0$. 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
\begin{equation}
\label{s1}
u'' + \lambda f(x,u)=0 \quad \text{for $-1__0$; 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
\begin{equation} \label{s9}
u''+f(\frac{1}{\sqrt{\lambda}} t,u)=0, \quad u'(0)=u(\sqrt{\lambda})=0 \,.
\end{equation}
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},
\begin{equation} \label{s10}
v''+f(\frac{1}{\sqrt{\lambda}} t,v) < 0 \,;
\end{equation}
i.e., $v(t)$ is a supersolution of \eqref{s9}. We may assume that
\begin{equation} \label{s11}
v(t) 0$ and small} \,.
\end{equation}
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
\begin{equation} \label{s12}
\frac{1}{2} {u'}^2(\xi) +\int_{u(0)}^{u(\xi)} f(\frac{1}{\sqrt{\lambda}}
t_2(u),u) \, du=0.
\end{equation}
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
\begin{equation} \label{s14}
\frac{1}{2} {v'}^2(\xi) +\int_{u(0)}^{u(\xi)}
f(\frac{1}{\sqrt{\lambda}} t_1(u),u) \, du > 0.
\end{equation}
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}
\begin{equation} \label{c1}
u''+\lambda \alpha (x) e^u=0, \quad -1__