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\markboth{ Infinitely many solutions at a resonance }
{ Philip Korman \&  Yi Li }
\begin{document}
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\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
Nonlinear Differential Equations, \newline
Electron. J. Diff. Eqns., Conf. 05, 2000, pp. 105--111\newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)}
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Infinitely many solutions at a resonance
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\thanks{ {\em Mathematics Subject Classifications:}  34B15.
\hfil\break\indent 
{\em Key words:} 
Bifurcation of solutions, the global solution curve.
\hfil\break\indent
\copyright 2000 Southwest Texas State University. 
\hfil\break\indent Published October 25, 2000.  \hfil\break\indent 
Partially supported by a Taft Faculty Grant at the University of Cincinnati } } 

\date{}
\author{ Philip Korman \&  Yi Li \\[12pt]
{\em Dedicated to Alan Lazer} \\ {\em on  his 60th birthday }}
\maketitle
\begin{abstract} 
We use bifurcation theory to show the existence of infinitely
many solutions  at the first eigenvalue for a class of Dirichlet 
problems in one dimension.
\end{abstract}


\newtheorem{thm}{Theorem}
\newtheorem{lma}{Lemma}



It has been observed that complexity of the solution curve for the boundary 
value problem
\begin{equation} \label{1}
u'' +\lambda f(u)=0 \quad \mbox{for $0<x<L$,} \quad u(0)=u(L)=0
\end{equation}
seems to mirror that of the nonlinearity $f(u)$, see e.g.  P.  Korman, Y.  Li
and T.  Ouyang \cite{KLO}.  Namely, if $f(u)$ is convex ($f=e^u$ is a prominent
example), $f(u)$ has at most one critical point, and correspondingly the
solution curve of (\ref{1}) admits only one turn.  Similarly for many functions
with two inflection points (like cubics, or modified Gelfand's equation) one can
show that solution curve admits exactly two turns, see \cite{KLO} and also P.
Korman and Y.  Li \cite{KL}.  It is natural to ask:  will solution curve admit
infinitely many turns if $f(u)$ changes concavity infinitely many times.  
It has been known for a while that this indeed might happen, 
see e.g.  R.  Schaaf and K.  Schmitt \cite{SS}, D.  Costa, H.  Jeggle, 
R.  Schaaf and K.  Schmitt \cite{SS1}, H.  Kielhofer 
and S.  Maier \cite{KM}.  Recent contributions include
Y.  Cheng \cite{C} and S.-H.  Wang \cite{W}.  This note was stimulated by an
example in Y.  Cheng \cite{C}, who has shown that for $f(u)=u+\sin{\sqrt u}$ the
problem (\ref{1}) admits infinitely many solutions at $\lambda =\lambda _1$, 
where $\lambda_1$ denotes as usual the principal eigenvalue of $-u''$ on $(0,L)$, $\lambda _1=
\frac{\pi^2}{L^2}$.  
The proof in \cite{C} used the quadrature method.  In this
note we use bifurcation theory to obtain a similar result.  The bifurcation
approach gives a clear understanding of the solution curve, and opens a way to
considering higher dimensions.  The papers \cite{SS} and \cite{SS1} also used
bifurcation approach, and they considered more general equations, as well as the
PDE case in two dimensions.  Our approach is different, we work with a smooth
curve of solutions (rather than a continuum of solutions bifurcating from
infinity), and we obtain an extension to more general nonlinearities.  Moreover,
we can estimate the number of turns the solution curve makes, until it reaches
any level $u_{\max}=c$.  \smallskip

We begin with a simple example.

\begin{thm}
The problem
\begin{equation} \label{2}
u'' +\lambda (u+\sin u)=0 \quad \mbox{for}  \quad 0<x<L, \quad u(0)=u(L)=0
\end{equation}
has a $C^1$ curve of positive solutions bifurcating (forward) off the trivial solution at $\lambda =\frac{\lambda _1}{2}$. This curve crosses infinitely many times the line
$\lambda =\lambda _1$, and tends to infinity at $\lambda =\lambda _1$. This curve exhausts
the set of all positive solutions of (\ref{2}). There are infinitely many
positive solutions at $\lambda =\lambda _1$, while at any other $\lambda $ the
number of positive solutions is at most finite.
\end{thm}


The proof will use the following lemma, which shows that there are no  positive solutions for $\lambda >0$
small, and for $\lambda$ large. We did not attempt to get the optimal bounds.
\begin{lma}
If the problem (\ref{2}) has a positive solution, then
\begin{equation} \label{3}
\frac{\lambda _1}{2} <\lambda < \frac{\pi}{\pi-1}\lambda _1.
\end{equation}
\end{lma}

\paragraph{Proof:} 
Set $f(u)=u+\sin u$. Observe that $0< f(u) < 2u$ for all $u > 0$.
Multiplying the equation (\ref{2}) by $u$, integrating by parts, and using
the Poincare inequality
\[
2\lambda \int_0^L u^2 \, dx > \lambda \int_0^L f(u)u \, dx =
\int_0^L {u'}^2 \, dx \geq \lambda _1 \int_0^L u^2 \, dx,
\]
from which the left inequality in (\ref{3}) follows.
\smallskip

On the other hand, multiplying the equation (\ref{3}) by $\phi _1
=\sin{\frac{\pi}{L}x}$, and integrating over $(0,L)$, we obtain
(with $\lambda _1 = \frac{\pi^2}{L^2}$)
\begin{equation} \label{4}
\int_0^L \left(-\lambda _1u+\lambda u+\lambda \sin u \right) \phi _1 \, dx=0.
\end{equation}
Since $\phi _1 >0$, we obtain a contradiction in (\ref{4}), provided that
\[
g(u) \equiv \frac{(\lambda -\lambda _1)}{\lambda } u+ \sin u >0
\]
for all $u>0$. This will certainly happen if $\lambda >\lambda _1$, and $g(\pi)
\geq 1$, which leads to the right inequality in (\ref{3}).


\paragraph{Proof of the Theorem 1.} It is well-known that there is a curve of positive
solutions of (\ref{2}) bifurcating off the trivial one (to the right)
at $\frac{\lambda _1}{2}$. Similarly there a curve of positive solutions bifurcating
from infinity at $\lambda =\lambda _1$, see e.g. p. 673 in E. Zeidler \cite{Z}. We claim that both branches link up,
giving a unique solution curve. Indeed, let us start with the branch bifurcating from infinity. This branch extends globally, since at each of its points either the implicit function theorem or a bifurcation theorem of Crandall-Rabinowitz applies, see \cite{KLO} for details. By Lemma 1 this branch is constrained to a strip $\frac{\lambda _1}{2}<\lambda <\frac{\pi}{\pi-1}\lambda _1$, and so it has to go to zero at $\lambda _1$. By uniqueness of the bifurcating solution, this branch has to link up with the lower one.
\smallskip

We show next that the solution curve
changes its direction infinitely many times.
Define $F(u)=\int_0^u f(t) \, dt$, $G(u)=uf(u)-2F(u)$. Notice that
$G'(u)=uf'(u)-f(u)$. Differentiating the problem (\ref{1}) with respect to $\lambda$, we get
\begin{equation} \label{5}
u''_{\lambda} +\lambda f'(u)u_{\lambda }+f(u)=0 \quad \mbox{for $0<x<L,$} \quad
u_{\lambda }(0)=u_{\lambda }(L)=0.
\end{equation}
Multiplying the equation (\ref{1}) by $u_{\lambda }$, and (\ref{5}) by $u$,
then subtracting and integrating, we obtain
\begin{equation} \label{6}
-\lambda \frac{d}{d \lambda} \int_0^L G(u) \, dx=\int_0^L uf(u) \, dx.
\end{equation}
Since the integral on the right in (\ref{6}) is positive, it
follows that $\int_0^L G(u(x)) \, dx$ is monotone in $\lambda $. If we can show
that this integral is not monotone as the solution curve goes to infinity,
it will follow that the solution curve is not monotone in $\lambda $, i.e. it
has to change its direction infinitely many times.
Let $s=u(\frac{L}{2})$. Clearly,
$u(\frac{L}{2}) \to \infty$ as $\lambda \to \lambda _1$, since $u(x)$ achieves its maximum at the midpoint of the interval $(0,L)$.  Define $\phi (x)$
by $u(x)=s \phi (x)$, and set
\begin{equation} \label{7}
I(s) \equiv \int_0^L G(u(x)) \, dx = \int _0^L G(s\phi (x)) \, dx.
\end{equation}
To show that $I(s)$ is not monotone in $s$ it suffices to show that
$I'(s)$ changes sign infinitely many times. Defining $J(s)=\int_0^L
\sin (s\phi (x)) \phi (x) \, dx$, we express
\begin{equation} \label{8}
I'(s)= \int_0^{L} \left[s \phi  (x) \cos (s\phi (x))-\sin (s\phi (x)) \right]
\phi (x) \, dx=sJ'(s)-J(s).
\end{equation}
We claim that the function $J(s)$ changes sign infinitely many times.
Indeed, performing a change of variables by letting $u= s\phi (x)$, we have
\begin{equation} \label{*}
J(s)=2 \int_0^{\frac{L}{2}} \sin (s\phi (x)) \phi (x) \, dx
=\frac{2}{s} \int _0^{u(\frac{L}{2})} \sin u \frac{dx}{du}u \, du.
\end{equation}
Here $\frac{dx}{du}$ is the derivative of the inverse function to $u(x)$.
Since $ \frac{dx}{du}u$ is an increasing function, it is easy to see
that
\[
J(s)\left\{ \begin{array}{ll}
>0, & \mbox{if $u(\frac{L}{2})=n\pi$, and $n$ is odd}  \\[2pt]
<0 & \mbox{if $u(\frac{L}{2})=n\pi$, and $n$ is even.}
\end{array} \right.
\]
Hence $J(s)$ changes sign infinitely many times. It follows that $J(s)$ has
infinitely many points of negative minimum, where by (\ref{8}) $I'(s)$ is positive, and infinitely many points of positive maximum where $I'(s)$ is
negative. It follows that $I'(s)$ changes sign infinitely many times, and so
$I(s)$ is not monotone,
and hence the solution curve changes its direction infinitely many times.
\smallskip

It remains to show that the solution curve
intersects the line $\lambda =\lambda _1$ infinitely many times. From (\ref{4}) we conclude
\begin{equation} \label{9}
(\lambda _1 - \lambda ) \int_0^L u\phi _1 \, dx= \lambda  \int_0^L \sin u \; \phi _1 \, dx =
\lambda  \int_0^L \sin s\phi \; \phi _1 \, dx.
\end{equation}
It follows from a standard analysis of bifurcation from infinity that $\phi(x)=\phi _1 (x) +o(1)$
for $s$ large.
We then have
\begin{equation} \label{10}
\int_0^L \sin s\phi \; \phi _1 \, dx=J(s)+o(1)\int_0^L \sin (s\phi)  \, dx.
\end{equation}
We now recall an asymptotic formula from  sec. 40 in Y.V. Sidorov et al \cite{S} (see also Corollary 1 in sec. 39).
\begin{lma}
Assume that the functions $f(x)$ and $g(x)$ are infinitely differentiable
on the interval $[0,1]$, and assume that
\[
g(x)<g(0) \quad \mbox{for all $x \in (0,1]$}, \quad
\mbox{and} \quad g'(0)=0, \quad g''(0)<0.
\]
Then as $s \to \infty$
\[
\int_0^1 f(x) e^{isg(x)} \, dx=\frac{1}{2} e^{i\left(sg(0)-\frac{\pi}{4}
\right)}
\sqrt{ \frac{2\pi}{s|g''(0)|}} f(0) +O(\frac{1}{s}).
\]
\end{lma}
  Using elliptic regularity, we notice that
\[
|\phi (x)-\phi _1(x)|_{C^2[0,1]} =o(1) \quad \mbox{as $s \to \infty$},
\]
which implies in particular that $\phi ''(\frac{L}{2})=-\frac{\pi^2}{L^2} +o(1)$.
Applying Lemma 2 to $J(s)=2\Im \int_0^{\frac{L}{2}} e^{is\phi(x)} \phi(x) \, dx=L\Im \int_0^1 e^{is\phi(\frac{L}{2}\xi)} \phi(\frac{L}{2}\xi) \, d \xi$, we conclude that (observe that $\phi (\frac{L}{2})=1$, $\phi '(\frac{L}{2})=0$)
\[
J(s) \sim \frac{c_0}{\sqrt s} \sin \left(s-\frac{\pi}{4} \right),
\]
for some positive constant $c_0$. Similarly
\[
\int_0^L \sin (s\phi)  \, dx \sim \frac{c_0}{\sqrt s} \sin \left(s-\frac{\pi}{4} \right).
\]
It follows from (\ref{10}) that the integral on the left changes sign infinitely many times, and then from (\ref{9}) we conclude that the solution curve crosses the line $\lambda =\lambda _1$ infinitely many times, completing the proof.

\smallskip

We now consider more general equations. Recall that we denote $F(u)=\int _0^u f(t) \, dt$.

We say that a function $f(u)$ satisfies Schaaf-Schmitt condition if there exist two monotone  sequences
$\{x_n \}$ and $\{y_n \}$ with $x_n \to \infty$, $y_n \to \infty$ such that
\begin{eqnarray}
\label{11}
& F(x_n)-F(x) \geq 0 \quad \mbox{for all $0 \leq x <x_n$} \\ \nonumber
& F(y_n)-F(x) \leq 0 \quad \mbox{for all $0 \leq x <y_n$}, \nonumber
\end{eqnarray}
both inequalities being strict on a set of positive measure. Setting $x=0$, we observe that (\ref{11}) implies in particular
\begin{equation} \label{12}
F(x_n) \geq 0, \quad F(y_n) \leq 0 \quad \mbox{for all $n$}.
\end{equation}
\begin{lma}
Let $f(u) \in C^1[0,\infty)$ satisfy Schaaf-Schmitt condition, and $g(u) \in C^1[0,\infty)$ be a positive function with $g'(u)>0$ for all $u>0$. Then
\begin{equation} \label{13}
\int _0^{x_n} f(u)g(u) \, du >0, \quad \int _0^{y_n} f(u)g(u) \, du <0.
\end{equation}
\end{lma}

\paragraph{Proof:} 
Using (\ref{11}) and (\ref{12}) we have
\begin{eqnarray*}
& \int _0^{x_n} f(u)g(u) \, du =\int _0^{x_n} g(u) d \left[F(u)-F(x_n) \right] \\
& = g(0)F(x_n)- \int _0^{x_n} g'(u) \left[F(u)-F(x_n) \right] \, du >0,
\end{eqnarray*}
and the second inequality in (\ref{13}) is proved similarly.
\hfill$\diamondsuit$ 

\begin{thm}
Consider the problem
\begin{equation} \label{14}
u'' +\lambda \left(u+f(u)\right)=0 \quad \mbox{for $0<x<L$,} \quad u(0)=u(L)=0.
\end{equation}
Assume that $f(u) \in C^2[0,\infty)$ satisfies
\begin{equation} \label{15}
f(0)=0, \quad f'(0) >-1,
\end{equation}
\begin{equation} \label{15a}
u+f(u)>0 \quad \mbox{for all $u>0$},
\end{equation}
\begin{equation} \label{16}
\frac{f(u)}{u} \to 0 \quad \mbox{as $u \to \infty$}.
\end{equation}
Assume finally that either $f(u)$ or $uf(u)$ satisfies Schaaf-Schmitt condition. Then the problem (\ref{14}) has a $C^1$ curve of positive solutions bifurcating off the trivial solution at $\lambda =\frac{\lambda _1}{1+f'(0)}$. This curve makes infinitely many turns, and tends to infinity at $\lambda =\lambda _1$. This curve exhausts
the set of positive solutions of (\ref{14}).
\end{thm}

\paragraph{Proof:} 
Similar to Theorem 1.   Examining our derivation of (\ref{*}), one sees that this time
\[
J(s)=\frac{2}{s} \int _0^{u(\frac{L}{2})} f(u) \frac{dx}{du}u \, du=
\frac{2}{s} \int _0^s f(u) \frac{dx}{du}u \, du.
\]
Since by (\ref{15a}) the function $\frac{dx}{du}$ is increasing, applying Lemma 3 we 
see that $J(s)$ changes sign infinitely many times, and hence the solution 
curve changes direction infinitely many times. Clearly, an analog of Lemma 1 
holds, so that the curves bifurcating from zero and infinity have to link up.
\hfill$\diamondsuit$ 

\paragraph{Remarks.}
\begin{enumerate}
\item One can have several variations of the Theorem 2. For example, 
we could drop the condition (\ref{15}),
and have a curve bifurcating from infinity, and making infinitely many turns. 
Alternatively, we could drop the condition (\ref{16})
and obtain a curve bifurcating from zero, and making infinitely many turns.
\item It follows from \cite{SS} that the problem (\ref{14}) has infinitely many solutions at $\lambda =\lambda _1$ if $f(u)$ satisfies Schaaf-Schmitt condition. Clearly, it may happen that $uf(u)$ satisfies Schaaf-Schmitt condition, while $f(u)$ does not.
\item Another novelty of our approach is that we can estimate the number of 
turns the solution curve makes until it reaches any level $u(0)= c$.
\item Arguing as in Lemma 1, we see that either the solution curve crosses 
the line $\lambda =\lambda _1$ infinitely many times, or else it makes infinitely many 
turns to the left of this line. (Assuming that $\lambda >\lambda _1$ for sufficiently 
large $s$, we multiply the equation (\ref{14}) by $u$, integrate by parts, 
and obtain a contradiction.)
\end{enumerate}


\paragraph{Acknowledgment.} We wish to thank  Y. Cheng and S.-H. Wang for sending us  preprints
of their work, and K.R. Meyer, T. Ouyang and J. Shi for useful comments.

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\end{thebibliography} \medskip

\noindent{\sc Philip Korman} \\
Institute for Dynamics and \\
Department of Mathematical Sciences \\
University of Cincinnati \\
Cincinnati Ohio 45221-0025 \\
e-mail: kormanp@math.uc.edu  \medskip

\noindent{\sc Yi Li } \\
Department of Mathematics \\
University of Iowa \\
Iowa City Iowa 52242 \\
e-mail: yi-li@uiowa.edu
\end{document}