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\markboth{\hfil Similarity of solution branches \hfil EJDE/Conf/10}
{EJDE/Conf/10 \hfil Philip Korman \hfil}

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\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
Fifth Mississippi State Conference on Differential Equations and
Computational Simulations, \newline
Electronic Journal of Differential Equations,
Conference 10, 2003, pp 187--191. \newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu (login: ftp)}
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  Similarity of solution branches for two-point semilinear problems
%
\thanks{ {\em Mathematics Subject Classifications:} 34B15.
\hfil\break\indent
{\em Key words:} Similarity of solution curves.
\hfil\break\indent
\copyright 2003 Southwest Texas State University. \hfil\break\indent
Published February 28, 2003.} }

\date{}
\author{Philip Korman}
\maketitle

\begin{abstract}
 For semilinear autonomous two-point problems, we show that all
 Neumann branches and all Dirichlet branches with odd number of
 interior roots  have the same shape. On the other hand, Dirichlet
 branches with even number of roots may look differently.
 While this result has been proved previously by Schaaf \cite{S},
 our approach appears to be simpler.
\end{abstract}

\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

We are interested in positive, negative and sign-changing solution
branches of the problem
\begin{equation}
\label{0.1}
u''+\lambda f(u)=0, \quad  \mbox{on }(0,1),
\end{equation}
subject to either Dirichlet or Neumann boundary conditions.
Here $\lambda$ is a positive parameter, and we are interested in how the solutions
evolve with $\lambda$. Typically the problem (\ref{0.1}) will have infinitely
many solution branches, each with a different number of interior roots
(or monotonicity changes). It turns out that these branches tend all to
have similar shapes, and there is a correspondence between the Dirichlet
and Neumann branches.


We begin with a simple ``warm up" example. Let us assume that $f(u)$ is an
odd function, and consider the problem (\ref{0.1}) together with the Dirichlet
conditions (\ref{2}) below. Then all solution curves are similar,
and moreover each sign-changing solution consists of self-similar parts.
This follows from two simple observations. First, any solution of (\ref{0.1})
is symmetric with respect to any of its critical points. Secondly,
if $u$ is a solution of (\ref{0.1}) and (\ref{2}), then so is $-u$. Hence,
for any sign-changing solution all positive humps are identical up to
translation, and they are identical to the negative humps via reflection
across the $x$-axis. So that after rescaling, we identify any sign-changing
solution with a positive one, at a different $\lambda$, and hence all solution
curves have the same form. For example, consider $f(u)=u-u^3$.
Then a familiar picture of parabola-like curves bifurcating to the right
from the zero solution at $\lambda _k= k^2 \pi^2$ (see e.g. \cite{CH}) follows
immediately, since the same is true for positive solutions bifurcating
at $\lambda _1$ (as can be easily seen using super- and subsolutions).
Exactly the same picture, with the same justification, holds for the
pendulum equation, when $f(u) = \sin u$.


It turns out that self-similarity holds for general $f(u)$. In particular,
it always holds for Neumann problems. For the Dirichlet problems, we distinguish
between the cases of odd and even number of interior roots.
It turns out that all Neumann branches and all Dirichlet branches with odd number
of interior roots have the same shape. On the other hand, Dirichlet branches
with even number of roots may behave differently. These results were
previously observed by R. Schaaf \cite{S}.

\section{Similarity and self-similarity of Dirichlet and Neumann branches}
We consider the  problem
\begin{equation}
\label{1}
u''+\lambda f(u)=0, \quad  \mbox{on }(0,1),
\end{equation}
subject to either Dirichlet
\begin{equation}
\label{2}
u(0)=u(1)=0,
\end{equation}
or Neumann
\begin{equation}
\label{3}
u'(0)=u'(1)=0
\end{equation}
boundary conditions. We shall consider the solutions as the positive
parameter $\lambda $ varies, and refer to the solution curves as either
Dirichlet or Neumann branches, depending on the boundary conditions used.
Observe that any solution of the equation (\ref{1}) is symmetric with
respect to any of its critical points. This implies, in particular,
that either minimum or maximum occurs at any critical point. It follows
that any solution of Neumann problem is determined by its values on any subinterval
$I \subset (0,1)$, whose end-points are two consecutive critical points of
$u(x)$. We can then obtain the solution on the entire interval $(0,1)$
through reflections and translations. We refer to $I$, and any other
interval uniquely determining the solution through reflections and
translations, as a {\em determining interval}. The interval $I$, joining
two consecutive critical points of $u(x)$, is also a
determining interval for the Dirichlet problem. Another determining interval for
Dirichlet problem is $(\xi, \eta)$, where $0 \leq \xi<\theta<\eta \leq 1$ are
three consecutive roots of $u(x)$. This interval contains both positive
and negative humps (and all positive (negative) humps are translations of
one another).


The natural way to distinguish the Dirichlet branches is by the number of
interior roots, and the Neumann branches can be identified by the number of
changes of monotonicity (both properties clearly remain constant on the
solution curves). Any solution of the Dirichlet problem with at least one
interior root contains a solution of the Neumann problem on a subinterval
of $(0,1)$. Indeed, just consider the solution between two consecutive
critical points. In order for solutions of the  Neumann problem to contain in
turn a solution of the Dirichlet problem, we need to impose some conditions on
$f(u)$. Namely, we assume that
\begin{equation}
\label{4}
f(0) = 0,
\end{equation}
and there exist two constants $-\infty \leq m<0<M \leq \infty$ so that
\begin{itemize}
\item[(f1)] $f(u)>0 $ for $u \in (0,M)$,

\item[(f2)] $f(u)<0 $ for $u \in (m,0)$.
\end{itemize}


\begin{lemma} \label{lm1}
Under the conditions (\ref{4}), (f1) and (f2) any solution of the Neumann
problem for (\ref{1}), satisfying
\begin{equation}
\label{41}
m<u(x)<M \quad \mbox{for all }x
\end{equation}
 has a root between any two critical points.
\end{lemma}

\paragraph{Proof:}
Follows immediately, by multiplying the equation (\ref{1}) by $u'$,
and integrating between any two consecutive critical points.
\hfill$\diamondsuit$

\paragraph{Definition.} We call two solution branches of (\ref{1}) to be
{\em similar} if for any solution on the either branch there is a determining
interval so that by stretching  of $x$, or by reflection $x \to 2a-x$,
for some $a \in (0,1)$, we obtain a solution from the other branch on a
(different) determining interval.


Clearly, if solution branches are similar then the corresponding solution
curves in $(\lambda,u)$ "plane" have the same shape.

\begin{theorem}\label{thm:1}
All Neumann branches of (\ref{1}) are similar, and if $f(u)$ satisfies the
conditions (\ref{4}), (f1) and (f2), while all solutions satisfy
(\ref{41}), then the Neumann branches are similar to the Dirichlet ones with an odd
number of interior roots (and these Dirichlet branches  are also all similar).
\end{theorem}

\paragraph{Proof:}
We begin with Neumann branches. If a Neumann solution changes monotonicity twice,
then its increasing part is a reflection of its decreasing part with respect
to $x=\frac{1}{2}$. If a Neumann solution changes monotonicity $n$ times, then
all critical points occur at $i/n$, $i=1, \ldots, n-1$, and the graphs of
solution on all intervals where it is increasing (decreasing) are
translations of one another. Since an  interval connecting any two critical
points is a determining interval, the equivalence of the Neumann branches follows
(via rescaling).
% \hfill$\diamondsuit$ \smallskip
\medskip

If a Dirichlet solution has $2k-1$ interior roots, it has $k$ identical positive
humps and $k$ identical negative humps. Assume for definiteness that solution
starts with a negative hump, followed by a positive one, and so on.
If $\xi$ is the first point of (negative) minimum of $u(x)$, then the first
interior root occurs at $2\xi$. If $2 \xi+\eta$ is the point of the first
(positive) maximum, then the second interior root occurs at $2 \xi+2\eta$.
 The last critical point, a positive maximum, occurs at $1-\eta$. Observe
 that  $k(2\xi+2\eta)=1$, i.e.
$\xi+\eta=\frac{1}{2k}$. So while both $\xi$ and $\eta$ vary with $\lambda$,
$u(x)$ solves the Neumann problem on the interval $(\xi, 1-\eta)$,
and this interval has a {\em fixed} length of
\[
1-\eta-\xi=\frac{2k-1}{2k}.
\]
So that any Dirichlet solution curve ``carries" inside it a solution of a
Neumann problem  on a fixed interval (which can be made to be $(0,1)$
by rescaling), and hence the Dirichlet branch cannot have any more complexity
(like extra turns) than any Neumann branch.


Conversely, consider the Neumann problem with  $2k+1$   changes of monotonicity.
Assume for definiteness that $u(0)<0$. Then $u(1)>0$. Assume that
$\xi=\xi(\lambda )$ is the smallest interior root, and $1-\eta$ is the largest one,
$\eta =\eta(\lambda )$. On the interval $(0,1)$ we then have
$2k+1$ negative half-humps, each of width $\xi $, and $2k+1$ positive ones,
each of width $\eta$. So that $\xi+\eta=\frac{1}{2k+1}$.
On the interval $(\xi, 1-\eta)$ we have a solution of the Dirichlet problem with
$2k-1$ interior roots, and the length of this interval is
\[
1-\eta-\xi=\frac{2k}{2k+1},
\]
which does not vary with $\lambda$. So that any Neumann branch "carries" inside it a
solution of a Dirichlet problem on a fixed interval, and hence the Neumann branch cannot
have any more complexity  than the corresponding Dirichlet branch with an odd
number of interior zeroes.


Finally, the Dirichlet branches with odd number of interior zeroes are all similar,
since any two such branches are similar to a pair of Neumann branches, but Neumann
branches are all similar.
\hfill$\diamondsuit$ \smallskip

The Dirichlet branches with even number of interior zeroes may behave differently,
as the following example due to Schaaf \cite{S} shows.


\paragraph{Example (\cite{S})} For the problem
\[
u''+\lambda (e^u-1)=0 \quad \mbox{on }(0,1), \quad u(0)=u(1)=0
\]
the branch bifurcating from the principal eigenvalue does not turn, while
all other branches have exactly one turn.

\begin{thebibliography}{99} \frenchspacing

\bibitem{CH}
S.-N. Chow and J. K. Hale, {\em Methods of Bifurcation Theory},
Springer-Verlag (1982).

\bibitem{S}
R. Schaaf, Global Solution Branches of Two Point Boundary Value Problems,
Lecture Notes in Mathemathics, no. {\bf 1458}, Springer-Verlag (1990).

\end{thebibliography}

\noindent\textsc{Philip Korman} \\
Department of Mathematical Sciences \\
University of Cincinnati \\
Cincinnati Ohio 45221-0025 \\
email: kormanp@math.uc.edu

\end{document}