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\AtBeginDocument{{\noindent\small
Sixth Mississippi State Conference on Differential Equations and
Computational Simulations,
{\em Electronic Journal of Differential Equations},
Conference 15 (2007), pp. 239--249.\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 2007 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document} \setcounter{page}{239}
\title[\hfilneg EJDE-2006/Conf/15\hfil On the exact multiplicity]
{On the exact multiplicity of solutions for boundary-value
problems via computing the direction of bifurcations}
\author[J. Rivera, Y. Li\hfil EJDE/Conf/15 \hfilneg]
{Joaquin Rivera, Yi Li}
\address{Joaquin Rivera \newline
Department of Mathematics\\
University of Iowa \\
Iowa City, Iowa 52242, USA}
\email{rvera@math.uiowa.edu}
\address{Yi Li \newline
Department of Mathematics \\
University of Iowa \\
Iowa City, Iowa 52242, USA\\
Hunan Normal University\\
Changsha 410081, Hunan, China }
\email{yi-li@uiowa.edu}
\thanks{Published February 28, 2007.}
\thanks{J. Rivera was supported by a grant from Sloan Fundation
and the Graduate College\hfill\break\indent
of the University of Iowa. \hfill\break\indent
Y. Li was supported by a Xiao-Xiang Grant from the Hunan Normal University}
\subjclass[2000]{34B15}
\keywords{Bifurcation points; direction of the turn;
multiplicity of solutions}
\dedicatory{Dedicated to Louis Nirenberg on his 80-th birthday}
\begin{abstract}
We consider positive solutions of the Dirichlet problem
\begin{gather*}
u''(x)+\lambda f(u(x))=0\quad\text{in }(-1,1), \\
u(-1)=u(1)=0.
\end{gather*}
depending on a positive parameter $\lambda $. We use two formulas derived in
\cite{KLO2} to compute all solutions $u$ where a turn may occur and to
compute the direction of the turn. As an application, we consider quintic a
polynomial $f(u)$ with positive and distinct roots. For such quintic
polynomials we conjecture the exact mutiplicity structure of positive
solutions and present computer assisted proofs of such exact bifurcation
diagrams for various distributions of the real roots. The limiting behavior
of the solutions on these bifurcation branches as $\lambda \to \infty $ and
their stabilities are also investigated.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{conjecture}[theorem]{Conjecture}
\section{Introduction}
We study exact bifurcation diagrams and exact multiplicity of the positive
solutions to the Dirichlet problem
\begin{equation} \label{0}
\begin{gathered} u''(x)+\lambda f(u(x))=0\quad\text{on }(-1,1),\\
u(-1)=u(1)=0, \end{gathered}
\end{equation}
depending on a positive parameter $\lambda $. We recall that solutions of
\eqref{0} are even functions, with $u'(x)<0$ for $x>0$, and hence
any solution is uniquely identified by $\alpha =u(0)$, see \cite{KLO}.
Actually, even more is true: the value of $u(0)=\alpha $ uniquely identifies
both $\lambda $ and $u(x)$, as follows easily by scaling $\lambda $ out of
\eqref{0}, and using uniqueness for initial value problems, see Dancer \cite
{D}. Hence the solution curves of \eqref{0} can be faithfully depicted by
two-dimensional curves on $(\lambda ,\alpha )$ plane. It is customary to
refer to these curves as \emph{bifurcation diagrams}. The Figure $1$ below
gives such a bifurcation diagram for a quintic polynomial with real roots at
$0.1,0.2,0.4$ and $0.5$. The shape of any bifurcation diagram is determined
by the \emph{turning points}. In \cite{KLO2} a necessary and sufficient
condition on $\alpha $ for the solution to be singular and thus a necessary
condition for the turning point to occur is given as follows:
\begin{equation}
G(\alpha )\equiv {F(\alpha )^{1/2}}\int_{0}^{\alpha }\frac{f(\alpha )-f(\tau
)}{\left[ F(\alpha )-F(\tau )\right] ^{3/2}}\,d\tau -2=0, \label{*}
\end{equation}
with $F(u)=\int_{0}^{u}f(t)\,dt$. This formula can be used to compute
numerically all turning points. At the turning points we are interested in
the \emph{turning direction}. It is shown in \cite{KLO2} that the curve tuns
to the right in $(\lambda ,\alpha )$ plane if
\begin{equation}
D(\alpha )\equiv \int_{0}^{\alpha }f''(u) \Big(\int_{u}^{\alpha
}f(s)\,ds\Big) \Big( \int_{0}^{u}\frac{ds}{\big( \int_{s}^{\alpha }f(t)\,dt
\big) ^{3/2}}\Big) ^{3}\,du<0, \label{**}
\end{equation}
and the turn is to the left if the opposite inequality is true.
In this paper we investigate the open problem of exact multiplicity in case
of a quintic function $f(u)$. See \cite
{A,BB,Ca,CHS,C1,C2,DL1,DL2,GKL,H,HKL,KL,MS,S1,SS,W1} for other multiplicity
results. We consider positive solutions in case
$f(u)=-(u-a)(u-b)(u-c)(u-d)(u-e)$, i.e. it is a quintic whose roots are five
distinct positive constants $00. \label{KLO1}
\end{equation}
Then there exists a critical $\lambda _{0}$, such that
problem \eqref{0} with $f(u)=-(u-a)(u-b)(u-c)$, $00\quad \text{and}\quad
\int_{c}^{e}f(t)\,dt>0, \label{LR1}
\end{equation}
has the following solution structure: There exist two critical
$0<\lambda _{1},\lambda _{2}$, such that the problem \eqref{0} with
$f(u)=-(u-a)(u-b)(u-c)(u-d)(u-e)$, $00\}; \\ P_{2}\equiv \{(a,b,c,d,e)\in
\mathbb{R}^{5}: (a,b,c,d,e)=(0.1t,0.2t,0.4t,0.6t,t),t>0\}; \\ P_{3}\equiv
\{(a,b,c,d,e)\in \mathbb{R}^{5}: (a,b,c,d,e)=(0.1t,0.2t,0.5t,0.6t,t),t>0\};
\\ \text{ }P_{4}\equiv \{(a,b,c,d,e)\in \mathbb{R}^{5}: (a,b,c,d,e) \text{
}=(0.1t,0.2t,0.5t,0.7t,t),t>0\};\ \\ P_{5}\equiv \{(a,b,c,d,e)\in
\mathbb{R}^{5}: (a,b,c,d,e)=(0.1t,0.2t,0.6t,0.7t,t),t>0\}; \\ P_{6}\equiv
\{(a,b,c,d,e)\in \mathbb{R}^{5}: (a,b,c,d,e)=(0.1t,0.2t,0.6t,0.8t,t),t>0\};
\\ P_{7}\equiv \{(a,b,c,d,e)\in \mathbb{R}^{5}:
(a,b,c,d,e)=(0.1t,0.2t,0.7t,0.8t,t),t>0\}; \\ P_{8}\equiv \{(a,b,c,d,e)\in
\mathbb{R}^{5}: (a,b,c,d,e)=(0.2t,0.3t,0.5t,0.6t,t),t>0\}; \\ P_{9}\equiv
\{(a,b,c,d,e)\in \mathbb{R}^{5}: (a,b,c,d,e)=(0.2t,0.3t,0.5t,0.7t,t),t>0\};
\\ P_{10}\equiv \{(a,b,c,d,e)\in \mathbb{R}^{5}:
(a,b,c,d,e)=(0.2t,0.3t,0.6t,0.7t,t),t>0\}; \\ P_{11}\equiv \{(a,b,c,d,e)\in
\mathbb{R}^{5}: (a,b,c,d,e)=(0.2t,0.3t,0.6t,0.8t,t),t>0\}; \\ P_{12}\equiv
\{(a,b,c,d,e)\in \mathbb{R}^{5}: (a,b,c,d,e)=(0.2t,0.3t,0.7t,0.8t,t),t>0\};
\\ P_{13}\equiv \{(a,b,c,d,e)\in \mathbb{R}^{5}:
(a,b,c,d,e)=(0.2t,0.3t,0.8t,0.9t,t),t>0\}; \\ P_{14}\equiv \{(a,b,c,d,e)\in
\mathbb{R}^{5}: (a,b,c,d,e)=(0.3t,0.4t,0.7t,0.8t,t),t>0\}. \end{gathered}
\label{paraset}
\end{equation}
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1}
\end{center}
\caption{ Solution with $u(0)=$alpha vs. $\protect\lambda =$lambda, where
$a=0.1$, $b=0.2$, $c=0.4$, $d=0.5$, $e=1$ with $10000$ points}
\end{figure}
\begin{theorem} \label{thm:11}
For each parameter group defined above in (\ref{paraset})
$P_{i},i=1\dots 14,$ there is an open neighborhood $U_{i}$ of
$P_{i},i=1\dots 14$ in $\mathbb{R}^{5}$ such that the above
conjecture holds for $(a,b,c,d,e)\in U_{i}$, $i=1\dots 14$.
\end{theorem}
Our proof of this theorem is based on two numerical computations, which
would constitute a ``traditional'' proof if their results could be
analytically justified. Our first computation shows that for each of the
above parameter groups, ranges of possible $\alpha =u(0)$, $G(\alpha )$ in (
\ref{*}) is very narrow and close to $0$ (including numerical errors), which
identify the ranges of $\alpha =u(0),$ where bifurcation might have
occurred. Our second computation shows that in these ranges of $\alpha =u(0)$
the function $D(\alpha )$ in (\ref{**}) is negative $,$ which means that
only turns to the right is possible. These two computations together prove
the conjecture for these parameter groups stated in the theorem above.
The paper is organized as follows: In \S 2 limiting behavior of the
solutions on these bifurcation branches and their stabilities are
investigated; (\ref{*}) and (\ref{**}) are derived and proved in \cite{KLO2}
but for the sake of completeness in 3 we will provide a derivation of (\ref
{**}) and in 4 we will provide a derivation of (\ref{*}) following \cite
{KLO2}.
\section{Limiting behavior and the stability of the branches}
We know from Theorem \ref{thm:11} that all the positive solutions for the
quintic problem \eqref{0} lie on three smooth solution curves for these open
sets $G_{i},i=1\dots 14$. These curves are denoted by $\Gamma _{0}$, which
is the curve that start at $(0,0)$ and it is increasing in $\lambda ;$the
other two curves, which we denoted by $\Gamma _{1}$, and $\Gamma _{2}$\ are
parabola-like curves, with exactly one turn to the right. We denote
$\Gamma_{i}^{+}$ and $\Gamma _{i}^{-}$ to be the upper and lower branch of
$\Gamma_{i}$, respectively for $\mathit{i}=1,2$. Although we provide a
computer assisted proof for this theorem, we also are able to prove the
limiting behavior results of the lower and upper branches of the solution
curves inspired by the results in \cite{KLO}.
\begin{theorem}
The first curve $\Gamma _{0}$, starts at $(\lambda =0,u=0)$, it is
increasing in $\lambda $, and
$\lim_{\lambda \to \infty }u(x,\lambda)=a$\ for $x\in (-1,1)$.
The second curve $\Gamma _{1}$\ has a turning point
at $\lambda _{1}$\ with $\lim_{\lambda \to \infty ,\text{ }\Gamma
_{1}^{+}}u(0,\lambda )=c$\ while $\lim_{\lambda \to \infty ,\Gamma
_{1}^{-}}u(0,\lambda )=\gamma _{1}.$\ The third curve $\Gamma _{2}$\ has a
turning point at $\lambda _{2}$\ with $\lim_{\lambda \to \infty
,\Gamma _{2}^{+}}u(0,\lambda )=e$\ while $\lim_{\lambda \to \infty ,
\text{ }\Gamma _{2}^{-}}u(0,\lambda )=\gamma _{2}.$\ Where $\gamma _{1}$\ is
the unique root of $\int_{a}^{\gamma _{1}}f(t)\,dt=0$\ in $(b,c)$\ and
$\gamma _{2}$\ is the unique root of $\int_{c}^{\gamma _{2}}f(t)\,dt=0$\ in
$(d,e)$.
\end{theorem}
\begin{proof}
We can view $\Gamma _{0}$ as a curve of solution emanating from the solution
$(u=0,\lambda =0)$ as a consequence of the Implicit Function Theorem. By the
standard result in sub-super solution (subsolution $\leq $\ supersolution)
we have that the solution curve stays below $a$, and since $f'(u)<0$
for $u,0$. Differentiating \eqref{0} with respect to $\lambda $ to
obtain that $u_{\lambda }>0$, as is was showed in \cite{KLO}. Hence, we
conclude that the curve tends to $a$ as $\lambda \to \infty $.
From our result above we have that at the points $( \lambda_{1},u_{1}) $,
and $( \lambda _{2},u_{2}) $ the curves of solutions turns to the right in
the $\lambda u$-plane, thus obtaining two parabolas-like curves. First, we
will show that $\Gamma _{1}^{+}$ is increasing for all $\lambda >\lambda
_{1} $. Assume the contrary, and let $\lambda ^{\ast }$\ be the first value
such that $u_{\lambda }(\lambda ^{\ast},x_{1})=0$ for some $x_{1}\in (0,1)$.
It is easy to see that $x_{1}$ is a minimum point of $u_{\lambda }$, and
$u_{\lambda }''(\lambda ^{\ast },x_{1})\geq 0$. Differentiating
\eqref{0} with respect to $\lambda $ and with respect to $x$:
\begin{gather}
u_{\lambda }''+f(u)+\lambda f'(u)u_{\lambda }=0,
\label{1} \\
u_{x}''+\lambda f'(u)u_{x}=0. \label{2}
\end{gather}
From \eqref{1} and the fact that $u_{\lambda }(\lambda ^{\ast },x_{1})=0$
and that $u_{\lambda }''(\lambda ^{\ast },x_{1})\geq 0$, we have
$f(u(\lambda ^{\ast },x_{1}))\leq 0,$ which implies that $u(\lambda ^{\ast
},x_{1})\in \lbrack a,b]$. Multiplying \eqref{1} by $u_{x}$ and \eqref{2} by
$u_{\lambda }$ and subtracting and integrating from $0$ to $x_{1}$ we
obtain:
\begin{equation}
(u_{\lambda }'u_{x}-u_{x}'u_{\lambda
})|_{0}^{x_{1}}+\int_{0}^{x_{1}}f(u)u_{x}dx=0 \label{3}
\end{equation}
Simplifying (\ref{3}) to obtain,
\begin{equation}
u''(0)u_{\lambda }(0)+\int_{u(0)}^{u(x_{1})}f(u)u_{x}dx=0
\label{4}
\end{equation}
Observe that the first term of (\ref{4}) is less or equal to 0, it follows
that the second term must be positive. As we have mentioned that a necessary
condition for $u(0)$ is $\int_{a}^{u(0)}f(u)du>0$,which implies that
$\int_{u(x_{1})}^{u(0)}f(u)du>\int_{a}^{u(0)}f(u)du>0$. Hence the second term
in (\ref{4}) is negative, which is a contradiction, and therefore the upper
branch is increasing. Furthermore, the upper branch is bounded by $c$, in
that case as $\lambda \to \infty $ the limit must exists. For $x\in (-1,1)$
this limit need to be $b$ or $c$, but $u$ is convex below $b$ and since the
upper branch is increasing cannot converge towards $b$. Hence the
$\lim_{\lambda \to \infty }\Gamma _{1}^{+}=c$. Similar argument show
the limiting behavior of the upper branch of the third curve.
For the lower branch of $\Gamma _{1}$, first recall that $\gamma _{1}$ is
the unique solution of equation $\int_{a}^{\gamma _{1}}f(u)du=0$ in $(b,c)$.
From the analysis in \cite{KLO2} it was shown that the bifurcation can only occur in
the interval $(\max \{\beta _{1},\gamma _{1}\},c)$, which implies that
$\Gamma _{1}^{-}$ is bounded below by $\gamma _{1}$, where $\beta _{1}$ is
the unique root of $f'(\beta _{1})=\frac{f(\beta _{1})}{\beta _{1}-a}
$ in $(b,c)$. By a similar argument used for the upper branch case, we will
show that the lower branch is decreasing in $\lambda $ at $x=0$. By assuming
the contrary, let $\lambda _{\ast }$ be the first value so that $u_{\lambda
}(0,\lambda _{\ast })=0$. By \eqref{0} we can conclude that $u_{\lambda
}''(0,\lambda _{\ast })<0$ thus $x=0$ is not a minimum for
$u_{\lambda }(x,\lambda _{\ast })$, so $u_{\lambda }(x,\lambda _{\ast })<0$
for $x>0$, but closed to 0.
Multiplying \eqref{1} by $u_{x}$ and \eqref{2} by $u_{\lambda }$ and
subtracting and integrating from $0$ to $1$ we obtain
\begin{equation}
(u_{\lambda }'u_{x}-u_{x}'u_{\lambda
})|_{0}^{1}+\int_{u(0)}^{0}f(u)du=0
\end{equation}
We proved earlier that the integral is negative, and the first term simplify
to $u_{\lambda }'(1)u'(1)$. Hence, $u_{\lambda
}'(1,\lambda _{\ast })<0$, thus $u_{\lambda }(x,\lambda _{\ast })$
is positive near 1. Then $u_{\lambda }(x,\lambda _{\ast })$ must have a zero
in $(0,1)$. Let $x_{1}$ be the smallest zero. Now, multiplying \eqref{1} by
$u_{x}$ and \eqref{2} by $u_{\lambda }$ and subtracting and integrating from
$0$ to $x_{1}$ we obtain:
\begin{equation}
(u_{\lambda }'u_{x}-u_{x}'u_{\lambda
})|_{0}^{x_{1}}+\int_{u(0)}^{u(x_{1})}f(u)du=0
\end{equation}
It is easy to observe, by the same arguments that were used before that both
terms of the equation are negative, which is a contradiction. Therefore the
lower branch $\Gamma _{1}^{-}$ is decreasing in $\lambda $ at $x=0$. And
similarly the lower branch $\Gamma _{2}^{-}$ can be shown to be decreasing
in $\lambda $ at $x=0$.
\end{proof}
\begin{theorem} \label{thm2.2}
Solution on the upper branch of $\Gamma _{i}$ are stable, while solution on
the lower branch of $\Gamma _{i}$ are unstable, for $i=1,2$.
\end{theorem}
\begin{proof}
First it is easy to show that $\Gamma _{0}$ is stable since $f'(u)<0$
in $(0,a)$. Next assume to the contrary that $u=u(\lambda ,x)$ is a solution
on the upper branch $\Gamma _{1}^{+}$ $\ $that is unstable. That is, we can
find a constant $\mu >0$, and $w(x)>0$, such that
\begin{equation}
\begin{gathered} w''+\lambda f'(u)w=\mu w\quad\text{on }(-1,1),\\
w(-1)=w(1)=0 \end{gathered} \label{6}
\end{equation}
We may assume that $\int_{0}^{1}w^{2}dx=1$. Observe that multiplying (\ref{0}
) by $u'$ and integrating over $(0,x)$ we obtain
\begin{equation}
| u'| \;\geq \sqrt{\lambda }K \label{7}
\end{equation}
for some $K>0$ when $\lambda $ is large, for all $x\in (\eta ,1)$, where
$u(\eta )=\alpha $ and $\alpha $ is the largest root of $f'(u)$ in
$(b,c)$. Recall that $u\to c$, we can find a constant $M$ and $\xi =\xi
(\lambda )$ near 0, such that $| u''(\xi)| \leq M$. Next,
differentiate \eqref{0} and multiply by $w$, and multiply (\ref{6}) by
$u'$, and subtracting the equations to obtain,
\begin{equation}
u_{x}''w-w''u'=\mu wu'\; \label{8}
\end{equation}
Integrating this equation over $\xi $ to 1, we have
\begin{equation}
-u'(1)w(1)-u''(\xi )w(\xi )+u'(\xi
)w'(\xi )+\mu \int_{\xi }^{1}wu'dx=0\,. \label{9}
\end{equation}
Recall that $w$ is bounded, and $w''>0$ on $(0,1)$, it follows
that $w'(\xi )>0$. Thus we know that the second term in (\ref{9}) is
bounded, and the third and fourth term are negative. Assume the first term
is positive and small, that is $| w'(1)| =O\big( \frac{1}{\sqrt{
\lambda }}\big) $. Since $w(x)$ is convex on $(0,\eta )$, it follows that it
must attain its maximum on $(\eta ,1)$, and since $\int_{0}^{1}w^{2}dx=1$
the maximum is at least 1. Changing the variable to $t=1-x$, we have from (
\ref{6}) the following estimate
\begin{equation}
w(t)\leq c\lambda w, w(0)=0, w'(0)=O\left( \frac{1}{ \sqrt{\lambda }}
\right) , \quad 00$ the curve turns to the right, and if $J<0$
to the left. We derive here a formula for $J$, which does not require a
detailed knowledge of $u(x)$, and any knowledge of $w(x)$. It depends only
on the maximal value of the critical solution $u(0)\equiv \alpha $.
\begin{theorem}[\cite{KLO2}]\label{thm:1}
At any critical solution $u(x)$, with $u(0)=\alpha $,
\[
J =-c\int_{0}^{\alpha }f''(u)\Big( \int_{u}^{\alpha
}f(s)\,ds\Big) \Big( \int_{0}^{u}\frac{ds}{\big( \int_{s}^{\alpha
}f(t)\,dt\big) ^{3/2}}\Big) ^{3}\,du
=-cD(\alpha ),
\]
where $c=\frac{1}{4\sqrt{2}}{u'}^{3}(1){w'}^{3}(1)>0$.
\end{theorem}
\begin{proof}
Differentiating equation \eqref{1},
\begin{equation}
u''(x)+f'(u(x))u'(x)=0\quad \text{in }(-1,1).
\end{equation}
Using this equation and \eqref{2}, we conclude that the function
$u''(x)w(x)-u'(x)w'(x)$ is constant, and hence
\begin{equation}
u''(x)w(x)-u'(x)w'(x)=-C,\quad \mbox{where }
C=u'(1)w'(1)>0.
\end{equation}
We rewrite this as
\begin{equation*}
\big( \frac{w}{u'}\big) '=\frac{C}{{u'}^{2}},
\end{equation*}
and then integrate, concluding that
\begin{equation}
w(x)=-{C}u'(x)\int_{x}^{1}\frac{1}{{u'}^{2}(t)}\,dt.
\end{equation}
This formula will allow us to exclude $w(x)$ in $J$. (Observe that
$\int_{x}^{1}\frac{1}{{u'}^{2}(t)}\,dt$ tends to infinity as $x\to 0$
, while $u'(x)$ tends to zero. Hence both terms ought to be kept
together in numerical computations.) Using (\ref{7}) in (\ref{3})
\begin{equation}
J=2C^{3}\int_{0}^{1}f''(u(x)){u'}^{3}(x)\Big(
\int_{x}^{1}\frac{1}{{u'}^{2}(t)}\,dt\Big) ^{3}\,dx.
\end{equation}
We now wish to exclude $u'(x)$ from (\ref{8}). Since the energy
$\frac{{u'}^{2}}{2}(x)+F(u(x))$ is constant,
\begin{equation}
\frac{{u'}^{2}}{2}(x)+F(u(x))=F(u(0))=F(\alpha ).
\end{equation}
On the interval $(0,1)$ we express
\begin{equation}
u'(x)=-\sqrt{2}\sqrt{F(\alpha )-F(u(x))}.
\end{equation}
We use this formula in the integral $\int_{x}^{1}\frac{1}{{u'(t)}
^{2}}\,dt$, and then we make a change of variables $t\to s$, by letting
$s=u(t)$. We have
\begin{equation*}
\int_{x}^{1}\frac{1}{{u'}^{2}(t)}\,dt=-\frac{1}{2^{3/2}}\int_{x}^{1}
\frac{u'(t)\,dt}{\left[ F(\alpha )-F(u(t))\right] ^{3/2}}=-\frac{1}{
2^{3/2}}\int_{u(x)}^{0}\frac{1}{\left[ F(\alpha )-F(s)\right] ^{3/2}}\,ds.
\end{equation*}
We then have
\begin{equation*}
J=c\int_{0}^{1}f''(u(x))u'(x)\left[ F(\alpha )-F(u(x))
\right] \Big( \int_{0}^{u(x)}\frac{1}{\left[ F(\alpha )-F(s)\right] ^{3/2}}
\,ds\Big) ^{3}\,dx,
\end{equation*}
with $c=\frac{1}{4\sqrt{2}}C^{3}$. Finally, replacing $F(\alpha )-F(u)$ by
$\int_{u}^{\alpha }f(s)\,ds$, making a change of variables $u=u(x)$, and
writing $\alpha $ for $u(0)$, we obtain (\ref{4}).
\end{proof}
Computing the bifurcation points
In the previous section we computed the direction of turn, assuming that
bifurcation occurs at $u(0)=\alpha$. We now provide a way to determine all
possible $\alpha$'s at which bifurcation may occur, i.e. the corresponding
solution of \eqref{1} is singular.
\begin{theorem}[\cite{KLO2}] \label{thm:5}
A solution of the problem \eqref{1} with the
maximal value $\alpha =u(0)$ is singular if and only if
\begin{equation}
G(\alpha )\equiv F(\alpha )^{1/2}\int_{0}^{\alpha }\frac{f(\alpha )-f(\tau )
}{\left[ F(\alpha )-F(\tau )\right] ^{3/2}}\,d\tau -2=0. \label{70}
\end{equation}
\end{theorem}
\begin{proof}
We need to show that the problem \eqref{2} has a non-trivial solution. As
follows by (\ref{7}) (or by direct verification) the function ${
\displaystyle w(x)=-u'(x)\int_{x}^{1} \frac{1}{{u'}^{2}(t)}
\, dt}$ satisfies the equation in \eqref{2}. Also $w(1)=0$. If we also have
\begin{equation}
w'(0)=0 \label{71}
\end{equation}
then since $u(x)$ is an even function, the same is true for $w(x)$ (by
uniqueness for initial value problems), and hence $w(-1)=0$, which gives us
a non-trivial solution of \eqref{2}. Conversely, every non-trivial solution
of \eqref{2} is an even function, and hence (\ref{71}) is satisfied.
Using the equation in \eqref{1}, we compute
\begin{equation*}
w'(x)=f(u(x))\int_{x}^{1}\frac{1}{{u'}^{2}(t)}\, dt+\frac{1}{
u'(x)}.
\end{equation*}
Using the formula \eqref{10} from the previous section and the one right
below it, we express
\begin{equation} \label{72}
2^{3/2}w'(x)=\int_{0}^{u(x)}\frac{f(u(x))}{\left[F(\alpha)-F(\tau)
\right]^{3/2}}\, d\tau- \frac{2}{\left[F(\alpha)-F(u(x))\right]^{1/2}}.
\end{equation}
If we try to set here $x=0$, then both terms on the right are infinite.
Instead, we observe that
\begin{equation} \label{73}
\begin{aligned}
-\frac{2}{\left[F(\alpha)-F(u)\right]^{1/2}}
&=-\int_{0}^{u}\frac{d}{d\tau}
\frac{2}{\left[F(\alpha)-F(\tau)\right]^{1/2}} \, d\tau-\frac{2}{
F(\alpha)^{1/2}} \\
&=-\int_{0}^{u}\frac{f(\tau)}{\left[F(\alpha)-F(\tau)\right]^{3/2}}\, d\tau -
\frac{2}{F(\alpha)^{1/2}}.
\end{aligned}
\end{equation}
Using (\ref{73}) in (\ref{72}), we obtain
\begin{equation}
2^{3/2}w'(x)=\int_{0}^{u(x)}\frac{f(u(x))-f(\tau)}{\left[
F(\alpha)-F(\tau)\right]^{3/2}}\, d\tau-\frac{2}{F(\alpha)^{1/2}}.
\end{equation}
The integral on the right is now non-singular as we let $x\to0$. At $x=0$ we
see that (\ref{71}) is equivalent to (\ref{70}).
\end{proof}
A computer assisted proof
For the quintic nonlinearity problem \eqref{11}, by letting $u=ev$, we may
assume that $e=1$, so that our nonlinearity is
$f(u)=-(u-a)(u-b)(u-c)(u-d)(u-1)$, with new $a,b,c$ and $d$, i.e. we consider
\begin{equation} \label{n2}
\begin{gathered} u''+\lambda(u-a)(u-b)(u-c)(u-d)(u-1) = 0\quad \mbox{in
}(-1,1), \\ u(-1) = u(1)=0. \end{gathered}
\end{equation}
This substitution allows us to ``compactify'' the parameter set, since now
$00$. Now let
$\gamma_{1}$ be the unique root of $\int_{a}^{\gamma_{1}}f(t)\, dt=0$ in
$(b,c)$ and $\gamma_{2}$ be the unique root of $\int_{c}^{\gamma_{2}}f(t)\,
dt=0$ in $(d,1)$\textit{.} Next let $\beta_{1}$ be the unique root of
$f'(\beta_{1})=\frac{f(\beta_{1})}{ \beta-a}$ in $(b,c)$ and
$\beta_{2}$ be the unique root of $f'(\beta_{2})=\frac{f(\beta_{2})}{
\beta-c}$ in $(d,1)$ (the point where a straight line through the point
$(a,0)$ touches the graph of $y=f(u)$ for $u\in(a,c)$ and where a straight
line through the point $(c,0)$ touches the graph of $y=f(u)$ for $u\in(c,1)$
.)
We parameterize the solutions of (\ref{n2}) by their maximum value $\alpha
=u(0)$. For each parameter group in Theorem \ref{thm:11} we compute
$\beta_{i}$ and $\gamma _{i},i=1,2$. Since the bifurcation could only occur
for $\alpha \in (\max (\beta _{1}, \gamma _{1}),c)$ and for $\alpha \in
(\max (\beta _{2}$,$\gamma _{2}),1)$, we compute $G$ defined by (\ref{*})
for $\alpha \in (\max (\beta _{1},\gamma _{1}),c)\cup (\max (\beta _{2}$,
$\gamma _{2}),1)$ and produce two ranges of $\alpha =u(0)$ in each case
(including numerical errors). Next for these ranges we compute the integral
$D$ defined by (\ref{**}). Actually we have computed
\begin{equation*}
II\equiv F(\alpha )^{5/2}\int_{0}^{\alpha }f''(u) \Big(
\int_{u}^{\alpha }f(s)\,ds\Big) \Big( \int_{0}^{u}\frac{ds}{\big(
\int_{s}^{\alpha }f(t)\,dt\big) ^{3/2}} \Big) ^{3}\,du,
\end{equation*}
where the extra term $F(\alpha )^{5/2}$ is introduced to make this integral
scaling invariant in $f$. Our computation shows that $II<0$, which implies
that only turns to the right are possible near in these ranges. Since both
$G $ and $D$ are continuous in the parameters $(a,b,c,d,e)$ and $D\neq 0$
implies that the zero of $D$ is not degenerate, there exists an open
neighborhood $U_{i}$ for each parameter group $P_{i},i=1\dots 14$.
As an example, when $(a,b,c,d,e)=(0.1,0.2,0.4,0.5,1)$, we find that
\[
\max(\beta_{1},\gamma_{1})=0.277989142,\quad
\max(\beta_{2},\gamma_{2})=0.83216494
\]
so that $G$ is evaluated in $(0.277989142,0.4)
\cup(0.83216494,1)$. We find the ranges of where $G$ is
``close'' to $0$ are $(0.32560,0.32562)\cup(0.91744,0.91746)$
and compute the
values of $D$ there to be negative in $100's$ and in $10's$.
\end{proof}
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\end{document}