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\markboth{\hfil A one dimensional Hammerstein problem \hfil}%
{\hfil Jun Hua \& James L. Moseley\hfil}
\begin{document} \setcounter{page}{47}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc 15th Annual Conference of Applied Mathematics, Univ. of Central Oklahoma},
\newline
Electronic Journal of Differential Equations, Conference~02, 1999, pp. 47--60. 
\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu (login: ftp)}
\vspace{\bigskipamount} \\
A one dimensional Hammerstein problem  
\thanks{ {\em 1991 Mathematics Subject Classifications:} 35P30.
\hfil\break\indent
{\em Key words and phrases:} Hammerstein problem, Nonlinear eigenvalue problem.
\hfil\break\indent
\copyright 1999 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Published November 24, 1999.} }

\date{}
\author{Jun Hua \& James L. Moseley}
\maketitle

\begin{abstract}
Nonlinear equations of the form $L[u]=\lambda g(u)$ where $L$ is a linear
operator on a function space and $g$ maps $u$ to the composition function 
$g\circ u$ arise in the theory of spontaneous combustion. If $L$ is
invertible, such an equation can be written as a Hammerstein equation, 
$u=B[u]$ where $B[u]=\lambda L^{-1}[g(u)]$. To investigate the importance of
the growth rate of $g$ and the sign and magnitude of $\lambda $ on the
number of solutions of such problems, in this paper we consider the
one-dimensional problem $L(x)=\lambda g(x)$ where $L(x)=ax$.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}

\section{Introduction}

We wish to investigate the number of solutions (and their computation) to problems of the form
\begin{equation}
L[u]=\lambda g(u)  \label{1.1}
\end{equation}
where $L:V\to W$ is a linear operator and $V$ and $W$ are
function spaces whose domains are the same set, say $D$, and whose codomains
are the real numbers ${\mathbb R}$. If $u\in V$ and $x\in D$, then the value
of the function $g(u)$ at $x$ is $g(u(x))$ where $g:{\mathbb R}\to 
{\mathbb R}$. Thus we use the symbol $g$ for a real valued function of a real
variable as well as for the (nonlinear Nemytskii) operator from $V$ to $W$
that this function defines by the composition $g\circ u$. An example is
\begin{eqnarray}
-\Delta u =\lambda e^u && \vec{x}=[x,y,z]^{T}\in \Omega \subseteq {\mathbb R}^3 
\label{1.2} \\ 
u(\vec{x})=0  && \vec{x}\in \partial \Omega \nonumber
\end{eqnarray}
Here $L$ is the negative of the Laplacian operator in three
spatial dimensions with homogeneous Dirichlet boundary conditions, $\vec{x}$
is a point in ${\mathbb R}^3$, $\Omega $ is an open connected region in 
${\mathbb R}^3$, $\partial \Omega $ is its boundary, $V=\{u\in V_{1}:u(\vec{x%
})=0$ $\forall $ $\vec{x}\in \partial \Omega \}$ where $V_{1}=C^{2}(\Omega ,%
{\mathbb R})\cap C(\bar{\Omega},{\mathbb R})$, and $W=C(\bar{\Omega},{\mathbb R}%
)$. \ Hence $D=\bar{\Omega}$. Such problems arise in the theory of
combustion (Deberness \cite{d1},  Moseley \cite{m1, m2, m3, m4}). 
For this problem in ${\mathbb R}^{2}$, it is known that for $\lambda
<0 $, there exists a unique solution. However, for $\lambda >0$ and large,
there is no solution. But for $\lambda >0$ and small, there are at least two
solutions. If $\lambda =0$, the solution set is the null space of $L$ and
since $L$ is invertible the problem has only the trivial solution $u=0$.
If $g(u)=u$, (\ref{1.1}) is a spectral problem for $L$. Hence (\ref{1.1}) is sometimes
referred to as a nonlinear eigenvalue problem.

If $L$ is invertible, then (\ref{1.1}) can be written as the Hammerstein equation 
\begin{equation}
u-\lambda L^{-1}[g(u)]=0.  \label{1.3}
\end{equation}
A solution of (\ref{1.3}) is a fixed point of the combined operator 
$B=\lambda (L^{-1}\circ g)$. The level of difficulty of problems of type
(\ref{1.1}) or (\ref{1.3}) varies greatly depending on the number of elements in $D$,
the value of $n$, and the operator $L$. We list several categories, starting
with the easiest.

\begin{enumerate}
\item  One dimensional problems (i.e., $D$ contains only one element).

\item  Multidimensional problems (i.e., $D$ is finite, but contains two or
more elements).

\item  Infinite dimensional problems with $D\subseteq {\mathbb R}$ $(n=1)$
and $L$ a first, second, or higher order differential operator.

\item  Infinite dimensional problems with $D$ $\subseteq {\mathbb R}^{n}$ , 
$n=2,3,4,\dots$ and $L$ a first, second, or higher order partial differential
operator. 
\end{enumerate}

\noindent Since $L$ is linear, we at most have linear coupling and often
this coupling is weak. The coupling of $L^{-1}$ may be stronger than the
coupling of $L$ and is a reason to examine (\ref{1.1}) directly even when $L$ is
invertible. To investigate the fundamental importance of the growth rate of 
$g$ and the sign and magnitude of $\lambda $ on the number of solutions to
problems of this type, in this paper we consider the one dimensional
nonlinear eigenvalue problem
\begin{equation}
ax=\lambda g(x).  \label{1.4}
\end{equation}
Here $L: {\mathbb R\to R}$ is $L(x)=ax$, $g:{\mathbb R }\to {\mathbb R}$ 
is a continuous function and $a$ and $\lambda $ are
parameters. (If $a\neq 0$, $L$ is invertible.) To this end, we first
consider two types of behavior for a continuous function $f: {\mathbb R}
\to {\mathbb R}$ (i.e., $f\in C({\mathbb R},{\mathbb R})=\{f:{\mathbb R}
\to {\mathbb R}:f$ is continuous $\forall x\in $ ${\mathbb R}\}$). If $L$ is
invertible $(a\neq 0)$, we take $f(x)=x-kg(x)$ where $k=\lambda /a.$
Although less restrictive conditions on $f$ can be obtained, for convenience
we assume that $f$ has a continuous second derivative for all $x$ in 
${\mathbb R}$; that is, $f\in C^{2}({\mathbb R, R})=\{f:{\mathbb R \to R} 
:f''(x)$ exists and is continuous $\forall x\in {\mathbb R}
$\}. We are interested in sufficient conditions on $f$ that will determine
the number of solutions of the equation
\begin{equation}
f(x)=0.  \label{1.5}
\end{equation}
The obvious advantage of considering the scalar equation (one dimensional
problem) (\ref{1.4}) or (\ref{1.5}) over an abstract Hammerstein equation or a
Hammerstein equation of the type (\ref{1.1}) where $D$ is finite or infinite
dimensional is that much more (often everything) can be said for many
functions $g(x)$ (and classes of functions). However, the techniques
investigated here are quite different from the standard fixed point theorems
(e.g., contraction mapping theorem and the Brouwer and Schauder fixed point
theorems) and are expected reveal distinctive new results when extended to
higher dimensions including methods for solving multidimensional problems.

\section{Linear and quadratic properties}

In addition to $f\in C^{2}({\mathbb R},{\mathbb R})$ we also assume some of the
following:
\begin{description}
\item{H1} $\lim_{x\to\infty}f(x)=+\infty$
\item{H2} $\lim_{x\to\infty}f(x)=-\infty$
\item{H3} $\lim_{x\to-\infty}f(x)=-\infty$
\item{H4} $\lim_{x\to-\infty}(x)=+\infty$
\item{H5} $f'(x)>0$
\item{H6} $f'(x)<0$
\item{H7} $f''(x)>0$
\item{H8} $f''(x)<0$
\end{description}
We consider properties that $f:{\mathbb R\to R}$ may have
that are also possessed by the linear function $F(x)=ax+b$ $(a\neq 0)$:

\paragraph{Definition}
If $f(x)$ satisfies H1 and H3, we say that it is \textbf{mainly increasing}.
If it satisfies H1, H3 and H5, we say that it is \textbf{consistently
increasing}. On the other hand, if $f(x)$ satisfies H2 and H4, we say it is 
\textbf{mainly decreasing}. If it satisfies H2, H4, and H6, we say it is 
\textbf{consistently decreasing}. If $f(x)$ is mainly increasing or
decreasing, then we say $f(x)$ is \textbf{mainly monotonic}. If $f(x)$ is
consistently increasing or decreasing, then we say $f(x)$ is \textbf{%
consistently monotonic}.
\medskip

For completeness we review a theorem that establishes the existence and
uniqueness of solutions to (\ref{1.5}) if $f(x)$ is mainly or consistently
monotonic. Like the Jordan Curve Theorem, it is geometrically obvious and an
analytic proof can be given \cite{m7}.

\begin{theorem}
If a function $f$ is mainly monotonic, then for any $c$ in ${\mathbb R}$,
there exists at least one $x$ in ${\mathbb R}$ such that $f(x)=c$. If the
function is consistently monotonic, then for any $c$ in ${\mathbb R}$, there
exists exactly one $x$ in ${\mathbb R}$ such that $f(x)=c$.
\end{theorem}

If $f(x)$ is consistently monotonic it is similar to the linear
function $F(x)=ax+b$ $(a\neq 0)$ in that (\ref{1.5}), like $F(x)=0$, has exactly
one solution. We say that $f(x)$ has the \textbf{linear property}.
We also consider some properties that $f(x)$ may have that are also
possessed by the quadratic function $G(x)=ax^{2}+bx+c$ $(a\neq 0).$

\paragraph{Definition}
If $f(x)$ satisfies H1,H4 (H2,H3), we say $f(x)$\textbf{ opens upwards (downwards)}.
If $f(x)$ satisfies H1, H4, H7 (H2, H3, H8), we say $f(x)$ is 
\textbf{completely concave up (down)}.
\medskip

Again for completeness, we review two theorems that establish when (\ref{1.5}) has
exactly zero, one, or two solutions. \ Again they are geometrically obvious
and can be proved analytically \cite{m7}.

\begin{theorem}
If $f$ opens upwards (downwards), then there exists at least one $x_{0}\in R$
such that $f(x_{0})=min f(x)\quad (f(x_{0})=\max f(x))$.
\end{theorem}

\begin{theorem}
If $f$ opens upwards (downwards) and $m=f(x_{0})=\min$ $f(x)$ 
$(m=f(x_{0})=\max f(x))$, then 
\begin{enumerate}
\item $c<m$ $(c>m)$  implies that there is no $x$  such that $f(x)=c$, 
\item $c=m$ implies that there is at least one $x$  such that $f(x)=c$, namely $x=x_{0}$, 
\item $c>m$ $(c<m)$ implies that there are at least two (distinct) values of $x$ 
for which $f(x)=c$.
\end{enumerate}
\end{theorem}

If $f(x)$ is completely concave up or down, we see that $f(x)$ has
similar properties to the quadratic function $G(x)=ax^{2}+bx+c$ $(a\neq 0)$
in that (\ref{1.5}), like $G(x)=0$, has exactly $0$, $1$, or $2$ solutions. We say 
$f(x)$ has the \textbf{quadratic property}.

\section{Zero and nonzero values for the parameters} \setcounter{equation}{0}

Returning to (\ref{1.4}) we consider whether the values for the parameters $a$ and 
$\lambda $ are zero or not. Recall that if $\lambda =0$ the solution set of
(\ref{1.1}) is just the null space of $L$. If $L$ is the zero operator and 
$\lambda \neq 0$, then solutions are those functions $u$ which map all values
of $x$ into the zeros of $g$. If $L$ is the zero operator and $\lambda =0$,
then all $u$ in $V$ are solutions. For completeness, we stand these results
explicitly for (\ref{1.4}).

\begin{theorem}
If $a=\lambda =0$, then $\forall x\in {\mathbb R}$, $x$ is a solution. If 
$a\neq 0$, $\lambda =0$, we have $x=0$ is the unique solution. If $a=0$, 
$\lambda \neq 0$, then $x$ is a solution of (\ref{1.4}) if and only if $x$ is a
solution of $g(x)=0$.
\end{theorem}

From now on we assume $a\neq 0$ and $\lambda \neq 0$. We may then
divide by $a$ and have one nonzero parameter.

\begin{theorem}
If $a\neq 0$, $\lambda \neq 0$, the solution set of $ax=\lambda g(x)$ is the
same as the solution set of
\begin{equation}
x=kg(x)  \label{3.1}
\end{equation}
where $k=\lambda /a$.
\end{theorem}

\section{Conditions for infinite limits as $x\to \pm\infty $}
\setcounter{equation}{0}
We let 
\begin{equation}f(x)=x-kg(x)(k\neq 0)  \label{4.1}
\end{equation}
which we may also write as 
\begin{equation}
f(x)=x-kg(x)=x\left( 1-\frac{kg(x)}{x}\right) 
=g(x)\left( \frac{x}{g(x)}-k\right),  \label{4.2}
\end{equation}
provided $x\neq 0$ and $g(x)\neq 0$. From this it is easy to
establish that $\lim_{x\to\infty}f(x)=\infty $ for
the following conditions. Although less restrictive conditions can be
stated, for convenience we consider only limit conditions on $g(x)$, $g(x)/x$,
and $x/g(x)$ that are finite or $\pm \infty $.

\begin{description}
\item{PP1} $\lim_{x\to\infty}g(x)=-\infty $, $\lim_{x\to\infty}x/g(x)=M$,
$M \in {\mathbb R}$ and $M<k$ (e.g. $g(x)=-1/3x+2, k=-1)$

\item{PP2} $\lim_{x\to\infty}g(x)=-\infty $ and $k>0.$
(e.g. $g(x)=-e^x$ or $g(x)=-x^{2}$ and $k=2$)

\item{PP3} $\lim_{x\to\infty}g(x)=M$,  (e.g. $g(x)=$ Arctan $x$)

\item{PP4} $\lim_{x\to\infty}g(x)=\infty $ and $k<0$.  (e.g. $g(x)=e^{x}$ or $g(x)=x^{2}$ and $k=-1$)

\item{PP5} $\lim_{x\to\infty}g(x)=\infty $ and $\lim_{x\to\infty}x/g(x)=\infty $.
(e.g. $g(x)=x^{1/3}$)

\item{PP6} $\lim_{x\to\infty}g(x)/x=M$, and $kM<1$. (e.g. $g(x)=3x+2$ and $k=-1$)

\item{PP7} $\lim_{x\to\infty}g(x)=\infty $,  $\lim_{x\to\infty}x/g(x)=M$, and $M>k.$
(e.g. $g(x)=3x+2$ and $k=-1$)
\end{description}

To obtain conditions such that $\lim_{x\to\infty}f(x)= -\infty $ we must completely counteract the effect of the linear term.

\begin{description}
\item{PN1}  $\lim_{x\to\infty}g(x)=-\infty $, $\lim_{x\to\infty}x/g(x)=M$, and $M>k$.
(e.g. $g(x)=-3x+2$ and $k=-2$)

\item{PN2} $\lim_{x\to\infty}g(x)/x=M$, and $kM>1$. (e.g. $g(x)=2x+$ Arctan $x$
and $k=1$)

\item{PN3} $\lim_{x\to\infty}g(x)=\infty $, $\lim_{x\to\infty}x/g(x)=M$ and 
$M<k$.  (e.g. $g(x)=3x^{2}+4$ and $k=1$)
\end{description}

\noindent Similar to the conditions above, we can obtain properties as $x$
approaches $-\infty $. It is easy to establish that $\lim_{x\to-\infty}f(x)=-\infty $ 
for the following conditions.

\begin{description}
\item{NN1} $\lim_{x\to-\infty}g(x)=\infty $, $\lim_{x\to-\infty}x/g(x)=M$ and $M<k$.
(e.g. $g(x)=-1/3x$, $k=-1$ or $g(x)=e^{-x}$, $k=1$)

\item{NN2} $\lim_{x\to-\infty}g(x)=\infty $ and $k>0$.  (e.g. $g(x)=e^{-x}$ or $g(x)=-x^{2}$ and $k=1$)

\item{NN3} $\lim_{x\to-\infty}g(x)=M$  (e.g. $g(x)=e^{x}$ or $g(x)=$ Arctan $x$)

\item{NN4} $\lim_{x\to-\infty}g(x)=-\infty $ and $k<0$. (e.g. $g(x)=-e^{-x}$ or 
$g(x)=-x^{2}$ and $k=-1$)

\item{NN5} $\lim_{x\to-\infty}g(x)=-\infty $ and $\lim_{x\to-\infty}x/g(x)=\infty $. 
(e.g. $g(x)=x^{1/3}$)

\item{NN6} $\lim_{x\to-\infty}g(x)/x=M$ and $kM<1$.  (e.g. $g(x)=3x+2$ and $k=-2$)

\item{NN7} $\lim_{x\to-\infty}g(x)=-\infty $,  $\lim_{x\to-\infty}x/g(x)=M$, and 
$M>k$.  (e.g. $g(x)=3x+2$ and $k=-1$)
\end{description}

\noindent Finally, we can establish that $\lim_{x\to -\infty }f(x)=\infty $ 
for the following conditions.

\begin{description}
\item{NP1} $\lim_{x\to-\infty}g(x)=-\infty $, $\lim_{x\to-\infty}x/g(x)=M$, 
and $M<k$. (e.g. $g(x)=3x+2$ and $k=2$)

\item{NP2} $\lim_{x\to-\infty}g(x)/x=M$ and $kM>1$.
(e.g. $g(x)=2x+$Arctan$(x)$ and $k=1$)

\item{NP3} $\lim_{x\to-\infty}g(x)=\infty $,
$\lim_{x\to-\infty}x/g(x)=M$ and $M>k$.
(e.g. $g(x)=3x^{2}+4$ and $k=-1$)
\end{description}

\section{Linear property} \setcounter{equation}{0}

For $f(x)$ given by (\ref{4.1}), we consider conditions such that$f(x)$ is mainly
increasing (and mainly decreasing). We then add conditions so that $f(x)$ is
consistently increasing (and consistently decreasing) so that it has the
linear property. We begin by looking at conditions so that $f(x)$ is mainly
increasing when $k>0$. Of those conditions such that
$\lim_{x\to -\infty}f(x)=\infty $, only PP4 explicitly
requires $k<0$ and must be eliminated. For PP1 we note

\begin{theorem}
If $\lim_{x\to\infty}g(x)=-\infty $ and $\lim_{x\to\infty}x/g(x)=M$,
then $M\leq 0$.
\end{theorem}

\paragraph{Proof.}
$\lim_{x\to\infty}g(x)=-\infty $, then there
exists $M_{0}>0$ such that $\forall x>M_{0}$, $g(x)<0$. Hence $\forall
x>M_{0}$, $x/g(x)<0$. Hence $\lim_{x\to\infty}x/g(x)=M\leq 0$.
\bigskip

Since we are assuming $k>0$, the requirement that $k>M$ in PP1 is
superfluous. Also assuming $k>0$ allows us to combine PP2 and PP3 into one
statement. For emphasis, we modify the new statement (PPKP2) as well as PP1
and PP7 (PPKP1 and PPKP5) to include the constraint $k>0$. \ The sign of $k$
has no impact on PP5 (PPKP3). The impact of $k>0$ on PP6 (PPKP4) is examined
later. We do a similar reorganization for those cases for which 
$\lim_{x\to -\infty }f(x)=-\infty $. Hence we have the
following sufficient conditions for the function $f(x)=x-kg(x)$ to be
mainly increasing.

\begin{theorem}
Suppose $k>0$ and one of the following hypothesis holds:
\begin{description}
\item{PPKP1} $\lim_{x\to\infty}g(x)=-\infty $, 
$\lim_{x\to-\infty}x/g(x)=M$, and $k>0$ $(\geq M)$. (PP1)

\item{PPKP2} $\lim_{x\to\infty}g(x)=-\infty $
or $M>0$  and $k>0$. (PP2 or PP3)

\item{PPKP3} $\lim_{x\to\infty}g(x)=\infty $ and 
$\lim_{x\to\infty}x/g(x)=\infty $. (PP5)

\item{PPKP4} $\lim_{x\to\infty}x/g(x)=M$ and $kM>1$. (PP6)

\item{PPKP5} $\lim_{x\to\infty}g(x)=\infty $,
$\lim_{x\to\infty}x/g(x)=M$ and $M>k>0$. (PP7).
\end{description}

\noindent Suppose further that one of the following hypothesis holds
\begin{description}
\item{NNKP1} $\lim_{x\to-\infty}g(x)=\infty$,
$\lim_{x\to-\infty}x/g(x)=M$, and $k>0$ ($k\geq M$). (NN1)

\item{NNKP2} $\lim_{x\to-\infty}g(x)=\infty $ or $M>0$ and $k>0$.
(NN2 or NN3)

\item{NNKP3} $\lim_{x\to-\infty}g(x)=-\infty $
and $\lim_{x\to-\infty}x/g(x)=\infty $. (NN5)

\item{NNKP4} $\lim_{x\to-\infty}x/g(x)=M$ and $kM<1$. (NN6)

\item{NNKP5} $\lim_{x\to -\infty}g(x)=\infty $, 
$\lim_{x\to-\infty}x/g(x)=M$, and $M>k>0$. (NN7)
\end{description}

\noindent Then $f$  given by (\ref{4.1}) is mainly
increasing.\end{theorem}

Note that by requiring $k>0$, PPKP1 is now contained in PPKP2 and
can be eliminated. Similarly  for NNKP1. To consider PPKP4 and PPKP5, we
note that these depend on the growth rate of $g(x)$ as compared with $x$. If 
$\lim_{x\to\infty}g(x)/x=M$ and $M\neq 0$, then the growth rate of $g(x)$
is essentially linear. In this case
the linear part of $g(x)$ can be moved to the other side of the equation and
combined with $x$. Hence we really wish to consider only the case when
$M=0$. But if $M=0$ and $k>0$, neither PPKP4 or PPKP5 is possible. We consider
the details. Suppose $\lim_{x\to\infty}g(x)/x=M$, $M\neq 0$ and
$\lim_{x\to-\infty}g(x)/x=N$, $N\neq 0$. We then consider
\begin{equation}
g_{1}(x)=\left\{ \begin{array}{ll}
g(x)-Mx & x>0 \\ 
g(x)-Nx & x<0
\end{array}
\right.   \label{5.1}
\end{equation}

\begin{theorem}
Suppose $\lim_{x\to\infty}g(x)/x=M$, $M\neq 0$ and
$\lim_{x\to-\infty}g(x)/x=N$, $N\neq 0$. Let $g_{1}(x)$ be given by (\ref{5.1}). \\
Then $\lim_{x\to\infty}g_{1}(x)/x=0$,
$\lim_{x\to -\infty}g_{1}(x)/x=0$, \\
$\lim_{x\to\infty}g_{1}(x)/g(x)=0$, and
$\lim_{x\to-\infty}g_{1}(x)/g(x)=0$. 
\end{theorem}

\paragraph{Proof.}
$\lim_{x\to\infty}g_{1}(x)/x=\lim_{x\to\infty }(g(x)-Mx)/x
=\lim_{x\to\infty}(g(x)/x-M)=M-M=0$.
Similarly for $\lim_{x\to-\infty}g_{1}(x)/x
=\lim_{x\to\infty}g_{1}(x)/g(x)=0$,\\
$\lim_{x\to\infty }(g(x)Mx)/g(x)
=\lim_{x\to \infty }(1-M/(g(x)/x))=\lim_{x\to\infty}(1-M/M)=0$.
Similarly for $\lim_{x\to-\infty}g_{1}(x)/g(x)$. \hfill$\diamondsuit$
\medskip

Thus when $\lim_{x\to\infty}g(x)/x=M$, $M\neq 0$ and/or
$\lim_{x\to-\infty}g(x)/x=N$, $N\neq 0$, we rewrite (\ref{3.1}) as
$$x+k(g_{1}(x)-g(x))=kg_{1}(x).$$
Hence $f$ can be written as
$f(x)=x+k(g_{1}(x)-g(x))-kg_{1}(x)=x(1+k(g_{1}(x)-g(x))/xkg_{1}(x)/x)
=g_{1}(x)(x/g_{1}(x)+k))$.
Conditions can now be obtained so that $f(x)$ approaches $\pm \infty $
as $x$ approaches $\pm \infty $ using $g_{1}(x)$ instead of $g(x)$. However,
care must be taken since $x+k(g_{1}(x)-g(x))$ need not be linear, but is
only guaranteed to be ``piecewise'' linear. Also $g_{1}(x)$ is continuous at 
$x=0$, but not differentiable unless $M=N$. This is only a minor
inconvenience and well worth the elimination of a linear component in $g$ by
moving it to the other side of the equation. This is a small price to pay to
reap the benefits of the function on the right hand side having a truly
nonlinear growth rate. We leave such issues to future work and from now on
only allow $\lim_{x\to \infty}g(x)/x=M$ where
$M=0$ or $M=\pm \infty $ and $\lim_{x\to -\infty }g(x)/x=N$
where $N=0$ or $N=\pm \infty $. We now modify Theorem 7 to consider
only these possibilities.

\begin{theorem}
Suppose $k>0$ and one of the following hypothesis holds: 
\begin{description}
\item{PPKPN1} $\lim_{x\to\infty}g(x)$ $=-\infty $ or 
$0$, (e.g., $g(x)=-e^x$ or $g(x)=-x^3$),\ 

\item{PPKPN2} $\lim_{x\to\infty}g(x)$ $=$ $\infty $
and $\lim_{x\to\infty}x/g(x)=\infty $. (e.g., $g(x)=x^{1/3}$). 
\end{description}

\noindent Suppose further that one of the following
hypothesis holds: 
\begin{description}
\item{NNKPN1} $\lim_{x\to-\infty}g(x)=\infty $ or 
$0$, (e.g., $g(x)=-e^x$or $g(x)=-x^3$),

\item{NNKPN2} $\lim_{x\to-\infty}$ $g(x)=-\infty $
and $\lim_{x\to-\infty}x/g(x)=\infty $, (e.g.,  $g(x)=x^{1/3}$). 
\end{description}
\noindent Then the function $f$ given by (\ref{4.1}) is mainly increasing.
\end{theorem}

\noindent Similarly, we consider mainly increasing functions $f(x)$ when 
$k<0 $. 

     
\begin{theorem}
Suppose $k<0$ and one of the following hypothesis holds: 
\begin{description}
\item{PPKNN1} $\lim_{x\to\infty}g(x)=\infty $ or $0$,
(e.g., $g(x)=e^x$ or $g(x)=x^3$),

\item{PPKNN2} $\lim_{x\to\infty}g(x)=-\infty $ and 
$x/g(x)=-\infty $. (e.g., $g(x)=-x^{1/3}$). 
\end{description}
\noindent Suppose further that one of the following
hypothesis holds:
\begin{description}
\item{NNKNN1} $\lim_{x\to-\infty}$ $g(x)=-\infty $
or $\lim_{x\to-\infty}$ $g(x)=0$, (e.g., $g(x)=e^x$or $g(x)=x^3$). 

\item{NNKNN2} $\lim_{x\to-\infty}$ $g(x)=\infty $
and $\lim_{x\to-\infty}x/g(x)=-\infty $ (e.g., 
$g(x)=-x^{1/3}$). 
\end{description}
Then the function $f$ given by (\ref{4.1}) is mainly increasing.
\end{theorem}


We now consider functions $f(x)$ that are mainly decreasing. Following the
previous procedure, we retain only the cases where 
$\lim_{x\to\infty }g(x)/x=M$ where $M=0$ or $\pm \infty $ and 
$\lim_{x\to -\infty }g(x)/x=N$ where $N=0$ or $\pm \infty $. We
first consider the case of $k>0$. This eliminates PN1 and PN2 as well as NP2
and NP3.

\begin{theorem}
Suppose $k>0$, if 
\begin{description}
\item{PNKPN1} $\lim_{x\to\infty}g(x)=\infty $, 
$\lim_{x\to\infty}x/g(x)=0$  (e.g., $g(x)=\sinh x$)
and

\item{NPKPN1} $\lim_{x\to-\infty}g(x)=-\infty$, 
$\lim_{x\to-\infty}x/g(x)=0$, (e.g., $g(x)=\sinh x$)
\end{description}
\noindent then $f(x)$ is mainly decreasing.
\end{theorem}

Now, we consider the case $k<0$. \ 

\begin{theorem}
Suppose $k<0$, if 
\begin{description}
\item{PNKNN1} $\lim_{x\to\infty}g(x)=-\infty$, 
$\lim_{x\to\infty}x/g(x)=0$ (e.g., $g(x)=\sinh x$) and 

\item{NPKNN1} $\lim_{x\to-\infty}g(x)=\infty $,
$\lim_{x\to-\infty}x/g(x)=0$, (e.g., $g(x)=-\sinh x$)
\end{description} 
\noindent then $f(x)$ is mainly decreasing.
\end{theorem} 


\begin{corollary}
If $a\neq 0$, $\lambda \neq 0$ and 

$k>0$, PPKPN1 or 2, and NNKPN1 or 2 (mainly increasing) \newline
or

$k<0$, PPKNN1 or 2, and NNKNN1 or 2 (mainly increasing)\newline
or

$k>0$, PNKPN1, and NPKPN1 (mainly decreasing) \newline
or

$k>0$, PNKNN1, and NPKNN1 (mainly decreasing)

\noindent so that $f(x)$ given by (\ref{4.1}) is mainly monotonic, then
(\ref{3.1}) (and (\ref{1.3})) has at least one solution.  
\end{corollary}

We now investigate sufficient conditions for $f(x)$ given by (\ref{4.1}) to be
consistently monotonic. Since we require $f'(x)>0,$ we obtain
conditions on $k$ and $g(x)$.

\begin{theorem}
If $a\neq 0$, $\lambda \neq 0$, $k=\lambda /a>0$, and 

PPKPN1 or 2, NNKPN1 or 2, and $g'(x)>1/k$(consistently increasing)\newline
or

PNKPN1, NPKPN1, and $g'(x)<1/k$ (consistently decreasing)

\noindent then the function $f(x)$ given by (\ref{4.1}) is consistently monotonic.
\end{theorem}

\begin{theorem}
If $a\neq 0$, $\lambda \neq 0$, $k=\lambda /a<0$, and 

PPKNN1 or 2 , NNKNN1 or 2, and $g'(x)>1/k$ (consistently increasing)\newline
or

PNKNN1 or 2, NPKNN1 and $g'(x)<1/k$ (consistently decreasing) \newline
then the function $f(x)$ given by (\ref{4.1}) is consistently monotonic.
\end{theorem}

\begin{corollary}
If $a\neq 0$, $\lambda \neq 0$ and \newline
 
$k>0$, PPKPN1 or 2, NNKPN1 or 2, and $g'(x)>1/k$ (consistently increasing)\newline
or

$k<0$. PPKNN1 or 2, NNKNN1 or 2, and $g'(x)>1/k$ (consistently increasing)\newline
or

$k>0$, PNKPN1, NPKPN1, and $g'(x)<1/k$ (consistently decreasing)\newline
or

$k<0$, PNKNN1, NPKNN1, and $g'(x)<1/k$ (consistently decreasing)\newline
so that the function $f(x)$ given by (\ref{4.1}) is consistently monotonic,
then (\ref{3.1}) (and (\ref{1.3})) has exactly one solution.
\end{corollary}

\section{Quadratic property} \setcounter{equation}{0}

We now wish to consider cases when $f$ opens upwards or opens downwards. As
with the linear cases, we consider the sign of $k$ separately and only cases
where the growth rate of $g(x)$ is not linear. We begin with the case where 
$f$ opens upwards and $k>0$.

\begin{theorem}
Suppose $k>0$ and one of the following hypothesis holds: 
\begin{description}
\item{PPKPN1} $\lim_{x\to\infty}g(x)=-\infty $ or
$0$, (e.g., $g(x)=-e^{-x})$.

\item{PPKPN2} $\lim_{x\to\infty}g(x)=\infty $  and
$\lim_{x\to\infty}x/g(x)=\infty $. (e.g., $g(x)=e^{-x}+x^{1/3}$). 
\end{description}

\noindent Suppose further that the following hypothesis holds: 

\begin{description}
\item{NPKPN1} $\lim_{x\to-\infty}g(x)=-\infty $  and
$\lim_{x\to-\infty}x/g(x)=0$. (e.g., $g(x)=-e^{-x}$ or
$g(x)=-e^{-x}+x^{1/3}$.
\end{description}

\noindent Then the function $f(x)$ given by (\ref{4.1}) opens
upwards. \end{theorem}

Now we consider the case where $f$ opens upwards and $k<0$.

\begin{theorem}
Suppose $k<0$ and one of the following hypothesis holds: 

\begin{description}
\item{PPKNN1} $\lim_{x\to\infty}g(x)=\infty $ or $0$,
(e.g., $g(x)=e^{-x}$) \newline
or 

\item{PPKNN2} $\lim_{x\to\infty}g(x)=-\infty $ and
$\lim_{x\to \infty}x/g(x)=-\infty $. (e.g., $g(x)=e^{-x}-x^{1/3}$
\end{description}

\noindent Suppose further that the following hypothesis holds: 

\begin{description}
\item{NPKNN1} $\lim_{x\to-\infty}g(x)=\infty $ and
$\lim_{x\to-\infty}x/g(x)=0$. (e.g., $g(x)=e^{-x}$ or
$g(x)=e^{-x}-x^{1/3}$). 
\end{description}

\noindent Then the function $f(x)$ given by (\ref{4.1}) opens upwards. 
\end{theorem}

Now we consider the case where $f$ opens downwards and $k>0$.

\begin{theorem}
Suppose $k>0$ and the following hypothesis holds: 

\begin{description}
\item{PNKPN1} $\lim_{x\to\infty}g(x)=\infty $ and
$\lim_{x\to\infty}x/g(x)=0$. (e.g., $g(x)=x^{2}$ or
$g(x)=e^x+x^{1/3}$). 
\end{description}

\noindent Suppose further that one of the following hypothesis holds: 

\begin{description}
\item{NNKPN1} $\lim_{x\to-\infty}g(x)=\infty $ or $0$,
(e.g., $g(x)=x^{2}$)

\item{NNKPN2} $\lim_{x\to-\infty}g(x)=-\infty $ and
$\lim_{x\to-\infty}x/g(x)=\infty $. (e.g., $g(x)=e^x+x^{1/3}$).
\end{description}

\noindent Then the function $f(x)$ given by (\ref{4.1}) opens downwards. 
\end{theorem}

Now we consider the case that $k<0$.

\begin{theorem}
Suppose $k<0$ and the following hypothesis holds:

\begin{description}
\item{PNKNN1} $\lim_{x\to\infty}g(x)=-\infty $ and
$\lim_{x\to\infty}x/g(x)=0$. (e.g., $g(x)=-x^{2}$ or
$g(x)=-e^x-x^{1/3}$.
\end{description}

\noindent Suppose further that one of the following hypothesis holds:

\begin{description}
\item{NNKNN1} $\lim_{x\to-\infty}g(x)=-\infty $ or $0$,
(e.g., $g(x)=-x^{2})$

\item{NNKNN2} $\lim_{x\to-\infty}g(x)=\infty $ and
$\lim{x\to -\infty }x/g(x)=-\infty $.  (e.g., $g(x)=-e^{-x}-x^{1/3}$).
\end{description}

\noindent Then the function $f(x)$ given by (\ref{4.1}) opens downwards.
\end{theorem}


We next state theorems when there exists no solution, at least one
solution, and at least two solutions.

\begin{theorem}
Suppose $m=\min_{x\in{\mathbb R}}f(x)=f(x_{0})$ where 
$f'(x_{0})=0$ and 
\begin{description}
\item  $k>0$, PPKPN1 or 2, and NPKPN1, (e.g., $g(x)=-e^{-x}$)\newline
or 
\item  $k<0$, PPKNN1 or 2, and NPKNN1. (e.g., $g(x)=e^{-x}$).
\end{description}
Then $f$ given by (\ref{4.1}) opens upward. Hence 
\begin{description}
\item{(C1)} $m>0$ implies (\ref{3.1}) (and (\ref{1.3})) has no solution, 

\item{(C2)} $m=0$ implies (\ref{3.1}) (and (\ref{1.3})) has at least
one solution, namely $x=x_{0}$.

\item{(C3)} $m<0$ implies (\ref{3.1}) (and (\ref{1.3})) has at least
two solutions, say $x_{1}$ and $x_{2}$. 
Furthermore, we may assume $x_{1}<x_{0}<x_{2}$.
\end{description}
\end{theorem}

Similarly,

\begin{theorem}
Suppose $m=\max{x\in {\mathbb R}}f(x)=f(x_{0})$ where $f'(x_{0})=0$
and 
\begin{description}
\item  $k>0$, PNKPN1, and NNKPN1 or 2, (e.g., $g(x)=x^{2}$)\newline
or 

\item  $k<0$, PNKNN1, and NNKNN1 or 2. (e.g., $g(x)=-x^{2}$).
\end{description}

\noindent Then $f$ given by (\ref{4.1}) opens downward. Hence 
\begin{description}
\item{(C1)} $m<0$ implies (\ref{3.1}) (and (\ref{1.3})) has no
solution, 

\item{(C2)} $m=0$ implies (\ref{3.1}) (and (\ref{1.3})) has at least
one solution, namely $x=x_{0}$.

\item{(C3)} $m>0$ implies (\ref{3.1}) (and (\ref{1.3})) has at least
two solutions, say $x_{1}$ and $x_{2}$.
Furthermore, we may assume $x_{1}<x_{0}<x_{2}$
\end{description}
\end{theorem}

Note that if $f$ is given by (\ref{4.1}), then $f''(x)=kg''(x)$.
Hence we can consider cases where $f$ is
completely concave up or down. We can then state conditions where there
exist no solutions, exactly one solution, or exactly two solutions.

\begin{theorem}
Suppose $m=\min_{x\in{\mathbb R}}f(x)=f(x_{0})$ where $f'(x_{0})=0$ and 

  $k>0$, PPKPN1 or 2, NPKPN1, and $\forall x\in {\mathbb R}$,
$g''(x)>0$\newline
or 

$k<0$, PPKNN1 or 2, NPKNN1 and $\forall x\in {\mathbb R}$,
$g''(x)<0$.


\noindent Then $f$ given by (\ref{4.1}) is completely concave up. Hence 
\begin{description}
\item{(C1)} $m>0$ implies (\ref{3.1}) (and (\ref{1.3})) has no
solution, 

\item{(C2)} $m=0$ implies (\ref{3.1}) (and (\ref{1.3})) has exactly
one solution, namely $x=x_{0}$. 
\item{(C3)} $m<0$ implies (\ref{3.1}) (and (\ref{1.3})) has exactly
two solutions, say $x_{1}$ and $x_{2}$. 
Furthermore, we may assume $x_{1}<x_{0}<x_{2}$.
\end{description}
\end{theorem}

Similarly,

\begin{theorem}
Suppose $m=\max_{x\in {\mathbb R}}f(x)=f(x_{0}$) where 
$f'(x_{0})=0$ and 

  $k>0$, PNKPN1, NNKPN1 or 2, and $\forall x\in {\mathbb R}$,
$g''(x)<0$\newline
or 

$k<0$, PNKNN1, NNKNN1 or 2, and $\forall x\in {\mathbb R}$,
$g''(x)>0$. 

\noindent Then $f$ given by (\ref{4.1}) is completely concave down. Hence 

\begin{description}
\item{(C1)} $m<0$ implies (\ref{3.1}) (and (\ref{1.3})) has no solution, 

\item{(C2)} $m=0$ implies (\ref{3.1}) (and (\ref{1.3})) has exactly
one solution, namely $x=x_{0}$. 

\item{(C3)} $m>0$ implies (\ref{3.1}) (and (\ref{1.3})) has exactly
two solutions, say $x_{1}$ and $x_{2}$.
Furthermore, we may assume $x_{1}<x_{0}<x_{2}$.
\end{description}
\end{theorem}

\section{Future work} \setcounter{equation}{0}

In the future, we will consider multidimensional systems beginning with the
scalar equations
\begin{eqnarray}
&ax+by=\lambda g(x)&  \label{7.1} \\
&cx+dy=\lambda g(y).&  \label{7.2}
\end{eqnarray}
These can be written as the vector or matrix equation
\begin{equation}
A\vec{x}=\lambda \vec{g}(\vec{x})  \label{7.3}
\end{equation}
where
\[
A=\left[ \begin{array}{cc} a & b \\ c & d\end{array} \right],\quad
\vec{x}=\left[\begin{array}{c} x \\ y\end{array} \right]\,,
\mbox{ and }\vec{g}(\vec{x})=\left[ 
\begin{array}{c} g(x) \\  g(y) \end{array}\right] \,.
\]
Extensions to $n$ dimensions should follow. The problem (\ref{7.3}) has
five parameters in addition to the function $g$. If $b\neq 0$, (\ref{7.1}) may be
solved for $y$ and substituted into (\ref{7.2}) to obtain a single equation. The
steps in the above process can then be followed to obtain conditions where
the linear property of the quadratic property obtains for (\ref{7.3}). This will
include direct applications of the theorems in this paper. Hence we will
obtain linear conditions where (\ref{7.3}) has exactly one solution as well as
quadratic conditions where there are no solutions or exactly one or two
solutions. Tridiagonal matrices should also be amenable to this process.

\section{Summary}

We give sufficient conditions to determine exactly the number of solutions
for the one dimensional nonlinear eigenvalue problem $ax=\lambda g(x)$. We
first considered the cases when $a$ or $\lambda $ (or both) are zero. We
then assumed $a\neq 0$, $\lambda \neq 0$, let $k=\lambda /a$ and focused on
the solution set of $f(x)=0$. For any function $f(x)$ we then defined
the linear and quadratic properties depending on the limits of $f$ as $x$
approaches $\pm \infty $, and on its monotonicity and concavity properties.
We then considered sufficient conditions on $k$ and $g(x)$ so that
$f(x)=x-kg(x)$ approaches $\pm \infty $ as $x$ approaches $\pm \infty $
and hence
obtained sufficient conditions for $f(x)$ to be mainly monotonic and
consistently monotonic. This allowed us to establish sufficient conditions
so that $f(x)$ has the linear property and hence conditions so that 
$ax=\lambda g(x)$ has at least one and exactly one solution. We also gave
sufficient conditions for $f(x)$ to open upwards and downwards and to be
completely concave up and down so that f has the quadratic property. This
leads to sufficient conditions so that $ax=\lambda g(x)$ has no solution, at
least one solution, and at least two solutions as well as sufficient
conditions where exactly one solution, and exactly two solutions exist.
Examples are given for both the linear and quadratic cases. Future work on
multidimensional systems includes two dimensional problems and tridiagonal
systems.


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\end{thebibliography} \bigskip

\noindent{\sc Jun Hua } \\ 
{\sc James L. Moseley} \\
West Virginia University\\
Morgantown, West Va 26506-6310, USA\\
(304) 293-2011, e-mail:  moseley@math.wvu.edu 

\end{document}