\documentclass[reqno]{amsart}
\usepackage{epic}
\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 32, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2004/32\hfil Solution curves]
{Solution curves of $2m$-th order boundary-value problems}
\author[Bryan P. Rynne\hfil EJDE-2004/32\hfilneg]
{Bryan P. Rynne}
\address{Bryan P. Rynne \hfill\break
Mathematics Department, Heriot-Watt University,
Edinburgh EH14 4AS, Scotland}
\email{bryan@ma.hw.ac.uk}
\date{}
\thanks{Submitted December 15, 2003. Published March 3, 2004.}
\subjclass[2000]{34B15}
\keywords{Ordinary differential equations, nonlinear boundary value problems}
\begin{abstract}
We consider a boundary-value problem of the form $L u = \lambda f(u)$,
where $L$ is a $2m$-th order disconjugate ordinary differential operator
($m \ge 2$ is an integer),
$\lambda \in [0,\infty)$,
and the function $f : \mathbb{R} \to \mathbb{R}$ is $C^2$ and
satisfies $f(\xi) > 0$, $\xi \in \mathbb{R}$.
Under various convexity or concavity type assumptions on $f$
we show that this problem has a smooth curve, $\mathcal{S}_0$, of solutions
$(\lambda,u)$, emanating from $(\lambda,u) = (0,0)$,
and we describe the shape and asymptotes of $\mathcal{S}_0$.
All the solutions on $\mathcal{S}_0$ are positive
and all solutions for which $u$ is stable lie on $\mathcal{S}_0$.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lemma}[thm]{Lemma}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{remark}[thm]{Remark}
\section{Introduction} \label{intro.sec}
For any integer $m \ge 2$, we consider the $2m$-th order boundary-value
problem
\begin{gather}
(-1)^m u^{(2m)}(x) + \sum_{i=0}^{m-1} (-1)^i p_{i} u^{(2i)}(x) = \lambda f(u(x)) , \quad x \in (-1,1),
\label{main.eq}\\
u^{(i)}(-1) = u^{(i)}(1) = 0, \quad i=0,\dots, m-1,
\label{bc.eq}
\end{gather}
where $p_i \ge 0$, $i=0,\dots, m-1$, are constants and
$u^{(i)}$ is the $i$th derivative of $u \in C^{2m}[-1,1]$,
the number $\lambda \in \mathbb{R}_+ := [0,\infty)$,
and the function
$f : \mathbb{R} \to \mathbb{R}$ is $C^2$ and
satisfies
\begin{equation} \label{posf.eq}
f(\xi) > 0, \quad \xi \in \mathbb{R}
\end{equation}
(we are only interested here in positive solutions so the
behaviour of $f(\xi)$ when $\xi < 0$ is irrelevant).
We assume that \eqref{posf.eq} holds throughout the paper and,
under various additional assumptions on $f$,
we show that \eqref{main.eq}--\eqref{bc.eq} has a curve of solutions
$(\lambda,u)$ in $\mathbb{R}_+ \times C^{2m}[-1,1]$,
emanating from $(\lambda,u) = (0,0)$,
and we describe the shape of this curve.
All the solutions on this curve are positive
(that is, $u$ is positive on $(-1,1)$),
and any solutions for which $u$ is stable lie on this curve.
In the second order case ($m=1$) this problem has been considered
in many papers, for example \cite{BIS}, \cite{CR}, \cite{DANSTR}, \cite{K1}, \cite{K2}, \cite{K3}, \cite{LAE}, \cite{WAN}.
Detailed results for this case are obtained in \cite{BIS} and \cite{LAE}
by using quadrature to derive explicit formulae for
$\lambda = \lambda(\rho)$, $u = u(\rho) \in C^2[-1,1]$ as functions of a parameter
$\rho \ge 0$, with $\rho = |u(\rho)|_0$, such that for each $\rho \ge 0$,
the pair $(\lambda(\rho),u(\rho))$ is a solution.
The results on the shape of the curve of solutions are then obtained by
investigating the function $\rho \to \lambda(\rho)$.
Such a formula for the solutions is not available in the higher order case.
Similar results, for the second order case, are also obtained in
Section~4 of \cite{CR}, and in \cite{K1}, \cite{K2}, \cite{K3},
where the strategy is to use the implicit
function theorem to construct a solution curve in $\mathbb{R} \times C^2[-1,1]$,
and then investigate the structure of this curve directly.
The approach we adopt is similar to this, for the higher order case,
and we obtain most of the results obtained in the above papers
for the second order case.
However, many of the standard tools for second order differential equations
used in these papers, such as the the maximum principle, the Sturm
comparison theorem and simplicity of the zeros of solutions of linear
equations, are not available in the higher order case.
This leads to considerable complication in some of the proofs here and forces
us to use the sophisticated theory of `disconjugate' differential
operators described
in \cite{ELEV}.
Higher order problems ($m > 1$) have also been
investigated recently.
For applications to elasticity see \cite{BW}, \cite{LO} and \cite{YAN},
and the references therein.
For general $n$th-order problems see, for example,
\cite{EHI} and \cite{EHII},
and the references therein
(these papers allow the order $n$ to be odd, and the boundary conditions are more general than here;
the boundary conditions considered here are of the type considered
in \cite{AO} and \cite{EHII}, with $n=2m$ and $k=p=m$).
\section{Preliminary results} \label{prelim.sec}
For any integer $r \ge 0$, let $C^r[-1,1]$ denote the standard Banach
space of real valued, $r$-times continuously differentiable functions
defined on $[-1,1]$, with the norm $|u|_r = \sum_{i=0}^r |u^{(i)}|_0$,
where $|\cdot|_0$ denotes the usual sup-norm on $C^0[-1,1]$.
For any $u,\,v \in C^0[-1,1]$, let
$\langle u,v \rangle = \int_{-1}^1 u v$.
Let $H^{2m}(-1,1)$ denote the standard Sobolev space of order $2m$ on
$(-1,1)$, with the standard norm, which will be denoted by $\|\cdot\|_{2m}$.
For any $u \in C^{2m}[-1,1]$,
let $S(u)$ denote the number of changes of sign of $u$ in $(-1,1)$,
and let $Z(u)$ denote the number of zeros of $u$ in $(-1,1)$
(in the applications below, $u$ will be a non-trivial solution of a
differential equation so these numbers will be finite).
If all the zeros of $u$ in $(-1,1)$ are simple then $S(u)=Z(u)$.
Let
$$
X = \{ u \in C^{2m}[-1,1] : \text{$u$ satisfies \eqref{bc.eq}} \},
\quad
Y = C^0[-1,1].
$$
We define the operator $L : X \to Y$ by
$$
L u := (-1)^m u^{(2m)} + \sum_{i=0}^{m-1} (-1)^i p_{i} u^{(2i)},
\quad u \in X.
$$
It can be verified that $\langle L u , v \rangle = \langle u , L v \rangle$,
for all $u,\,v \in X$ and
$$
\langle L u , u \rangle > 0, \quad 0 \ne u \in X,
$$
that is, $L$ is formally self-adjoint and positive definite on $X$
with respect to the inner product
$\langle \cdot\,,\cdot \rangle$.
It is shown in Remark~2.1 of \cite{RYN} that this positive
definiteness of $L$ implies that the disconjugacy condition imposed on
$L$ in \cite{ELEV} holds, and hence all the results of \cite{ELEV}
hold for the above $L$
(in \cite{ELEV}, $L$ need not be formally self-adjoint,
and the order of $L$ is denoted by $n$, and may be odd).
In particular, $L$ has the following factorisation:
there exists functions
$\rho_i \in C^{2m-i}[-1,1]$, with $\rho_i > 0$ on $[-1,1]$,
$i = 0,\dots,2m$,
such that, if we define
\begin{align*}
L_0 w &:= \rho_0 w,
\quad
L_i w := \rho_i (L_{i-1} w)', \quad i=1,\dots,2m,
\end{align*}
for any $w \in C^{2m}[-1,1]$,
then $L$ has the form
$$
L u = (-1)^m L_{2m} u , \quad u \in X.
$$
We note that the term $(-1)^m$ is not included in the definition of $L$
in \cite{ELEV}.
This sign factor is convenient here (in particular, for the spectral
properties of $L$), but must be borne in mind when results from
\cite{ELEV} are quoted.
The functions $L_i u$, $i=0,\dots,2m-1$, will be called
{\em quasi-derivatives} of $u$.
For any $w \in C^{(2m)}[-1,1]$ we let $\nu(\pm 1,w)$ denote the total
number of quasi-derivatives $L_i w(\pm 1)$, $i=0,\dots,2m-1$,
which are zero.
If $u \in X$ then the boundary conditions
\eqref{bc.eq} imply that $L_i u(\pm 1)=0$, $i=0,\dots,m-1$,
so that $\nu(\pm 1,u) \ge m$.
Furthermore, Corollary~3 of \cite{ELEV} shows that the operator
$L : X \to Y$ is non-singular.
We also need some results from \cite{ELEV} regarding the eigenvalue problem
\begin{equation} \label{lin_eval_prob.eq}
L u = \mu p u,
\end{equation}
for functions $p \in C^0[-1,1]$ with $Z(p)=0$.
For convenience we state these results in the following lemma.
\begin{lemma} \label{lin_eval_prob.lem}
There exists a strictly increasing sequence of eigenvalues of
\eqref{lin_eval_prob.eq}, denoted by $\mu_k(p) > 0$, $k = 1,\,2,\dots.$
Each eigenvalue $\mu_k$ has multiplicity one
$($both geometric and algebraic$)$,
and any corresponding eigenfunction $\phi_k$ has only simple zeros in
$(-1,1)$, and $Z(\phi_k) = k-1$.
Also, $\phi_k^{(m)}(\pm 1) \ne 0$.
\end{lemma}
For any $u \in X$, define $f(u) \in Y$ by
$f(u)(x) = f(u(x))$, $x \in [-1,1]$.
Then \eqref{main.eq}--\eqref{bc.eq} can be rewritten as
\begin{equation} \label{bif.eq}
L u = \lambda f(u).
\end{equation}
Let $\mathcal{S}$ denote the set of solutions $(\lambda,u)$ of \eqref{bif.eq} in
$\mathbb{R}_+ \times X$.
Since $L$ is non-singular,
there are positive constants $b_1,\,b_2$ such that
\begin{equation} \label{lafuest.eq}
b_1 |u|_{2m} \le |L u|_0 = \lambda |f(u)|_0 \le b_2 |u|_{2m} , \quad (\lambda,u) \in \mathcal{S}.
\end{equation}
Also, the only solution of \eqref{bif.eq}
with $\lambda=0$ is $(0,0)$.
Let $\mathcal{S}_0$ denote the connected component of $\mathcal{S}$ which
contains $(0,0)$.
We will be primarily interested in the structure of $\mathcal{S}_0$,
and the stability of the solutions on $\mathcal{S}_0$.
We say that a function $u \in X$ is {\em positive} if $u(x) > 0$
for $x \in (-1,1)$;
we say that a solution $(\lambda,u) \in \mathcal{S}$ is {\em positive} if $u$
is positive.
\begin{lemma} \label{pos.lem}
Every solution $(\lambda,u) \in \mathcal{S} \setminus \{(0,0)\}$ is positive
and
\begin{equation} \label{um_non_zero.eq}
u^{(m)}(\pm 1) \ne 0.
\end{equation}
\end{lemma}
\begin{proof}
If $u \equiv 0$ then since $f(0)>0$ it follows from \eqref{bif.eq}
that $\lambda=0$.
Now suppose that $u \not\equiv 0$ and $\lambda > 0$.
Corollary~1 of \cite{ELEV} shows that
$Z(u) \le S(L u) = S(f(u)) = 0$ (since $f(u)$ is positive),
that is, $u$ has no zeros in $(-1,1)$.
If $u < 0$ on $(-1,1)$ then
$0 < \langle L u , u \rangle = \lambda \langle f(u) , u \rangle \le 0$,
which is impossible.
Finally, since $u$ and $Lu$ are positive we have $S(u) = S(Lu) = 0$,
so setting $h=2m$ in (6) in \cite{ELEV} yields,
$$
0 \ge S(u) + \nu(-1,u) + \nu(1,u) - 2m - S(Lu)
=
\nu(-1,u) + \nu(1,u) - 2m
$$
(it follows easily from its definition, which we will not repeat here,
that the quantity $N_{2m}(u)$ occurring in \cite{ELEV}
satisfies
$N_{2m}(u) \ge \nu(-1,u) + \nu(1,u)$).
This inequality, together with \eqref{bc.eq} and the definition of
$\nu(\pm 1,u)$, shows that \eqref{um_non_zero.eq} must hold.
\end{proof}
We now define a $C^2$ mapping $F : \mathbb{R} \times X \to Y$ by
$F(\lambda,u) = L u - \lambda f(u) ,$
$(\lambda,u) \in \mathbb{R} \times X.$
Clearly, \eqref{bif.eq} is equivalent to the equation
$F(\lambda,u) = 0.$
At any $(\lambda,u) \in \mathbb{R} \times X$ the Fr\'echet derivative,
$D_{(\lambda,u)}F(\lambda,u) : \mathbb{R} \times X \to Y$,
is given by
$$
D_{(\lambda,u)}F(\lambda,u) (\mu,v) = (L - \lambda f'(u)) v - \mu f(u) ,
\quad (\mu,v) \in \mathbb{R} \times X,
$$
and this operator is Fredholm with index 1.
We now show that if this derivative satisfies a suitable
condition everywhere on $\mathcal{S}_0$ then $\mathcal{S}_0$ is a $C^2$ curve with a global
$C^2$ parametrisation
(this condition will be verified in the following sections under
various hypotheses on $f$).
\begin{lemma} \label{globcurve.lem}
Suppose that at every $(\lambda,u) \in \mathcal{S}_0$ the operator
$D_{(\lambda,u)}F(\lambda,u)$ is surjective.
Then $\mathcal{S}_0$ has a $C^2$ parametrisation
$s \to (\lambda(s),u(s)) : \mathbb{R}_+ \to \mathbb{R}_+ \times X$,
such that
$(\lambda(0),u(0)) = (0,0)$, $\lambda_s(0) > 0$,
\begin{align}
\lim_{s\to\infty} \lambda(s) |f(u(s))|_0 = \infty, \label{lafutoinfty.eq}
\end{align}
and, for any $s \ge 0$, the $s$-derivative $(\lambda_s(s),u_s(s))$ satisfies
\begin{gather}
\lambda_s^2(s) + \sum_{i=0}^{2m} \langle u_s^{(i)}(s),u_s^{(i)}(s) \rangle = 1 ,
\label{paramprops.eq}
\\
\bigl(L - \lambda(s) f'(u(s)) \bigr) u_s(s) = \lambda_s(s) f(u(s)) .
\label{sderiv.eq}
\end{gather}
\end{lemma}
\begin{proof}
Since $L$ is non-singular we can construct
(using the implicit function theorem)
a $\delta_0 > 0$ and a parametrisation
$s \to (\lambda(s),u(s)) : [0,\delta_0) \to \mathbb{R}_+ \times X$
such that
$(\lambda(0),u(0)) = (0,0)$, $\lambda_s(0) > 0$,
and \eqref{paramprops.eq}
holds.
Furthermore, the surjectivity hypothesis in the lemma implies
that the set
$\mathcal{S}_0 \setminus \{(0,0)\}$
is a $C^2$ connected curve in $\mathbb{R}_+ \times X$
(see Sections~4.15 and~4.18 in \cite{ZEI} for details)
and a local parametrization with the property
\eqref{paramprops.eq}
can be constructed for this curve near
any $(\lambda,u) \in \mathcal{S}_0 \setminus \{(0,0)\}$
(by the implicit function theorem).
Thus the above local parametrisation near $(0,0)$
can be extended to a maximal interval
$[0,s_{\rm max})$.
Suppose that $s_{\rm max} < \infty$, and for each $n = 1,2,\dots,$
let $s_n = s_{\rm max}-1/n$,
and let $\lambda_n = \lambda(s_n)$, $u_n = u(s_n)$.
Then, by \eqref{paramprops.eq},
$|\lambda_n| + \|u_n\|_{2m} \le C$, for some $C>0$,
and so, by compactness of the embedding of $H^{2m}(-1,1)$ in $C^{2m-1}[-1,1]$,
we may suppose that
$\lambda_n \to \lambda_\infty$ and $u_n \to u_\infty$ in $C^{2m-1}[-1,1]$,
and hence, by \eqref{bif.eq}, $u_n \to u_\infty$ in $C^{2m}[-1,1]$,
and $(\lambda_\infty,u_\infty) \in \mathcal{S}_0$.
But now, by the above local result, the parametrisation can be
extended to the right of $s_{\rm max}$, which contradicts the maximality
of the interval $[0,s_{\rm max})$ and shows that the parametrisation
extends to $[0,\infty)$.
A similar argument shows that
\begin{equation} \label{laplusufourtoinfty.eq}
\lim_{s\to\infty} \bigl( |\lambda(s)| + \|u(s)\|_{2m} \bigr) = \infty .
\end{equation}
Now suppose that there exists a sequence $(s_n)$ in $\mathbb{R}_+$ such that
$s_n \to \infty$
and the sequence $(|u_n|_{2m})$ is bounded.
Then $\lambda_n \to \infty$ and $u_n \to u_\infty$ in $C^{2m-1}[-1,1]$
(after taking a subsequence if necessary),
and hence $|f(u_n)|_0 \to |f(u_\infty)|_0 > 0$
(by \eqref{posf.eq}).
But these results contradict \eqref{lafuest.eq},
which proves that
$\lim_{s\to\infty} |u(s)|_{2m} = \infty$; \eqref{lafutoinfty.eq}
then follows from \eqref{lafuest.eq}.
Finally, differentiating the equation
$
F(\lambda(s),u(s)) \equiv 0
$
with respect to $s$, at any $s \ge 0$, yields \eqref{sderiv.eq}.
\end{proof}
Derivatives with respect to $s$ will always be denoted by
a subscript $s$, to avoid confusion with derivatives with respect to $x$.
\begin{remark} \rm
The condition \eqref{paramprops.eq} says that the chosen parametrisation
of the curve $C_0$ is a `unit speed'
parametrisation in the space $\mathbb{R} \times H^{2m}(-1,1)$.
\end{remark}
\begin{remark} \label{determined.rem} \rm
Equation \eqref{sderiv.eq} says that the derivative
$(\lambda_s(s),u_s(s))$ lies in the null space of $D_{(\lambda,u)}F(\lambda(s),u(s))$,
and the surjectivity condition in Lemma~\ref{globcurve.lem} implies
that this null-space is 1-dimensional,
so \eqref{sderiv.eq} determines $(\lambda_s(s),u_s(s))$ uniquely, up to
a scale factor, whose magnitude is determined by the unit speed
condition \eqref{paramprops.eq}.
\end{remark}
We will also consider the stability of the solutions
on $\mathcal{S}_0$ and relate this to the shape of $\mathcal{S}_0$.
A solution $(\lambda,u) \in \mathcal{S}$ is said to be {\em stable} if all the eigenvalues
of the operator $D_u F(\lambda,u) = L - \lambda f'(u)$ are strictly positive.
Suppose that Lemma~\ref{globcurve.lem} holds
and, for $s \ge 0$, let $\sigma(s)$ denote the principal
(that is, the least) eigenvalue of
the operator $L - \lambda(s) f'(u(s))$.
By definition $(\lambda(s),u(s)) \in \mathcal{S}_0$ is stable
if and only if $\sigma(s) > 0$.
By standard continuous dependence results
the function $\sigma(\cdot)$ is continuous on $\mathbb{R}_+$.
Also, letting $\sigma_0$ denote the principal eigenvalue of $L$,
we have $\sigma(0) = \sigma_0$, and our assumptions ensure that $L$ is positive
definite, that is $\sigma_0 > 0$.
Thus, if $s$ is sufficiently small, $(\lambda(s),u(s))$ is stable.
\section{Increasing $f$} \label{incf.sec}
Throughout this section we suppose that
\begin{equation} \label{fdpos.eq}
f'(\xi) > 0, \quad \xi > 0.
\end{equation}
\begin{lemma} \label{notonezero.lem}
If $(\lambda,u) \in \mathcal{S}$ and
\begin{equation} \label{v.eq}
(L - \lambda f'(u)) w =0, \quad 0 \ne w \in X,
\end{equation}
then
$Z(w) \ne 1$.
\end{lemma}
\begin{proof}
Differentiating \eqref{main.eq} with respect to $x$ and letting
$v = u' := u^{(1)}$ yields
\begin{align}
\widetilde L v &= \lambda f'(u) v ,
\label{udequn.eq}
\\
v^{(i)}(-1) = v^{(i)}(1) &= 0, \quad i=0,\dots,m-2,
\label{udbc.eq}
\end{align}
where $\widetilde L$ is defined in the same way as $L$, except that we apply it to
functions in $C^{2m}[-1,1]$, not just in $X$.
From \eqref{bc.eq} and \eqref{v.eq}--\eqref{udbc.eq} we obtain, by
integration by parts,
$$
\lambda \langle f'(u) v , w \rangle = (-1)^m [v^{(m-1)} w^{(m)}]_{-1}^1 + \lambda \langle v,f'(u) w \rangle
$$
and hence,
\begin{equation} \label{vwz.eq}
u^{(m)}(-1) w^{(m)}(-1) = u^{(m)}(1) w^{(m)}(1) .
\end{equation}
Now, by \eqref{fdpos.eq}, $f'(u) > 0$ on $(-1,1)$, so the
results in Lemma~\ref{lin_eval_prob.lem} hold for
the eigenvalue problem $L w = \lambda f'(u) w$
and show that
if $Z(w)=1$ then $S(w)=1$,
and so (after multiplying $w$ by $-1$ if necessary)
$w^{(m)}(-1) > 0$, $(-1)^m w^{(m)}(1) < 0$.
Also, since $u$ is positive and satisfies
\eqref{bc.eq} and \eqref{um_non_zero.eq},
we must have $u^{(m)}(-1) > 0$, $(-1)^m u^{(m)}(1) > 0$.
However, combining these results contradicts \eqref{vwz.eq},
which completes the proof.
\end{proof}
\begin{lemma} \label{mutwopos.lem}
If $(\lambda,u) \in \mathcal{S}_0$ then
$\lambda < \mu_2(f'(u)) .$
Hence, if
\eqref{v.eq} holds then $\lambda = \mu_1(f'(u))$ and $Z(w)=0$.
\end{lemma}
\begin{proof}
It follows from
Lemmas~\ref{lin_eval_prob.lem} and \ref{notonezero.lem}
that if $(\lambda,u) \in \mathcal{S}_0$ then $\lambda \ne \mu_2(f'(u))$.
Since $0 < \mu_2(0))$,
the first result follows from the continuity of the eigenvalues with respect
to $(\lambda,u)$ on $\mathcal{S}_0$.
The second result now follows from this and Lemma~\ref{lin_eval_prob.lem},
since the hypothesis implies
that $\lambda = \mu_k(f'(u))$, for some $k \ge 1$, and $w$ is a corresponding
eigenfunction.
\end{proof}
\begin{thm} \label{fincglobcurve.thm}
$\mathcal{S}_0$ has the global parametrisation described in Lemma~\ref{globcurve.lem}.
\end{thm}
\begin{proof}
We must show that if $(\lambda,u) \in \mathcal{S}_0$ then $D_{(\lambda,u)}F(\lambda,u)$
is surjective.
Firstly, this is clearly true if the operator $L - \lambda f'(u)$ is non-singular.
On the other hand, if $L - \lambda f'(u)$ is singular then,
by Lemma~\ref{mutwopos.lem},
$\lambda = \mu_1(f'(u))$ and
$(L - \lambda f'(u)) w = 0$ for some positive $w \in X$.
So, by Lemma~\ref{lin_eval_prob.lem},
$$
1 = \dim N(L - \lambda f'(u)) = {\rm codim} R(L - \lambda f'(u))
$$
(here, $N$ and $R$ denote null space and range respectively).
Thus $D_{(\lambda,u)}F(\lambda,u)$ is surjective if
$f(u) \not\in R(L - \lambda f'(u))$,
which by standard spectral theory is equivalent to
$\langle f(u) , w \rangle \ne 0$.
However, this is clearly true since $w$ and $f(u)$ are positive
(by \eqref{posf.eq}),
which proves that $D_{(\lambda,u)}F(\lambda,u)$ is surjective.
\end{proof}
The stability of solutions on the curve $\mathcal{S}_0$ is related to the shape
of the curve as described in the following theorem.
We note that \eqref{fdpos.eq}, together with the results
of \cite{ELEV}, ensure that if $\sigma(s) \ge 0$ then $\sigma(s)$
is a simple eigenvalue, with a corresponding normalised, positive
eigenfunction $\psi(s)$,
satisfying the equation
\begin{equation} \label{sieval.eq}
(L - \lambda(s) f'(u(s))) \psi(s) = \sigma(s) \psi(s).
\end{equation}
\begin{thm} \label{laszero_stab.thm}
For $s \ge 0$,
we have $\lambda_s(s) = 0$ if and only if $\sigma(s) = 0$.
If
$\lambda_s(s) = 0$ then $Z(u_s(s)) = 0$.
If $\lambda_s(s) \ne 0$ then
$\mathop{\rm sgn} \lambda_s(s) = \mathop{\rm sgn} \sigma(s)$.
If $\lambda_s(s) > 0$ then $u_s(s)$ is positive.
\end{thm}
\begin{proof}
If $\lambda_s(s) = 0$ then, by \eqref{paramprops.eq}, $u_s(s) \ne 0$
and, by \eqref{sderiv.eq},
\begin{equation} \label{szderiv.eq}
(L - \lambda(s) f'(u(s))) u_s(s) = 0,
\end{equation}
so by Lemma~\ref{mutwopos.lem},
$Z(u_s(s)) = 0$,
and hence $\sigma(s) = 0$.
Conversely, if $\sigma(s) = 0$ then comparing \eqref{sieval.eq}
with \eqref{sderiv.eq} shows that
$\lambda_s(s) = 0$
(see Remark~\ref{determined.rem}).
Thus the set of zeros of $\lambda_s$ and $\sigma$ coincide;
we denote the complement of this set in $\mathbb{R}_+$ by $M$.
This proves the first part of the theorem.
To prove the second part we require the following result,
see Theorem~3.6 in \cite{CR}: if $\lambda_s(s_0) = 0$
then there exists $\delta(s_0)>0$ such that if
$|s-s_0| < \delta(s_0)$ and $s \in M$ then the quantities
$\sigma(s) \langle u_s(s),u_s(s) \rangle$
and
$\lambda_s(s) \langle u_s(s),f(u(s)) \rangle$
are both nonzero and have the same sign.
Define
$t_1$ (respectively $t_2$) to be the supremum of the set of
$t \in M$ (respectively $t \in \mathbb{R}_+$)
such that
$\mathop{\rm sgn} \lambda_s(s) = \mathop{\rm sgn} \sigma(s)$ for all $s \in [0,t] \cap M$.
Suppose that $t_2 < \infty$.
Since $\lambda_s(0)>0$, $\sigma(0)>0$, we have $0 < t_1 \le t_2$.
By continuity, $t_2 \not\in M$, but by definition
there exists $s \in M$ arbitrarily close to $t_2$ with $s > t_2$,
and
$\mathop{\rm sgn} \lambda_s(s) = - \mathop{\rm sgn} \sigma(s)$.
Thus, by Theorem~3.6 of \cite{CR} (quoted above), the function $u_s(t_2)$
is negative.
By a similar argument, $u_s(t_1)$ is positive.
Thus, $t_1 < t_2$, and $[t_1,t_2] \cap M = \emptyset$.
However, by the first part of the proof,
for all $s \in [t_1,t_2]$ we have $Z(u_s(s)) = 0$,
so by continuity the sign of $u_s(s)$ is constant for $s \in [t_1,t_2]$,
which is a contradiction.
Thus we deduce that $t_2 = \infty$, which proves the desired result.
Now suppose that $\lambda_s(s) > 0$ on some interval $(t_3,t_4)$,
with either $t_3=0$ or $\lambda_s(t_3) = 0$,
and either $t_4=\infty$ or $\lambda_s(t_4) = 0$.
The above results show that $u_s(t_3)$ is positive.
Let
$t_5$ be the supremum of the set of
$t \in [t_3,t_4]$
such that
$u_s(s)$ is positive for all $s \in [t_3,t]$.
Suppose that $t_5 < t_4$.
Then, by \eqref{sderiv.eq} and continuity, there exists $\delta>0$ such that if
$|s-t_5| < \delta$
then
$S(L u_s(s)) = 0$
and so, by Corollary~1 of \cite{ELEV},
$Z(u_s(s)) = 0$.
However, this contradicts the definition of $t_5$, and so proves that
$t_5 = t_4$, which completes the proof of the theorem.
\end{proof}
The following theorem shows that
for any $s > 0$,
$u(s)$ is even,
with a single local maximum at $x=0$
(this is easy to prove in the second order case).
\begin{thm} \label{symm.thm}
For all $s > 0$, $u(s)$ is even and $u'(s)$ has exactly one zero
in $(-1,1)$ $($at $x=0$, since $u'(s)$ must be odd$)$,
so that $|u(s)|_0 = u(s)(0)$.
\end{thm}
\begin{proof}
Clearly, $u(0)=0$ is even and the set
$$\Sigma := \{ s \in \mathbb{R}_+: \text{$u(s)$ is even}\}$$
is closed.
Now suppose that $s_0 \in \Sigma$ and there exists a sequence $(\delta_n)$
such that $\delta_n \to 0$ and $0 \le s_0 + \delta_n \not\in \Sigma$.
For any $s \ge 0$, define $\widetilde u(s)$ by
$\widetilde u(s)(x) := u(s)(-x)$, $x \in [-1,1]$.
Then the curve $s \to (\lambda(s),\widetilde u(s))$, $s \ge 0$, is a curve of solutions
of \eqref{bif.eq} satisfying the properties in Lemma~\ref{globcurve.lem},
with $(\lambda(s_0),\widetilde u(s_0)) = (\lambda(s_0),u(s_0))$,
but which is distinct from the curve $s \to (\lambda(s),u(s))$ near
$(\lambda(s_0),u(s_0))$.
However, this contradicts the implicit function theorem construction of
the local curve in the proof of Lemma~\ref{globcurve.lem}.
This shows that such a sequence cannot exist for any $s_0$,
and hence the set $\Sigma$ is also open in $\mathbb{R}_+$,
which implies that $\Sigma=\mathbb{R}_+$.
Next, for any $s > 0$, let $v(s) := u'(s)$.
Then $v(s)$ is odd and satisfies \eqref{udequn.eq}--\eqref{udbc.eq}
(with $\lambda = \lambda(s)$, $u = u(s)$)
and, by \eqref{um_non_zero.eq},
\begin{equation} \label{w_bcs.eq}
v^{(m-1)}(s)(\pm 1) \ne 0.
\end{equation}
We will show that $S(v(s)) = 1$ for all $s > 0$.
Firstly, when $s$ is small,
$(\lambda(s),u(s)) \simeq s \gamma (1, f(0) \eta)$,
in $\mathbb{R}_+ \times X$,
where $\gamma$ is a suitable scaling factor
and $\eta \in X$ satisfies $L \eta = 1$
($\eta$ is even since, if not, the function $\widetilde \phi$ defined
by $\widetilde \phi(x) = \eta(-x)$, $x \in [-1,1]$, would provide a second
solution of this equation, contradicting the non-singularity of $L$).
Thus, for small $s$, we have $S(v(s)) = S(\zeta)$, where $\zeta := \eta'$,
and since $\zeta$ is non-trivial and odd we must have $S(\zeta) \ge 1$.
Now, differentiating the equation $L \eta = 1$ yields $\widetilde L \zeta = 0$,
and hence
$L_{2m-1} \zeta = c_1$, for some constant $c_1$.
If $c_1 \ne 0$ then (6) in \cite{ELEV} (with $h=2m-1$) shows that
$$
0 \ge S(\zeta) + N_{2m-1}(\zeta) - (2m - 1) - S(L_{2m-1} \zeta)
\ge
S(\zeta) - 1
$$
(again, we omit the definition of $N_{2m-1}(\zeta)$ given in \cite{ELEV},
but we note that \eqref{udbc.eq} implies that $N_{2m-1}(\zeta) \ge 2m-2$),
and hence $S(\zeta) = 1$ in this case.
Now suppose that $c_1=0$, and so
$L_{2m-2} \zeta = c_2$, for some constant $c_2$.
By Corollary~3 in \cite{ELEV} it follows from this,
together with \eqref{udbc.eq},
that $c_2=0$ implies that $\zeta = 0$,
so we must have $c_2 \ne 0$.
On the other hand, repeating the above argument
(with $h=2m-2$)
yields
$0 \ge S(\zeta)$, which contradicts $S(\zeta) \ge 1$,
so we conclude that $c_1 \ne 0$,
and hence $S(\zeta)=1$.
Now suppose that there exists $s_0 > 0$ such that $v(s_0)$ has a double
zero in $(-1,1)$.
We consider the quantity $N(v(s_0))$ defined in \cite{ELEV}
(the definition is somewhat complicated so we will not repeat it here;
essentially, $N(v(s_0))$ is a count of the multiple zeros of $v(s_0)$).
Since $v$ satisfies \eqref{udequn.eq}
it follows from Lemma~1 in \cite{ELEV} and the definition of $N(v(s_0))$ that
\begin{equation} \label{Nletwom.eq}
\nu(-1,v(s_0)) + \nu(1,v(s_0)) + 2 \le N(v(s_0)) \le 2m
\end{equation}
(the 2 here is a lower bound on the contribution from the assumed double zero
of $v(s_0)$).
However, \eqref{udbc.eq} implies that $\nu(\pm 1,v(s_0)) \ge m-1$,
so we must have $N(v(s_0)) = 2m$.
Lemma~1 in \cite{ELEV} now
shows that $\nu(\pm 1,v(s_0))$ are even (odd) if $m$ is even (odd),
which implies that we must actually have $\nu(\pm 1,v(s_0)) \ge m$
(this parity argument relies on the positivity of $f'$,
that is, on \eqref{fdpos.eq}).
But then \eqref{Nletwom.eq} is contradictory, so we conclude that $v(s)$
cannot have a double zero in $(-1,1)$ for any $s > 0$.
It follows from this that $Z(v(s)) = S(v(s))$
and that $S(v(s)) = S(\zeta) = 1$, for all $s > 0$,
since if $S(v(s))$ changes
at some $s=s_0$ then, by \eqref{w_bcs.eq}, $v(s_0)$ would have a
double zero in $(-1,1)$.
\end{proof}
\begin{remark} \label{param_by_norm.rem} \rm
In certain second order problems $\mathcal{S}_0$ can be parametrised
by $|u|_0$, that is, the function $s \to |u(s)|_0$ is strictly increasing
on $\mathbb{R}_+$,
see for instance, \cite{DANSTR}.
We cannot show this here for the whole of $\mathcal{S}_0$, but
Theorem~\ref{laszero_stab.thm} shows that this holds on the stable
portions of $\mathcal{S}_0$, that is, $|u(\cdot)|_0$ is strictly increasing on any
interval on which $\lambda_s(\cdot) > 0$.
\end{remark}
We now examine the behaviour of
$\mathcal{S}_0$ as $s \to \infty$.
\begin{thm} \label{asymptga.thm}
We have
\begin{equation} \label{asymptga.eq}
\lim_{s\to\infty} |u(s)|_0 = \infty.
\end{equation}
If, in addition, the limit $\gamma_\infty := \lim_{\xi\to\infty} f(\xi)/\xi$
exists then $\lambda_\infty := \lim_{s\to\infty} \lambda(s)$
exists,
and $\lambda_\infty = \sigma_0/\gamma_\infty$
$($we allow $\gamma_\infty = \infty$ $($or $\gamma_\infty = 0)$ here,
in which case $\lambda_\infty = 0$ $($or $\lambda_\infty = \infty)$.
\end{thm}
\begin{proof}
We follow the proof of Lemma~4 in \cite{ELEV} to a certain extent,
but for brevity we merely describe the necessary changes.
As in \cite{ELEV}, for any $u \in X$ let
$$
r(u,x) = \Big( \sum_{i=0}^{2m-1} (L_i u)^2 (x) \Big)^{1/2},
\quad x \in [-1,1].
$$
Now let $(s_n)$ be an arbitrary sequence in $\mathbb{R}_+$ with
$s_n \to \infty$,
and for each $n$ let $\lambda_n = \lambda(s_n)$, $u_n = u(s_n)$.
Following the proof in \cite{ELEV}, without the normalization (11)
used there, (16) in \cite{ELEV} becomes
\begin{equation} \label{eliassix.eq}
\lambda_n \big| \int_{-1/2}^{1/2} \frac{f(u_n(x))}{\rho_{2m}(x)} \,dx \big|
\le B r(u_n,-1/2),
\end{equation}
while (17) becomes
\begin{equation} \label{eliassev.eq}
|(L_q u_n)(x)| \le C r(u_n,-1/2),
\quad
-1/2 \le x \le 1/2, \quad 0 \le q \le 2m-1,
\end{equation}
for positive constants $B,\,C$
(replacing $\alpha$, $\delta$, $y_\lambda$, $n$ in \cite{ELEV} with
$-1/2$, $1/2$, $u_n$, $2m$,
respectively, here).
Now, if the sequence $(r(u_n,-1/2))$ were bounded
then it would follow from \eqref{eliassix.eq} and \eqref{eliassev.eq}
(together with Theorem~\ref{symm.thm})
that the sequences $(\lambda_n)$ and $(|u_n|_0)$ are bounded,
which would contradict \eqref{lafutoinfty.eq}.
Thus we may suppose that $r(u_n,-1/2) \to \infty$.
Let $v_n = u_n/r(u_n,-1/2)$, $n=1,2,\dots.$
The argument in \cite{ELEV} shows that $v_n \to v_\infty$ in
$C^{2m-2}[-1/2,1/2]$
and $L_{2m-1} v_n \to L_{2m-1} v_\infty$ pointwisely in $(-1/2,1/2]$
(after choosing a subsequence if necessary),
and so either
$v_\infty \not\equiv 0$
or, by definition,
$$
\frac{r(u_n,1/2)}{r(u_n,-1/2)} \to 0.
$$
A similar argument using the functions
$\widetilde v_n = u_n/r(u_n,1/2)$, $n=1,2,\dots,$
and the limit $\widetilde v_n \to \widetilde v_\infty$ on $[-1/2,1/2]$, shows that either
$\widetilde v_\infty \not\equiv 0$
or,
$$
\frac{r(u_n,-1/2)}{r(u_n,1/2)} \to 0.
$$
From these alternatives we conclude that we must have
$v_\infty \not\equiv 0$ or $\widetilde v_\infty \not\equiv 0$
(or both).
The first result of the theorem now follows immediately from this.
To prove the second result we now suppose that the limit $\gamma_\infty$ exists,
and that $v_\infty \not\equiv 0$
(the case $\widetilde v_\infty \not\equiv 0$ is similar),
and hence $v_\infty \ge \epsilon$, for some $\epsilon>0$, on some non-trivial interval
$J \subset [-1/2,1/2]$.
If $\gamma_\infty = \infty$ then
$f(u_n(x))/r(u_n,-1/2) \to \infty$, uniformly for $x \in J$,
and so it follows from \eqref{eliassix.eq} that $\lambda_n \to 0$.
If $\gamma_\infty = 0$ then
$|f(u_n)|_0/|u_n|_0 \to 0$
(using $|u_n|_0 \to \infty$),
so by \eqref{lafuest.eq},
$\lambda_n \to \infty$.
If $0 < \gamma_\infty < \infty$ then, for $n$ sufficiently large,
$f(u_n) \ge \epsilon \gamma_\infty / 2r(u_n,-1/2)$ on $J$,
so by \eqref{eliassix.eq} the sequence $(\lambda_n)$ must be bounded
and, after taking a subsequence if necessary,
we have $\lambda_n \to \lambda_\infty$, for some $\lambda_\infty$.
Also,
$|f(u_n)|_0/|u_n|_0 \to \gamma_\infty$,
so by \eqref{lafuest.eq}, $|u_n|_{2m}/|u_n|_0 \le c$
for some constant $c$.
Now, defining the functions
$w_n = u_n/|u_n|_{2m}$, $n=1,2,\dots,$
we may suppose that $w_n \to w_\infty \ne 0$ in $C^{2m-1}[-1,1]$,
and
$f(u_n)/|u_n|_{2m} \rightharpoonup \gamma_\infty w_\infty$ in $L^2(-1,1)$
(by the argument on p.~648 of \cite{DI}).
Thus, taking the limit in \eqref{bif.eq}, it follows readily that
$w_\infty$ is a non-trivial, weak solution of the equation
$
L w_\infty = \lambda_\infty \gamma_\infty w_\infty ,
$
and hence $\lambda_\infty \gamma_\infty = \sigma_0$
(since $w_\infty \ge 0$ on $[-1,1]$).
This completes the proof.
\end{proof}
\section{Convex $f$} \label{convexf.sec}
Throughout this section we suppose that \eqref{fdpos.eq} holds and,
in addition,
\begin{equation} \label{fddpos.eq}
f''(\xi) > 0, \quad \xi > 0
\end{equation}
(thus $f$ is convex).
Hence the results of Sections~\ref{prelim.sec} and~\ref{incf.sec} hold.
Also, by \eqref{fdpos.eq} and \eqref{fddpos.eq} the limits
$\gamma_\infty$ (see Theorem~\ref{asymptga.thm})
and
$f'_\infty := \lim_{\xi\to\infty} f'(\xi)$
exist
(we allow $f'_\infty = \infty$),
and it can be verified that $\gamma_\infty = f'_\infty > 0$.
Thus, by Theorem~\ref{asymptga.thm}, we have the following
result.
\begin{lemma} \label{asympt.thm}
The limit $\lambda_\infty$ exists and $\lambda_\infty = \sigma_0/f'_\infty < \infty$.
\end{lemma}
We now study the shape of $\mathcal{S}_0$ further.
\begin{lemma} \label{lass.lem}
If, for some $s > 0$, $\lambda_s(s) = 0$ then $\lambda_{ss}(s) < 0$.
\end{lemma}
\begin{proof}
Differentiating \eqref{sderiv.eq}
with respect to $s$ and using $\lambda_s(s) = 0$ yields
\begin{equation} \label{ssderiv.eq}
(L - \lambda(s) f'(u(s))) u_{ss}(s) - \lambda(s) f''(u(s)) (u_s(s))^2
=
\lambda_{ss}(s) f(u(s)).
\end{equation}
Taking the inner product of this with $u_s(s)$,
integrating by parts, and using \eqref{sderiv.eq} yields
$$
- \lambda(s) \langle f''(u(s)) (u_s(s))^2 , u_s(s) \rangle
= \lambda_{ss}(s) \langle f(u(s)) , u_s(s) \rangle ,
$$
and so it follows from Theorem~\ref{laszero_stab.thm},
\eqref{posf.eq} and \eqref{fddpos.eq}
that $\lambda_{ss}(s) < 0$.
\end{proof}
Since $\lambda_s(0) > 0$, Lemma~\ref{lass.lem} shows that there is at most one
`turning point' $s_{\rm t} > 0$ such that $\lambda_s(s_{\rm t}) = 0$,
so Theorem~\ref{laszero_stab.thm} and Lemma~\ref{asympt.thm} give
the following result on the shape of $\mathcal{S}_0$.
\begin{thm} \label{shape.thm}
$\mathcal{S}_0$ must look like one of the curves
$(a)$--$(c)$ in Fig.~1.
Case $(c)$ occurs if and only if $f'_\infty = \infty$.
Furthermore, all the solutions on $\mathcal{S}_0$ are stable in
case $(a)$, while in cases $(b)$ and $(c)$ only the
solutions on the `lower' portion of the curve before the turning
point are stable.
\end{thm}
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(100,100)(0,0)
%
\put(10,60){\line(1,0){35}}
\put(9.15,95){$\uparrow$}
\put(10,60){\line(0,1){35}}
\put(44,59.15){$\rightarrow$}
\qbezier(10,60)(33,70)(38,93)
%\spline(10,60)(33,75)(38,93)
\dashline{3}(40,60)(40,93)
\put(2,93){$|u|_0$}
\put(38,55){$\lambda_\infty$}
\put(45,55){$\lambda$}
\put(28,50){(a)}
%
\put(60,60){\line(1,0){35}}
\put(59.15,95){$\uparrow$}
\put(60,60){\line(0,1){35}}
\put(94,59.15){$\rightarrow$}
\qbezier(60,60)(96,68)(80,78)
\qbezier(80,78)(71,85)(70,93)
%\spline(60,60)(94,68)(73,83)(70,90)(70,93)
\dashline{3}(68,60)(68,93)
\put(52,93){$|u|_0$}
\put(67,55){$\lambda_\infty$}
\put(95,55){$\lambda$}
\put(78,50){(b)}
%
\put(10,10){\line(1,0){35}}
\put(9.15,45){$\uparrow$}
\put(10,10){\line(0,1){35}}
\put(44,9.15){$\rightarrow$}
\qbezier(10,10)(46,18)(30,28)
\qbezier(30,28)(14,35)(12,45)
%\spline(10,10)(44,18)(20,33)(16,40)(15,45)
\put(2,43){$|u|_0$}
\put(8,5){$\lambda_\infty$}
\put(45,5){$\lambda$}
\put(28,0){(c)}
%
\put(60,10){\line(1,0){35}}
\put(59.15,45){$\uparrow$}
\put(60,10){\line(0,1){35}}
\put(94,9.15){$\rightarrow$}
\qbezier(60,10)(88,20)(98,45)
%\spline(60,10)(90,20)(100,45)
\put(52,43){$|u|_0$}
\put(95,5){$\lambda$}
\put(78,0){(d)}
%
\end{picture}
\end{center}
\caption{Possible forms of the solution curve $\mathcal{S}_0$}
\end{figure}
Note that, as mentioned in Remark~\ref{param_by_norm.rem}, we have not
shown that $\mathcal{S}_0$ can be parametrised
by $|u|_0$, so we cannot preclude `vertical oscillations' in the curves
in Fig.~1, that is, multiple solutions for a given $|u|_0$.
If the turning point $s_{\rm t}$ exists we have the following simple
estimate of the location of $\lambda(s_{\rm t})$
(analogous to Lemma~4.3 in \cite{CR}).
\begin{lemma} \label{turn_pt.lem}
If $s_{\rm t}$ exists, then
$\lambda(s_{\rm t}) < \sigma_0 / f'(0)$.
\end{lemma}
\begin{proof}
By definition, the operator $L - \lambda(s_{\rm t}) f'(u(s_{\rm t}))$
is positive semi-definite on $X$, and by \eqref{fddpos.eq},
$f'(u(s_{\rm t})) > f'(0)$ on $(-1,1)$,
so the operator $L - \lambda(s_{\rm t}) f'(0)$
is positive definite, and hence
$\lambda(s_{\rm t}) f'(0) < \sigma_0$.
\end{proof}
We now give a necessary and sufficient condition for a turning point to
exist,
that is, to distinguish between cases (a) and (b).
Defining $g(\xi) := f(\xi) - f'(\xi) \xi$, $\xi \ge 0$,
it is clear that
$g(0) > 0$ and, by \eqref{fddpos.eq}, $g'(\xi) < 0$, $\xi > 0$.
\begin{thm} \label{turningpt.thm}
Suppose that $f'_\infty < \infty$.
Then $\mathcal{S}_0$ has a turning point if and only if
$g(\xi_0) = 0$ for some $\xi_0 > 0$.
\end{thm}
\begin{proof}
Suppose that $\lambda_s(s) = 0$ for some $s>0$.
Taking the inner product of \eqref{sderiv.eq} with $u(s)$ yields
\[
\lambda(s) \langle f'(u(s)) u_s(s) , u(s) \rangle
=\langle L u_s(s) , u(s) \rangle
=\langle u_s(s) , L u(s) \rangle
=\lambda(s) \langle u_s(s), f(u(s)) \rangle,
\]
and hence
$
\langle u_s(s) , g(u(s)) \rangle = 0.
$
Now, by Theorem~\ref{laszero_stab.thm}, $Z(u_s(s))=0$ so $g$ must change
sign, and hence we must have $g(\xi_0) = 0$ for some $\xi_0 > 0$.
Now suppose that $g(\xi_0) = 0$ for some $\xi_0 > 0$,
and let $-\delta := g(\xi_0+1) < 0$, so that $g(\xi) \le -\delta$
for all $\xi \ge \xi_0 + 1$.
We can choose a sequence $(s_n)$ in $\mathbb{R}_+$
such that $u_n / |u_n|_{2m} \to w_\infty$ in $C^{2m-1}[-1,1]$,
with $w_\infty$ positive
(see the final part of the proof of Theorem~\ref{asymptga.thm}),
and hence
$\lim_{n\to\infty} u_n(x) = \infty$ for $x \in (-1,1)$.
Letting
$$
G_n = \{ x \in [-1,1] : u_n(x) > \xi_0 + 1 \},
\quad
B_n = \{ x \in [-1,1] : u_n(x) \le \xi_0 + 1 \},
$$
it follows that $\lim_{n\to\infty} |B_n| = 0$
(where $|B_n|$ denotes the Lebesgue measure of $B_n$).
Hence, for sufficiently large $n$,
\begin{align*}
\langle f(u_n) , u_n \rangle
&\le
\int_{B_n} f(u_n) u_n + \int_{G_n} \bigl( f'(u_n) (u_n)^2 - \delta \bigr)\\
&\le f(\xi_0 + 1)(\xi_0 + 1) |B_n| - \delta |G_n|
+ \langle f'(u_n) u_n , u_n \rangle\\
&< \langle f'(u_n) u_n , u_n \rangle ,
\end{align*}
and so, by Theorem~\ref{asympt.thm} and \eqref{fddpos.eq},
\begin{align*}
\lambda_\infty f'_\infty \langle u_n,u_n \rangle
&= \sigma_0 \langle u_n,u_n \rangle
\le \langle Lu_n,u_n \rangle
=\lambda_n \langle f(u_n) , u_n \rangle \\
&< \lambda_n \langle f'(u_n) u_n , u_n \rangle
< \lambda_n f'_\infty \langle u_n , u_n \rangle .
\end{align*}
Thus $\lambda_\infty < \lambda_n$ for sufficiently large $n$, so
$\mathcal{S}_0$ must have a turning point.
\end{proof}
\begin{remark} \rm
In the second order case it is shown in Theorem~3.2 of \cite{LAE} that
if $g$ has a zero then $\mathcal{S}_0$ has a turning point;
the discussion in \cite{BIS} proves the reverse implication.
\end{remark}
\begin{remark} \rm
The above proof that the existence of a turning point implies that $g$
has a zero did not use condition \eqref{fddpos.eq}.
Thus, if we merely suppose that \eqref{fdpos.eq} holds, and that $g$ does not have a zero, then $\mathcal{S}_0$ must look like one of the curves
(a) or (d) in Fig.~1, depending on whether
$\gamma_\infty < \infty$ or $\gamma_\infty = \infty$
(positivity of $g$ implies that the function $f(\xi)/\xi$ is
increasing, so $\gamma_\infty$ exists).
\end{remark}
In the second order case it can be shown that there is only one curve of
solutions, see Theorem~1 of \cite{DANSTR}
and the argument at the bottom of p.~1016 of \cite{K2}.
Although we cannot prove here that the only solutions of
\eqref{bif.eq} lie on $\mathcal{S}_0$, the following theorem shows that
all the stable solutions do.
\begin{thm} \label{stab_solns_on_sz.thm}
If $(\lambda,u) \in \mathcal{S}$ is stable then $(\lambda,u) \in \mathcal{S}_0$.
\end{thm}
\begin{proof}
Suppose, on the contrary, that there exists a stable solution
$(\lambda_1,u_1) \in \mathcal{S} \setminus \mathcal{S}_0$.
We may also suppose, without loss of generality, that
$0 < f'_\infty < \infty$
(by suitably redefining $f$ on the interval $[1 + |u_1|_0, \infty)$,
if necessary, which does not affect the solution $(\lambda_1,u_1)$).
Now, by stability,
$\langle (L - \lambda_1 f'(u_1)) v , v \rangle > 0$ for all non-zero $v \in X$,
and hence, $\lambda_1 < \mu_1(f'(u_1))$.
Thus the proof of Theorem~\ref{fincglobcurve.thm} shows that there is a
global, connected $C^2$ solution curve $\mathcal{S}_1 \subset \mathbb{R}_+ \times X$
with $(\lambda_1,u_1) \in \mathcal{S}_1$,
and that we may apply the implicit function theorem at any point
$(\lambda,u) \in \mathcal{S}_1$
(the proof relies on the point $(\lambda_1,u_1) \in \mathcal{S}_1$
satisfying $\lambda_1 < \mu_2(f'(u_1))$,
which follows from the stability assumption).
All the other results proved above for $\mathcal{S}_0$ also hold for $\mathcal{S}_1$.
Since $\mathcal{S}_0$ and $\mathcal{S}_1$ are closed and connected, but not equal to each
other, they must be disjoint.
In particular, $\mathcal{S}_1$ must be bounded away from the point $(0,0)$.
Also, Lemma~\ref{lass.lem} shows that $\mathcal{S}_1$ cannot be a `loop'
(homeomorphic to the unit circle),
so it must be an `open curve' (homeomorphic to $\mathbb{R}$),
and there is a global
unit speed parametrisation
$t \to (\lambda^1(t),u^1(t)) : \mathbb{R} \to \mathbb{R}_+ \times X$,
with $(\lambda^1(0),u^1(0)) = (\lambda_1,u_1)$,
and
$$
\lim_{t\to\pm\infty} |u^1(t)|_0 = \infty,
\quad
\lim_{t\to\pm\infty} \lambda^1(t) = \sigma_0/f'_\infty.
$$
However, the results in \cite{DANINF} on bifurcation from infinity at a
`simple' eigenvalue show that this cannot happen
(there cannot be two curves with the same `limit at infinity'),
and hence $(\lambda_1,u_1)$ cannot exist.
\end{proof}
\section{Concave $f$} \label{concf.sec}
Throughout this section we suppose that \eqref{fdpos.eq} holds and,
in addition,
\begin{equation} \label{fddneg.eq}
f''(\xi) < 0, \quad \xi > 0
\end{equation}
(thus $f$ is concave).
Again, the results of Sections~\ref{prelim.sec} and~\ref{incf.sec} hold,
and the limits $\gamma_\infty$, $f'_\infty$ exist,
with $0 \le \gamma_\infty = f'_\infty < \infty$ here.
\begin{thm} \label{concf_dholds.thm}
For all $s \ge 0$, $\lambda_s(s) > 0$ and $\sigma(s) > 0$,
and $\lambda_\infty = \sigma_0/f'_\infty$
Theorem~\ref{stab_solns_on_sz.thm} also holds.
\end{thm}
\begin{proof}
Changing the sign of $f''$ in the proof of
Lemma~\ref{lass.lem} shows that if $\lambda_s(s) = 0$ then
$\lambda_{ss}(s) > 0$ (by \eqref{fddneg.eq}).
However, since $\lambda_s(0)>0$, this precludes
the existence of a point $s>0$
for which $\lambda_s(s) = 0$, so we must have $\lambda_s(s) > 0$ for all $s$.
It now follows from Theorem~\ref{laszero_stab.thm} that $\sigma(s) > 0$ for all $s$.
The value of $\lambda_\infty$ follows from Theorem~\ref{asymptga.thm},
and the proof of Theorem~\ref{stab_solns_on_sz.thm} also holds here.
\end{proof}
\begin{cor} \label{asymptconc.cor}
$\mathcal{S}_0$ must look like one of the curves
$(a)$ or $(d)$ in Fig.~1,
depending on whether
$f'_\infty < \infty$ or $f'_\infty = 0$.
All the solutions on $\mathcal{S}_0$ are stable, and all stable solutions lie on $\mathcal{S}_0$.
\end{cor}
\section{Decreasing $f$} \label{decf.sec}
Throughout this section we suppose that
\begin{equation} \label{decf.eq}
f'(\xi) \le 0, \quad \xi > 0.
\end{equation}
Here, $f'_\infty$ may not exist, but $\gamma_\infty$ exists,
with $\gamma_\infty = 0$.
The results of Section~\ref{prelim.sec} hold.
\begin{thm} \label{decf_dholds.thm}
$\mathcal{S}_0$ has the global parametrisation described in Lemma~\ref{globcurve.lem}.
Also, $\lambda_s(s) > 0$ and $\sigma(s) > 0$ for all $s \ge 0$,
and $\lambda_\infty = \infty$.
Hence, for each $\lambda > 0$, equation \eqref{bif.eq} has a solution
$(\lambda,u) \in \mathcal{S}_0$;
for fixed $\lambda$ this solution is unique
$($thus, $\mathcal{S} = \mathcal{S}_0)$.
\end{thm}
\begin{proof}
If $(\lambda,u) \in \mathcal{S}_0$ then it follows from \eqref{decf.eq} and the
positivity of $L$ that the operator $L - \lambda f'(u)$ is positive definite,
and hence non-singular,
so the surjectivity hypothesis in Lemma~\ref{globcurve.lem}
holds.
Next, if we had $\lambda_s(s) = 0$, for some $s>0$,
then \eqref{szderiv.eq} would hold, with
non-zero $u_s(s)$, which would contradict the non-singularity of
$L - \lambda f'(u(s))$.
Thus $\lambda_s(s) \ne 0$ for all $s \ge 0$, so by continuity we
must have $\lambda_s(s) > 0$
(since $\lambda_s(0) > 0$).
Similarly, $\sigma(s) > 0$ for all $s \ge 0$.
Next, by \eqref{decf.eq}, $|f(u(s))|_0 = f(0)$,
so it follows from \eqref{lafutoinfty.eq} that $\lambda_\infty = \infty$.
Finally, suppose that for some $\lambda > 0$ there exists $u_i \in X$
such that $L u_i = \lambda f(u_i)$, $i=1,\,2.$
If $u_1-u_2 \ne 0$ then by \eqref{decf.eq}
$$
0 < \langle L(u_1-u_2) , u_1-u_2 \rangle =
\lambda \langle f(u_1) - f(u_2) , u_1-u_2 \rangle \le 0,
$$
and this contradiction shows that we must have $u_1-u_2=0$.
\end{proof}
The following result shows that in this case the whole of $\mathcal{S}_0$
can be parametrised by $|u|_0$,
see Remark~\ref{param_by_norm.rem}.
\begin{cor} \label{param_by_norm.cor}
The function $|u(\cdot)|_0$ is strictly increasing on $\mathbb{R}_+$.
\end{cor}
\begin{proof}
For any $s \ge 0$ the operator $L - \lambda f'(u(s))$ is positive definite so
the results of
\cite{ELEV} apply to this operator (see Section~\ref{prelim.sec}),
and hence, by \eqref{sderiv.eq} and Corollary~1 of \cite{ELEV},
$Z(u_s(s)) = 0$.
Since $u_s(0)$ is positive, $u_s(s)$ must be positive for all $s \ge 0$,
which proves the result.
\end{proof}
\begin{cor} \label{asymptdec.cor}
$\mathcal{S}_0$ must look like the curve $(d)$ in Fig.~1.
All the solutions on $\mathcal{S}_0 = \mathcal{S}$ are stable.
\end{cor}
\begin{remark} \label{asymptdec.rem} \rm
It is not clear if
$\lim_{s \to \infty} |u(s)|_0 = \infty$,
in general in this case,
although this is true if $f(\xi) \ge \delta > 0$ for all $\xi \ge 0$.
To see this, choose a positive function $\phi \in X$, and observe that
$$
\langle u(s) , L \phi \rangle
=
\langle L u(s) , \phi \rangle
=
\lambda(s) \langle f(u(s)) , \phi \rangle
\ge
\delta \lambda(s) \langle 1 , \phi \rangle
\to \infty.
$$
\end{remark}
\section{S-shaped $\mathcal{S}_0$} \label{sshaped.sec}
Throughout this section we suppose that
\eqref{fdpos.eq} holds, so that the results of
Sections~\ref{prelim.sec} and~\ref{incf.sec} hold on $\mathcal{S}_0$,
and we will give a sufficient condition for $\mathcal{S}_0$
to be `S-shaped', that is, to have at least two turning points.
We first give a sufficient condition for a turning point to
occur in $\mathcal{S}_0$ which gives more explicit information on where the turning
point occurs than the necessary and sufficient condition given in
Theorem~\ref{turningpt.thm}.
This result is based on Theorem~1 in \cite{K3},
and the proof here is an adaptation of the proof in \cite{K3} to deal with
the higher order equation.
We will require the following notation.
For $\xi \ge 0$, let $F(\xi) = \int_0^\xi f(t)\,dt$
and
$G(\xi) = 2F(\xi) - \xi f(\xi)$.
Note that $G'(\xi) = g(\xi)$
(where $g(\xi) := f(\xi) - f'(\xi) \xi$
was used in Theorem~\ref{turningpt.thm}),
and $g'(\xi) = - f''(\xi) \xi$.
Clearly, $G(0) = 0$ and $G'(0) = g(0) >~0$.
\begin{lemma} \label{suffturningpt.lem}
Suppose that there exist numbers $\xi_0$, $\alpha$ such that $0 < \xi_0 < \alpha$
and
\begin{gather}
g(\xi) > 0, \quad 0 < \xi < \xi_0,
\quad \text{and}\quad
g(\xi) < 0, \quad \xi_0 < \xi < \alpha,
\label{scondone.eq}
\\
G(\alpha) \le 0.
\label{scondtwo.eq}
\end{gather}
Then for any $s_\alpha$ satisfying $|u(s_\alpha)|_0 = \alpha$,
we have $\lambda_s(s_\alpha) \le 0$
$($by \eqref{asymptga.eq}, there exists at least one such $s_\alpha)$.
\end{lemma}
\begin{proof}
Suppose, on the contrary, that $\lambda_s(s_\alpha) > 0$,
and so, by Theorem~\ref{laszero_stab.thm}, $\sigma(s_\alpha) > 0$.
For any $\tau > 0$
let
$\Phi_\tau := \tau \psi(s_\alpha) + u'(s_\alpha)$.
\begin{lemma} \label{sshaped.lem}
For any $\tau > 0$,
$Z(\Phi_\tau) = 1$,
and the zero of $\Phi_\tau$ is simple.
\end{lemma}
\begin{proof}
From \eqref{bc.eq}, \eqref{um_non_zero.eq}
and Theorem~\ref{symm.thm},
$\Phi_\tau$ satisfies
\begin{gather}
\Phi_\tau^{(i)}(\pm 1) = 0, \quad i=0,\dots,m-2,
\label{Phibcz.eq}
\\
\Phi_\tau^{(m-1)}(\pm 1) = u^{(m)}(s_\alpha)(\pm 1) \ne 0,
\label{Phibcnonz.eq}
\\
\Phi_\tau(0) = \tau \psi(s_\alpha)(0) > 0.
\label{Phiznonz.eq}
\end{gather}
It follows from \eqref{Phibcnonz.eq} that there exists $\delta_\tau > 0$
such that
\begin{equation} \label{Phi_sign_chnge.eq}
\pm \Phi_\tau(x) > 0, \quad 0 < |x \pm 1| < \delta_\tau.
\end{equation}
Hence, by \eqref{Phiznonz.eq}, $\Phi_\tau$ changes sign in $(0,1)$ and
$S(\Phi_\tau) \ge 1$.
Next, from \eqref{sieval.eq} and \eqref{udequn.eq},
$\Phi_\tau$ satisfies the differential equation
$$
\widetilde L_\alpha \Phi_\tau := (\widetilde L - \lambda(s_\alpha) f'(u(s_\alpha))) \Phi_\tau
=
\tau \sigma(s_\alpha) \psi(s_\alpha).
$$
Furthermore, since $\sigma(s_\alpha) > 0$, the operator $\widetilde L_\alpha$ is disconjugate,
so the results of \cite{ELEV} apply to this equation
(see Section~\ref{prelim.sec} above).
In particular, by (6) in \cite{ELEV}, we have
\begin{equation} \label{sSinequ.eq}
S(\Phi_\tau) \le S(\widetilde L_\alpha \Phi_\tau) + 2m - N_{2m}(\Phi_\tau)
= 2m - N_{2m}(\Phi_\tau)
\le 2
\end{equation}
(since $\psi(s_\alpha)$ is positive,
and the boundary conditions \eqref{Phibcz.eq} imply that
$N_{2m}(\Phi_\tau) \ge 2m-2$, see \cite{ELEV}).
Now, if $\Phi_\tau$ had a double zero in $(-1,1)$ then this would
contribute a further 2 to $N_{2m}(\Phi_\tau)$, so that
$S(\Phi_\tau) = 0$, but we already know
that $S(\Phi_\tau) \ge 1$, so $\Phi_\tau$ can only have simple zeros
in $(-1,1)$.
Thus, $Z(\Phi_\tau) = S(\Phi_\tau) \ge 1$ and, by \eqref{Phi_sign_chnge.eq},
$Z(\Phi_\tau)$ must be odd,
so by \eqref{sSinequ.eq}
we have $Z(\Phi_\tau) = 1$.
\end{proof}
By Theorem~\ref{symm.thm} we can now define $x_\alpha \in (0,1)$ by
$u(s_\alpha)(x_\alpha) = \xi_0$,
and let
$\tau_\alpha := - u'(s_\alpha)(x_\alpha) / \psi(s_\alpha)(x_\alpha) > 0$.
Then
$\Phi_{\tau_\alpha}(x_\alpha) = 0$
and
$g(u(s_\alpha)(x_\alpha)) = 0$,
so from \eqref{scondone.eq} and Lemma~\ref{sshaped.lem},
we have
\begin{equation*}
\begin{split}
g(u(s_\alpha)) < 0\ \text{ and }\ \tau_\alpha \psi(s_\alpha) + u'(s_\alpha) &> 0, \quad
\text{on $(0,x_\alpha)$},
\\
g(u(s_\alpha)) > 0\ \text{ and }\ \tau_\alpha \psi(s_\alpha) + u'(s_\alpha) &< 0, \quad \text{on $(x_\alpha,1)$}.
\end{split}
\end{equation*}
Hence,
\begin{align*}
\frac{\tau_\alpha}{2} \langle g(u(s_\alpha)) , \psi(s_\alpha) \rangle
&=
\tau_\alpha \int_0^1 g(u(s_\alpha)) \psi(s_\alpha) \,dx
<
\int_0^1 g(u(s_\alpha)) (-u'(s_\alpha)) \,dx
\\&=
- \int_0^1 \frac{d}{dx} G(u(s_\alpha)) \,dx
=
G(\alpha) \le 0
\end{align*}
(using Theorem~\ref{symm.thm}).
On the other hand, taking the inner product of \eqref{sieval.eq} with
$u(s_\alpha)$ and the inner product of \eqref{bif.eq} with $\psi(s_\alpha)$,
and subtracting, yields
$$
\lambda(s_\alpha) \langle g(u(s_\alpha)) , \psi(s_\alpha) \rangle
=
\sigma(s_\alpha) \langle u(s_\alpha) , \psi(s_\alpha) \rangle \ge 0.
$$
This contradiction proves that we must have $\lambda_s(s_\alpha) \le 0$.
\end{proof}
We now give conditions for $\mathcal{S}_0$ to be S-shaped,
in the sense of the following theorem.
\begin{thm} \label{sshaped.thm}
Suppose that \eqref{scondtwo.eq} holds, for some $\alpha > 0$,
and that
\begin{equation}
f''(\xi) > 0, \quad 0 < \xi \le \alpha.
\label{scondthree.eq}
\end{equation}
Suppose also that $\gamma_\infty$
exists, with $0 \le \gamma_\infty \le f'(0)$.
Then there exists
$t_1,\,t_2$ such that:
$(i)$ $0 < t_1 < t_2;$
$(ii)$ $\lambda_s(t_1) = \lambda_s(t_2) = 0;$
$(iii)$ $0 < \lambda(t_2) < \lambda(t_1) < \sigma_0/f'(0);$
$(iv)$ $\lambda_\infty = \sigma_0/\gamma_\infty > \lambda(t_1)$.
\end{thm}
\begin{proof}
It follows from \eqref{scondtwo.eq} and \eqref{scondthree.eq}
that $g(\alpha) < 0$ and there exists $\xi_0 \in (0,\alpha)$ such that
\eqref{scondone.eq} holds, and hence the result of
Lemma~\ref{suffturningpt.lem} holds.
Thus we can define $t_1 = \inf \{ s \ge 0 : \lambda_s(s) \le 0 \}$,
and we have $|u(s)|_0 \le \alpha$ for $s \in [0,t_1]$.
It now follows from \eqref{scondthree.eq} that
\eqref{fddpos.eq} holds for $0 < \xi \le \alpha$,
and so the results of Section~\ref{convexf.sec} hold for $s \in [0,t_1]$.
Thus $\lambda_s(t_1) = 0$, and it follows from Lemma~\ref{turn_pt.lem}
that $\lambda(t_1) < \sigma_0/f'(0)$,
and from Lemma~\ref{lass.lem} that $\lambda_s(s) < 0$
for $s-t_1>0$ sufficiently small.
Also, Theorem~\ref{asymptga.thm} shows that $\lambda_\infty = \sigma_0/\gamma_\infty$.
Hence, letting $t_2$ be the point at which $\lambda(\cdot)$ attains its
minimum on the interval
$[t_1,\infty)$,
the results (i)--(iv) follow immediately.
\end{proof}
\begin{remark} \rm
Figs. 1~(a) and~(b) on p.~1012 of \cite{K2} illustrate the above result
when $\mathcal{S}_0$ is `exactly S-shaped'
(that is, when $\mathcal{S}_0$ has exactly two turning points).
Fig.~(a) corresponds to $\gamma_\infty = 0$,
while Fig.~(b) corresponds to $\gamma_\infty > 0$.
Exact S-shapedness is also obtained in \cite{WAN}.
We have been unable to obtain exact S-shapedness of $\mathcal{S}_0$ here --- the
arguments in \cite{K2} and \cite{WAN} rely on the problem being second order.
\end{remark}
It follows from Theorem~\ref{sshaped.thm} that if
$\lambda \in (\lambda(t_2), \lambda(t_1))$ then equation \eqref{bif.eq}
has at least three solutions $(\lambda,u(s_i)) \in \mathcal{S}_0$, $i=1,\,2,\,3,$
with $s_1 < t_1 < s_2 < t_2 < s_3$ and,
by Theorem~\ref{laszero_stab.thm} and the geometry of $\mathcal{S}_0$,
$(\lambda,u(s_1))$ will be stable,
$(\lambda,u(s_2))$ will not be stable,
and $(\lambda,u(s_3))$ (on the `upper' branch) will be stable if $\lambda_s(s_3)>0$
(that is, the curve $\mathcal{S}_0$ is moving to the right at this point).
We cannot rule out the possibility of more turning points, or
`vertical' points where $\lambda_s = 0$
(so that the corresponding solution is not stable).
However, we have the following corollary.
\begin{cor} \label{sshaped.cor}
If the hypotheses of Theorem~\ref{sshaped.thm} hold
then, for almost all $\lambda \in (\lambda(t_2), \lambda(t_1))$,
equation \eqref{bif.eq}
has at least two stable solutions.
\end{cor}
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\end{document}