip A special case of the Theorem \ref{thm:3} is when$f''(u)>0$for$u>\alpha$, then (\ref{11a}) is satisfied in that case. For that nonlinearity and the higher dimensional analog of (\ref{1}), T. Ouyang and J. Shi proved the results in the Theorem \ref{thm:3}. (See Theorem 6.21 in \cite{OS}.) Here we do not need to assume the convexity of$f$for$u>\alpha$. On the other hand,$f(0)>0$can be replaced by$f(0)=0$and$f'(0)\ge 0$. (See details in \cite{OS}.) \vspace{0.3in} \begin{texdraw} \arrowheadtype t:F \arrowheadsize l:0.08 w:0.04 \def\Rtext #1{\bsegment \textref h:L v:C \htext ( 0.08 0){#1} \esegment} \def\Ltext #1{\bsegment \textref h:R v:C \htext ( -0.08 0){#1} \esegment} \def\Ttext #1{\bsegment \textref h:C v:B \htext (0 0.06){#1} \esegment} \def\Btext #1{\bsegment \textref h:C v:T \htext (0 -0.06){#1} \esegment} \move (-1.3 0) \bsegment \move (0 0.5) \avec (1.8 0.5) \Rtext{$u$} \move (0 0) \avec (0 1.1) \Ttext{$h(u)$} \move (0 0.5) \clvec (0.7 0.52)(0.9 2)(1.5 0.25) \move (1.5 0.5) \Btext{$\alpha$} \lvec (1.5 0.55) \move (0.57 0.5) \Btext{$\beta$} \lvec (0.57 0.55) \move (0.96 0.5) \Btext{$u_1$} \lvec (0.96 0.55) \move (0.9 -0.1) \Btext{Figure 5: The graph of$h(u)$} \esegment \move (1.2 0) \bsegment \move (0 0.5) \avec (1.8 0.5) \Rtext{$u$} \move (0 0) \avec (0 1.1) \Ttext{$f(u)$} \move (0 0.5) \clvec (0.3 -0.5)(0.7 1.8)(1.5 0.5) \Btext{$c$} \move (0.47 0.5) \Btext{$b$} \move (0.9 -0.1) \Btext{Figure 6: The graph of$f(u)$} \esegment \end{texdraw} \vspace{0.3in} \section{A dual version of instability result} The instability result in Theorem \ref{thm:1} has a dual counterpart in the following theorem: \begin{thm}\label{thm:8} Assume that$f \in C^1[0,\infty)$and for some$\alpha> \beta >0$we have: \begin{eqnarray} \label{3.4} &h'(u)\leq 0 \quad \mbox{for$ 0 0$in (\ref{2})). \end{thm} The proof of Theorem \ref{thm:8} is exactly same as Theorem \ref{thm:1} except switching all$\leq$and$<$by$\geq$and$>$. As an application, we prove a result which generalizes one of the main results of \cite{KLO}. \begin{thm} \label{thm:9} Assume$f \in C^2[0,\infty)$,$f(0)=0$,$f(u)<0$for$u\in (0,b)\cup (c,\infty)$, and$f(u)>0$for$u \in (b,c)$, where$c>b>0$. Assume that for some$c>\alpha>\beta>b$we have \begin{eqnarray} \label{3.6} &f''(u)>0 \quad \mbox{for$0 0,& \\ \label{3.8} &h'(u)=f(u)-uf'(u)>0 \quad \mbox{for all $u>\alpha$}.& \end{eqnarray} (See the graph of $f(u)$ in Figure 6.) Then there exists a constant $\lambda_0>0$, so that the problem (\ref{1}) has no solution for $\lambda <\lambda _0$, exactly two solutions for $\lambda >\lambda _0$, and exactly one solution for $\lambda = \lambda_0$. Moreover, all solutions lie on a unique smooth solution curve. (See Figure 8.) \end{thm} \vspace{0.3in} \begin{texdraw} \arrowheadtype t:F \arrowheadsize l:0.08 w:0.04 \def\Rtext #1{\bsegment \textref h:L v:C \htext ( 0.08 0){#1} \esegment} \def\Ltext #1{\bsegment \textref h:R v:C \htext ( -0.08 0){#1} \esegment} \def\Ttext #1{\bsegment \textref h:C v:B \htext (0 0.06){#1} \esegment} \def\Btext #1{\bsegment \textref h:C v:T \htext (0 -0.06){#1} \esegment} \move (-1.3 0) \bsegment \move (0 0.5) \avec (1.8 0.5) \Rtext{$u$} \move (0 0) \avec (0 1.1) \Ttext{$h(u)$} \move (0 0.5) \clvec (0.7 0.48)(0.9 -1)(1.5 0.75) \move (1.5 0.5) \Btext{$\alpha$} \lvec (1.5 0.55) \move (0.57 0.5) \Btext{$\beta$} \lvec (0.57 0.55) \move (0.96 0.5) \Btext{$u_1$} \lvec (0.96 0.55) \move (0.9 -0.1) \Btext{Figure 7: The graph of $h(u)$} \esegment \move (1.2 0) \bsegment \move (0 0) \avec (1.8 0) \Rtext{$\lambda$} \move (0 0) \avec (0 1.1) \Ttext{$u(0)$} \move (1.8 0.4) \clvec (0.1 0.5)(0 0.87)(1.8 0.95) \lpatt (0.033) \move (0 0.38) \lvec (1.8 0.38) \move (0 0.98) \lvec (1.8 0.98) \move (0 0.79) \lvec (1.8 0.79) \move (0 0.79) \Ltext{$\alpha$} \move (0 0.98) \Ltext{$c$} \move (0.9 -0.1) \Btext{Figure 8: $\subset$-shaped curve} \esegment \end{texdraw} \vspace{0.3in} \paragraph{Proof:} Our proof combines the main ingredients of the proof in \cite{KLO} and that of the Theorem \ref{thm:2}. Since $h(0)=0$, $h'(0)=0$ and $h''(u)=-uf''(u)<0$ for $u$ near $0$, then $h(u)<0$ and $h'(u)<0$ near $u=0$. Since $h$ is concave for $u\in (0,\beta)$, we have $h'(u)<0$ for $u \in (0,\beta)$. But $h(\alpha)>0$, then $h$ must have a local minimum in $(\beta,\alpha)$, which is unique since $h''(u)>0$ in $(\beta,\alpha)$. Let the unique minimum of $h$ in $(\beta,\alpha)$ be $u_1$. Then $h'(u)>0$ in $(u_1,\alpha)$, and $h(u)>0$, $h'(u)>0$ for $u>\alpha$ by (\ref{3.8}). In particular, for any $u\geq \alpha$, we have (\ref{3.4}), and so the Theorem \ref{thm:8} applies. On the other hand, in \cite{KLO}, it is proved that for large $\lambda>0$, (\ref{1}) has a positive solution $u(\lambda, \cdot)$ which satisfies $\lim_{\lambda \to \infty} u(\lambda,0)=c$. Thus there exists a small $\delta>0$ such that for any $d \in (c-\delta,c)$, (\ref{1}) has a positive solution $u$ with $u(0)=d$. From Theorem \ref{thm:8}, as long as $c-\delta>\alpha$, that solution is strictly stable. So the solutions with $u(0)\in (c-\delta,c)$ are on a smooth curve, which continues to left as $\lambda$ and $u(0)$ decrease. We can continue the curve without any turns to the level $u(0)=\alpha$, since all the solutions with $\alpha\le u(0)0$ in those intervals. We would leave the details to readers. \section{A class of symmetric nonlinearities} For the autonomous equation (\ref{1}) both phase-plane analysis and bifurcation theory apply. If we allow explicit dependence of the nonlinearity on $x$, i.e. consider \label{2.1} u'' + \lambda f(x,u)=0 \quad \mbox{for $-10$},&\\ \label{2.1b} &f_x(x,u) < 0 \quad \mbox{for all $00$}.& \end{eqnarray} Recall that under the above conditions any positive solution of (\ref{2.1}) is an even function, with $u'(x)<0$ for all $x \in (0,1]$, see \cite{GNN}. As before the linearized problem \label{2.1c} w'' + \lambda f_u(x,u)w=0 \quad \mbox{for $-1 0 \quad \mbox{for all$-10$}. Then the set of positive solutions of (\ref{2.1}) can be parameterized by their maximum values$u(0)$. (I.e.$u(0)$uniquely determines the pair$(\lambda ,u(x))$.) \end{thm} \paragraph{Proof:} Let on the contrary$v(x)$be another solution of (\ref{2.1}) corresponding to some parameter$\mu \geq \lambda$, but$u(0)=v(0)$. The case of$\mu =\lambda$is not possible in view of uniqueness of initial value problems, so assume that$\mu >\lambda$. Then$v(x)$is a supersolution of (\ref{2.1}), i.e. \label{2.1e} v'' + \lambda f(x,v)<0 \quad \mbox{for$-10$small. Let$0 < \xi \leq 1$be the first point where the graphs of$u(x)$and$v(x)$intersect (i.e.$v(x)0. Subtracting (\ref{2.1g}) from (\ref{2.1f}), noticing that $x_2(u)>x_1(u)$ for all $u \in (u(\xi), u(0))$, and using the condition (\ref{2.1b}), we have $$\label{2.1h} \frac{1}{2}\left[ {u'}^2(\xi)-{v'}^2(\xi)\right] +\lambda\int_{u(\xi)}^{u(0)} \left[f(x_1(u),u)-f(x_2(u),u) \right] \, du<0.$$ Since both terms on the left are positive, we obtain a contradiction. \hfill$\diamondsuit$\smallskip Next we consider positive solutions of the boundary-value problem \label{2.2} u'' + \lambda b(x) f(u)=0 \quad \mbox{for $-10$ for $x\in [0,1]$, $b(x)=b(-x)$, $b'(x)<0$ for $x\in (0,1)$, and $f(u)>0$, so that this problem belongs to the class discussed above. For any solutions $u(x)$ let $(\mu, w(x))$ denote the principal eigenpair of the corresponding linearized equation, i.e. $w(x)>0$ satisfies \begin{eqnarray} \label{2.3} & w'' + \lambda b(x)f'(u)w+\mu w=0 \quad \mbox{for $-10$, $f'(u)>0$ for all $u>0$, and for some $\alpha >0$ the conditions (\ref{5}) and (\ref{10.1}) are satisfied. Then the solution of (\ref{2.2}) with $u(0)=\alpha$ is unstable if it exists. \end{thm} \paragraph{Proof:} In the proof of Theorem \ref{thm:1}, (\ref{6}) and (\ref{7}) are still true. Assume now that $u(x)$ is stable, i.e. $\mu \geq 0$ in (\ref{2.3}). Then $w(x)$ is a positive solution of the problem \label{+-} w''+ g(x,w)=0 \quad \mbox{for $-10$, $b'(x)<0$ in $(0,1)$ using (\ref{6}) and (\ref{10}), we have \begin{eqnarray}\label{16} \lefteqn{ \int_0 ^1 b(x)\left[f(u)-uf'(u) \right]w(x) \, dx }\\ &=& \int_0 ^{x_0}b(x)\left[f(u)-uf'(u) \right]w(x) \, dx +\int_{x_0}^1 b(x)\left[f(u)-uf'(u) \right]w(x) \, dx \nonumber \\ &<& \int_0 ^{x_0}b(x_0)\left[f(u)-uf'(u) \right]w(x) \, dx +\int_{x_0}^1 b(x_0)\left[f(u)-uf'(u) \right]w(x) \, dx \nonumber\\ &=&b(x_0)\int_0 ^1 \left[f(u)-uf'(u) \right]w(x) \, dx\le 0\,. \nonumber \end{eqnarray} On the other hand, multiplying the equation (\ref{2.3}) by $u$, the equation (\ref{2.2}) by $w$, subtracting and integrating over $(0,1)$, we have $$\label{15} \int_0 ^1 b(x)\left[f(u)-uf'(u) \right]w(x) \, dx=\frac{\mu }{\lambda} \int_0 ^1 uw \, dx \geq 0.$$ We reach a contradiction by combining (\ref{16}) and (\ref{15}). \hfill$\diamondsuit$\smallskip As an application we have the following exact multiplicity result. It extends the corresponding result in \cite{KO1} by not restricting the behavior of $f(u)$ at infinity. Its proof is similar to that of the Theorem \ref{thm:2}. Theorem \ref{thm:5} above allows us to conclude the uniqueness of the solution curve. \begin{thm}\label{thm:7} Assume $f \in C^2[0,\infty)$, $f(u)>0$, $f'(u)>0$ and $f''(u)>0$ for all $u>0$, while $h(\alpha ) \leq 0$ for some $\alpha>0$. Then there exist two constants $0 \leq \bar \lambda <\lambda _0$, so that the problem (\ref{2.2}) has no solution for $\lambda >\lambda _0$, exactly two solutions for $\bar \lambda < \lambda < \lambda _0$, and in case $\bar \lambda >0$ it has exactly one solution for $0<\lambda <\bar \lambda$. Moreover, all solutions lie on a unique smooth solution curve. If we moreover assume that (\ref{11c}) holds then $\bar \lambda =0$. \end{thm} \paragraph{Example.} Theorem \ref{thm:7} applies (with $\bar \lambda =0$) to an example from combustion theory \[ u'' + \lambda b(x)e^u=0 \quad \mbox{for \$-1