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\markboth{ Instability and exact multiplicity }
{ Philip Korman \& Junping Shi }
\begin{document}
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\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 $\lambda >0$)
\begin{equation}
\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
\begin{equation}
\label{1}
u'' + \lambda f(u)=0 \quad \mbox{for $-10$ satisfies
\begin{equation}
\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}),
\begin{equation}
\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.
\end{equation}
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
\begin{equation}
\label{8}
(-u')''+\lambda f'(u)(-u')=0 \quad \mbox{on $(0,1)$},
\end{equation}
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
\begin{equation}
\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.
\end{equation}
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
\begin{equation}
\label{10.1}
f''(u)> 0 \quad \mbox{for $0 < u < \alpha$},
\end{equation}
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}
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\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
\begin{equation}
\label{11a}
h'(u)=f(u)-uf'(u)<0 \quad \mbox{for all $u>\alpha$},
\end{equation}
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
\begin{equation}
\label{11c}
\lim _{u \to \infty} \frac{f(u)}{u}=\infty
\end{equation}
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
\begin{equation}
\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
\begin{equation}
\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
\begin{equation}
\label{13a}
u_x^{''}+\lambda f'(u)u_x=0.
\end{equation}
Similarly, differentiating the linearized equation for (\ref{1})
\begin{equation}
\label{13b}
w_x^{''}+\lambda f'(u)w_x+\lambda f''(u)u_xw=0.
\end{equation}
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
\begin{equation}
\label{13c}
\int_0^1 f''(u)u_x^2w \, dx=-f(0)w'(1)>0.
\end{equation}
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
\begin{equation}
\label{13d}
\int_0^1 f''(u)w^3 \, dx >\int_0^1 f''(u)u_x^2w \, dx>0.
\end{equation}
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}.)
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\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}
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\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
\begin{equation}
\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
\begin{equation}
\label{2.1c}
w'' + \lambda f_u(x,u)w=0 \quad \mbox{for $-1 0 \quad \mbox{for all $-10$}.
\end{equation}
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.
\begin{equation}
\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.
\end{equation}
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
\begin{equation}
\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.
\end{equation}
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
\begin{equation}
\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
\begin{equation}
\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
\begin{equation}\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.
\end{equation}
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__