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\title{Mixed principal eigenvalues in dimension one}

\author[Yu-Hui Zhang]{Yu-Hui Zhang}

\date[October 3, 2012]{\wuhao 2$^{\text{nd}}$ BNU-Swansea University Workshop \\Stochastic Processes and Applications(Oct.1-3, 2012) \\
$ $\\ This talk is based on the joint work with Mu-Fa Chen and
Ling-Di Wang.}

\institute[BNU]{\sihao Beijing Normal University}

\frame{\titlepage}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%            目录            %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\section*{目录}
  %  \frame{\frametitle{目录}\tableofcontents}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%         背景及问题        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\large
%----------------------------------------------------------------------------------------1

\begin{frame}
 \frametitle{Outline}
\vskip0.5cm
\begin{enumerate}
\item Problem
\vskip0.3cm
\item Background
\vskip0.3cm
\item Our work\\\small{
\qquad $(1)$ Analysis\\
\qquad $(2)$ Notations\\
\qquad $(3)$ Main results: (I), (II), (III)\\
\qquad $(4)$ From ND to DN \\
\qquad $(5)$ The property of eigenfunction}
\vskip0.3cm
\item Other related research
\item Main references
\end{enumerate}
\end{frame}



%----------------------------------------------------------------------------------------2
\frame{ \frametitle{1. Problem}\small
\begin{itemize}
\item
Consider one dimensional diffusion process: $$L=a(x)\frac{{\rm d}^{2}}{{\rm d} x^{2}}+b(x)\frac{\rm d}{{\rm d} x},\qquad x\in (0, D),\quad D\leqslant\infty,$$
where {\red $a$ and $b$ are Borel measurable, $a>0$ on $(0,D)$}.\\
\vskip0.2cm Boundary conditions: Neumman {\red (`N')}; Dirichlet {\red(`D')}. \\ Four cases: \quad ND\quad DN\quad NN\quad DD
\item
 Eigenequation: $$Lg=-\lambda g,\qquad {\red \text{`$+$'} }\qquad \text{ boundary conditions}.$$ $g$ is called the eigenfunction corresponding to the eigenvalue $\lambda$.
\end{itemize}}
%----------------------------------------------------------------------------------------3
\frame{\frametitle{1. Problem }\small
\begin{itemize}
\item<1> Let $C(x)=\int_0^xb/a$. Define two important measures:
\\ \hskip2cm{\blue $\text{Speed measure: } \mu({\rm d} x)=e^{C(x)}/a(x){\rm d} x$,}
\\ \hskip2cm{\blue $\text{Scale measure: } \nu({\rm d} x)=e^{-C(x)}{\rm d} x$.}
\item<1> {\red ND case}. Let $\lambda_0^{(ND)}$ be the minimal solution to the following eigenequation:
    $$Lg=-\lambda g,\qquad g'(0)=0,\quad g(D)=0 \text{ if }D<\infty.$$
\item $\lambda_0^{(ND)}$ has a classical variational formula:
{\small \begin{equation*}\label{f1.1}
 \lambda_0^{(ND)}=\inf\{D(f):\mu(f^{2})=1,\ f\in\mathscr{C}_{K}[0, D],\ f(D)=0 \text{ if } D<\infty\},
\end{equation*}
where $D(f)=\int_{0}^{D}af'^2{\rm d}\mu$,\\
 $\mathscr{C}_{K}[0,D]=\{f:f\in \mathscr{C}^{1}(0, D)\cap\mathscr{C}[0, D],\  \text{supp} (f)\, \text{is compact}\}.$}
\end{itemize}
}
%----------------------------------------------------------------------------------------4
\frame{\frametitle{1. Problem}\small

$$\aligned
{\mathscr C}[0, D]&=\{f:f\;\text{\rm is continuous on}\;[0, D]\},\\
 {\mathscr C}^k(0, D)&=\{f: f \text{ has continuous derivatives of order $k$ on $(0, D)$}\}, k\geqslant 1.
 \endaligned$$

 We focus on
\begin{block}{}
 \centerline\sihao{How to estimate on $\lambda_0^{(ND)}$ and $\lambda_0^{(DN)}$?}
\end{block}
Consider $\|P_tf\|\leqslant \|f\|e^{-\varepsilon t}$. Then $\lambda_0=\varepsilon_{\max}$, i.e.,$\lambda_0$ is the maximal {\red decay(stability) speed }of diffusion process.
}
%----------------------------------------------------------------------------------------5
\frame{\frametitle{2. Background}\small
There are a great number of publications for the background and motivation of the study on the principal eigenvalue or the Poincar\'e-type inequalities. Now we recall some results of our group as follows.

\begin{block}{}
 \small{In 1997, a variational formula for the lower bound of spectral gap of an elliptic operator was presented by M.F. Chen and F.Y. Wang. The eigenfunction $g$ is monotone. (NN)}
\end{block}
\begin{block}{}
 \small{In 1999, Chen obtained again some dual variational formulas for the lower and upper bounds of eigenvalue by analytic method and proved that the eigenfunction $g$ in the DN case is increasing. (NN and DN)}
\end{block}
\begin{block}{}
 \small{In 2000, Chen obtained variational formulas and basic estimates for eigenvalue in the DN case. (DN)}
\end{block}
}

%----------------------------------------------------------------------------------------6
\frame{\frametitle{2. Background}

\begin{block}{}
 \small{In 2002, Chen,  Y.H. Zhang and X.L. Zhao obtained variational formulas and explicit estimates for the processes with DD boundaries including continuous and discrete cases.  (DD)}
\end{block}
\begin{block}{}
\small{In 2003, using the variational formulas, M.F. Chen obtained approximating procedure for one dimensional diffusion process with NN boundary. (NN)}
\end{block}
\begin{block}{}
\small{In 2010, Chen obtained variational formulas and approximating procedure for birth-death process.(Four cases)}
\end{block}
\small Our work is a continuation of the above paper (in the
continuous situation). }
%---------------------------------------------------------7
\frame{\frametitle{3(1). Our work ---\small{Analysis}}\small
Another form of the operator: $ L=\frac{\rm d}{{\rm d} \mu}\frac{\rm d}{{\rm d} \nu}$.
$$\aligned{\green \Rightarrow}&\frac{\rm d}{{\rm d} \nu}g(\beta)-\frac{\rm d}{{\rm d} \nu}g(\alpha)=-\lambda_0\int_{\alpha}^{\beta}g{\rm d}\mu,\quad \alpha<\beta.\\
 &{\green\big\Downarrow} {\text{ by }{\blue g'(0)=0}}\\
 &\frac{\rm d}{{\rm d}\nu}g(x)=-\lambda_0 \int_0^x g{\rm d} \mu{\green\Rightarrow} {\blue I(g)(x):=-\frac{e^{-C(x)}}{g'(x)}\int_0^xg{\rm d} \mu}=\lambda_0^{-1}\\
 &{\green\big\Downarrow}\\
 &g(x)-g(D)=\lambda_0\int_x^D\nu({\rm d} s)\int_0^s g{\rm d}\mu.\\
 &{\green\big\Downarrow} {\text{ if }{\red g(D)=0 }}\\
 & g(x)\!=\!\lambda_0\!\!\int_x^D\!\!\!\nu({\rm d} s)\!\!\int_0^s \!\!g{\rm d}\mu {\green\Rightarrow}{\blue I\!I(g)(x)\!:=\!\frac{1}{g(x)}\!\!\int_x^D\!\!\!\nu({\rm d} s)\!\!\int_0^s\!\!g{\rm d}\mu}\!=\!\lambda_0^{-1}.\endaligned$$
}

%--------------------------------------------------------------------8
\frame{\frametitle{3(1). Our work---\small{Analysis}}{\small
\begin{itemize}
\item If we `know' the eigenfunction $g$, then $\blue{ \lambda_0=I(g)(x)^{-1}=I\!I(g)(x)^{-1}}.$
Moreover, ($h=g'/g$)
$${\blue \lambda_0=}-Lg(x)/g(x)=-(ah^2+bh+ah')(x)=:{\blue R(h)(x)};$$
and also  ($h=g'$)
$${\blue\lambda_0=}-(Lg)'(x)/g'(x)=-\big(ah'+bh\big)'/h(x)=:{\blue \overline R(h)(x)}.$$
 \item   However, it is very hard to know the explicit expression of eigenfunction in general case.
\item The property of eigenfunction: {\red either positive or negative; monotone; $g(D):=\lim_{x\to D}g(x)=0$ (Essential)}\\
 Lemmas 4, 5, 6 consist of the basis of the test functions used in the following definitions of ${\mathscr F}_\#$ and ${\mathscr H}$ $\to$ how to choose the test function.
\end{itemize}}}
%----------------------------------------------------------------------------9
\begin{frame}
\frametitle{3(2). Our work---\small{Notations}}
{\small
$$\aligned
&\mathscr{F}_{I}=\big\{f: f\in \mathscr{C}^1(0, D)\cap \mathscr{C}[0, D],\,{\red f|_{(0, D)}>0,\text{ and }f'|_{(0,D)}<0}\big\},\\
&\mathscr{F}_{I\!I} =\{f: f\in \mathscr{C}[0,D]\text{ and } f|_{(0, D)}>0\},\\
&\mathscr{H}\!=\!\big\{h: h\in \mathscr{C}^1(0, D)\cap \mathscr{C}[0, D],\, h(0)=0,\,h|_{(0, D)}<0\text{ if }\nu(0, D)\!\!<\!\!\infty,\\
&\qquad\qquad\, \text{and } h|_{(0, D)}\leqslant 0\text{ if }\nu(0, D)=\infty\big\}.
\endaligned$$
$$\aligned
&\mathscr{{\widetilde F}}_{I}\!=\big\{f\!:\! f\!\in\! {\mathscr C}^1\!(x_{0}, x_{1}) \cap {\mathscr C}[x_{0},x_{1}], {\red f'|_{(x_{0},x_{1})}\!<0}\, \text{for some}\, x_0,x_1\!\in\![0,D)\\&\text{\hskip 4em} \text{ with }x_0\!<\!x_1
\text{ and } {\red f=f(\cdot\vee x_{0}) \mathbf{1}_{[0, x_{1})}}\big\},\\
&\mathscr{{\widetilde F}}_{I\!I}\!=\big\{f: \exists x_{0}\in(0, D)\text{ such that } f=f \mathbf{1}_{[0, x_{0})}\, \text{ and }\, f\in  {\mathscr C}[0,x_{0}]\big\},\\
&\mathscr{\widetilde{H}}\!=\!\Big\{h\!:\! \exists x_{0}\!\in\!(0, D)\text{ such that } h\!\in\!{\mathscr C}^1(0,x_{0})\cap {\mathscr C}[0,x_{0}],\, h|_{(0, x_0)}\!<\!0,\\
&\text{\hskip 4em}h(0)=0,\, h|_{[x_0,D]}=0,\, \text{ and }\sup_{(0, x_0)}\big(ah^{2}+bh+ah'\big)<0\Big\}.
\endaligned$$}

\end{frame}
%----------------------------------------------------------------------------------10
\frame{\frametitle{3(2). Our work---\small{Notations}}\small
\begin{block}{Hypothesis}
\begin{equation}e^{C}/a \text{ and }  e^{-C} \text{ are locally integrable.}\label{f1}\end{equation}
\end{block}

\begin{itemize}
\item \text{single integral form}
$$I(f)(x)=-\frac{e^{-C(x)}}{f'(x)}\int_{0}^{x}f\,{\rm d}\mu,\quad x\in(0,D);$$
\item double integral form
$$I\!I(f)(x)=\frac{1}{f(x)}\int_{(x, D)\cap{\rm supp}(f)}\!\!\nu({\rm d} s)\!\int_{0}^{s}f{\rm d}\mu,\quad x\in {\rm supp}(f);$$
\item differential forms
$$\aligned &R(h)(x)=-\big(ah^{2}+bh+ah'\big)(x), \quad x\in(0,D);\\
&\overline R(h)(x)=-\big(ah'+bh\big)'(x)/h(x),\quad x\in (0,D).\endaligned$$
\end{itemize}}

%----------------------------------------------------------11
\begin{frame}
\frametitle{3(3). Main results: (I)}\small
\begin{block}{Theorem 1}
(1)Single integral form:
$$\aligned
\inf_{f\in\widetilde{\mathscr{F}}_{I}}\sup_{x\in E}I(f)(x)^{-1}=\lambda_{0}=\sup_{f\in\mathscr{F}_{I}}\inf_{x\in E}I(f)(x)^{-1}.
\endaligned$$
(2) Double integral form:
$$\aligned
\inf_{f\in\widetilde{\mathscr{F}}_{I\!I}}\sup_{x\in \rm supp(f)}I\!I(f)(x)^{-1}=\lambda_{0}=\sup_{f\in\mathscr{F}_{I\!I}}\inf_{x\in E}I\!I(f)(x)^{-1},
\endaligned$$
 Moreover, if $a, b\in\mathscr{C}[0,D]$, then we additionally have a\\
(3) differential form:
$$\aligned
\inf_{h\in\widetilde{\mathscr{H}}}\sup_{x\in E}R(h)(x)=\lambda_{0}=\sup_{h\in\mathscr{H}}\inf_{x\in E}R(h).
\endaligned$$
When $\exists$ a solution $(\lambda_{0},g)$ for the eigeneq., the sup. can be attained.
\end{block}
\end{frame}
%----------------------------------------------------------12
\begin{frame}
\frametitle{3(3). Main results: (I)}\small
\begin{block}{Proposition 1}
$$\aligned
\lambda_{0}&=\inf\Big\{D(f):\mu\big(f^{2}\big)=1,\,f\in \mathscr{C}^1(0, D)\cap \mathscr{C}[0,D]\text{ and }f(D)=0\Big\}\\
&{\red =:\tilde{\lambda}_{0}},\ \text{where}\ f(D)=\lim_{x\to D}f(x)\ \text{in the case of}\ D=\infty.
\endaligned$$
\vskip-0.4cm
$$\aligned
&\inf\limits_{f\in\mathscr{{\widetilde F}'}_{I}}\,\sup_{x\in(0, D)}I(f)(x)^{-1}=\lambda_{0}=\sup\limits_{f\in\mathscr{F}_{I}}\,\inf_{x\in(0, D)}I\!I(f)(x)^{-1},\\
&\inf\limits_{f\in\mathscr{{\widetilde F}}_{I\!I}\bigcup\mathscr{{\widetilde F}}_{I\!I}'}\,\sup_{x\in {\rm supp}(f)}I\!I(f)(x)^{-1}=\lambda_{0}=\inf\limits_{f\in\mathscr{{\widetilde F}}_{I}}\,\sup_{x\in
{\rm supp}(f)}I\!I(f)(x)^{-1},
\endaligned$$
$$\aligned
\text{where}\ &\mathscr{{\widetilde F}}_{I}'= \big\{f: \exists x_{0}\in(0, D)\text{ such that }f=f \mathbf{1}_{[0, x_{0})},\, f\in \mathscr{C}^1(0, x_{0})\\
&\text{\hskip 4em}\cap \mathscr{C}[0,x_{0}],\text{ and } f'|_{(0, x_{0})}<0\big\},\\
&\mathscr{{\widetilde F}'}_{I\!I}=\big\{f: f>0,\, f\in \mathscr{C}[0,D],\text{ and } fI\!I(f)\in L^{2}(\mu)\big\}.
\endaligned$$
Besides, the supremum over $\{f\in\mathscr{F}_{I}\}$ can be attained.\end{block}
\end{frame}
%----------------------------------------------------------13
\begin{frame}
\frametitle{3(3). Main results: (I)}\small

\begin{block}{}
$$\aligned
\lambda_{0}\geqslant\tilde{\lambda}_{0}&
\geqslant\sup\limits_{f\in\mathscr{F}_{I}}\inf\limits_{x\in(0,D)}I(f)(x)^{-1}\\
&=\sup\limits_{f\in\mathscr{F}_{I}}\inf\limits_{x\in(0, D)}I\!I(f)(x)^{-1}=\sup\limits_{f\in\mathscr{F}_{I\!I}}\inf\limits_{x\in(0, D)}I\!I(f)(x)^{-1}\\&{\blue \geqslant\sup\limits_{h\in\mathscr{H}}\inf\limits_{x\in(0, D)}R(h)(x)\geqslant\lambda_{0}\ \ \text{provided that}\ \lambda_0=\tilde\lambda_0.}
 \endaligned$$
\end{block}
\begin{block}{}
$$\aligned
\lambda_{0}&\leqslant\inf\limits_{f\in\mathscr{\widetilde{F}}_{I\!I}\bigcup\mathscr{\widetilde{F}}_{I\!I}'}
\sup\limits_{x\in \text{supp}(f)}I\!I(f)(x)^{-1}
=\inf\limits_{f\in\mathscr{\widetilde{F}}_{I\!I}}\sup\limits_{x\in
\text{supp}(f)}I\!I(f)(x)^{-1}
\\&= \inf\limits_{f\in\mathscr{\widetilde{F}}_{I}}\sup\limits_{x\in
\text{supp}(f)}I\!I(f)(x)^{-1}=
\inf\limits_{f\in\mathscr{\widetilde{F}}_{I}}\sup\limits_{x\in(0, D)}I(f)(x)^{-1}
\\&=\inf\limits_{f\in\mathscr{\widetilde{F}'}_{I}}\sup\limits_{x\in(0,
D)}I(f)(x)^{-1}({\red\leqslant \tilde\lambda_0\Rightarrow \lambda_0=\tilde\lambda_0})
\\&\leqslant \inf\limits_{h\in\mathscr{\widetilde{H}}}\sup\limits_{x\in(0,
D)}R(h)(x)\leqslant\lambda_{0}.
\endaligned$$
\end{block}

\end{frame}
%----------------------------------------------------------14
\begin{frame}
\frametitle{3(3). Main results: (I)}\small Sketch of proof for the
lower estimates:
\begin{block}{}
$$\lambda_{0}\geqslant\tilde{\lambda}_{0}\geqslant\sup\limits_{f\in\mathscr{F}_{I\!I}}\inf\limits_{x\in(0,D)}I\!I(f)(x)^{-1}.$$
\end{block}
(a) {\blue $\lambda_{0}\geqslant\tilde{\lambda}_{0}$} by definitions.

(b) Let $g$ be a test function of $\tilde{\lambda}_{0}$. Then for every $h$ with $h|_{(0,D)}>0$, by Cauchy-Schwarz's inequality and Fubini's Theorem,
$$1=\mu(g^{2})\leqslant  D(g)\sup\limits_{t\in(0, D)}\frac{1}{h(t)}\int_{0}^{t}\frac{e^{C(x)}}{a(x)}\text{d} x\int_{x}^{D}\frac{h(s)}{e^{C(s)}}\text{d} s=:D(g)\sup\limits_{t\in(0, D)}H(t).$$
For $f\in \mathscr{F}_{I\!I}$, we specify $h(t)=\int_{0}^{t}{a(s)}^{-1}{e^{C(s)}}f(s) \text{d} s$. Then
by Cauchy's mean value theorem,
$
\sup\limits_{t\in(0, D)}H(t)\leqslant\sup\limits_{x\in(0, D)}I\!I(f)(x)\Rightarrow$
$$\inf_{x\in(0, D)}I\!I(f)(x)^{-1}\leqslant\inf_{t\in(0, D)}H(t)^{-1}\leqslant D(g)\Rightarrow \cdots\leqslant \tilde{\lambda}_{0}.$$

\end{frame}
%----------------------------------------------------------15
\begin{frame}
\frametitle{3(3). Main results: (I)}\small
\begin{block}{}
$$\sup\limits_{f\in\mathscr{F}_{I}}\inf\limits_{x\in(0,D)}I(f)(x)^{-1}=\sup\limits_{f\in\mathscr{F}_{I}}\inf_{x\in(0,D)}I\!I(f)(x)^{-1}
=\sup\limits_{f\in\mathscr{F}_{I\!I}}\inf\limits_{x\in(0, D)}I\!I(f)(x)^{-1}.$$
\end{block}
(c) For $f\in\mathscr{F}_{I}$,  by using Cauchy's mean value theorem, we have
$$\sup\limits_{x\in(0, D)}I\!I(f)(x)\leqslant \sup\limits_{t\in(0, D)}-\frac{e^{-C(t)}}{f'(t)}\int_{0}^{t}f\text{d}\mu= \sup\limits_{x\in(0, D)}I(f)(x).$$
Since $\mathscr{F}_{I}\subset\mathscr{F}_{I\!I}$,
$$\sup\limits_{f\in\mathscr{F}_{I}}\inf\limits_{x\in(0,D)}I(f)(x)^{-1}\leqslant \sup\limits_{f\in\mathscr{F}_{I}}\inf\limits_{x\in(0, D)}I\!I(f)(x)^{-1}\leqslant
\sup\limits_{f\in\mathscr{F}_{I\!I}}\inf\limits_{x\in(0, D)}I\!I(f)(x)^{-1}$$
$${\red ?\leqslant\sup\limits_{f\in\mathscr{F}_{I}}\inf\limits_{x\in(0,D)}I(f)(x)^{-1}}$$
\end{frame}
%----------------------------------------------------------16
\begin{frame}
\frametitle{3(3). Main results: (I)}\small

(c)(continued) Let $f\in \mathscr{F}_{I\!I}$. Put $g(x)=fI\!I(f)(x)$.
Then $g\in\mathscr{F}_{I}$ and $I(g)(s)^{-1}\geqslant\inf_{x\in(0, D)}I\!I(f)(x)^{-1}\Rightarrow$
$$\inf_{s\in (0, D)}I(g)(s)^{-1}\geqslant\inf_{x\in(0, D)}I\!I(f)(x)^{-1}\Rightarrow\cdots.$$

In the following, we pre-assume that {\red $\lambda_{0}=\tilde\lambda_{0}$}.

A different way to prove the equalities:

By {\red the property of eigenfunction} $g$, we have seen that $\lambda_{0}=I(g)(x)^{-1}$ for $x\in(0, D)$ and $g\in\mathscr{F}_{I}\Rightarrow$
$$\lambda_{0}=\inf\limits_{x\in(0, D)}I(g)(x)^{-1}\leqslant
\sup\limits_{f\in\mathscr{F}_{I}}\inf\limits_{x\in(0, D)}I(f)(x)^{-1}\leqslant\cdots\leqslant \lambda_{0}.$$
({\blue Lemmas 4, 5})
\end{frame}
%----------------------------------------------------------17
\begin{frame}
\frametitle{3(3). Main results: (I)}\small

Now we assume that $a, b\in {\mathscr C}[0,D]$.
\begin{block}{}
$$\sup_{f\in\mathscr{F}_{I\!I}}\inf_{x\in(0, D)}I\!I(f)(x)^{-1} \geqslant
\sup_{h\in\mathscr{H}}\inf_{x\in(0, D)}R(h)(x).$$
\end{block}

(d) For each $h\in\mathscr{H}$ with $h={g'}/{g}$, we have $R(h)=-L g/g$.

If $R(h)>0$ for a positive $g$ with $g'(0)=0$ and $h=g'/g$, then $g$  must be strictly decreasing.

Let $f=gR(h)$. Then $Lg=-f$ and $f\in\mathscr{F}_{I\!I}$. Since $Lg=-f$ and $g'(0)=0\Rightarrow$
 $g(x)\geqslant f(x)I\!I(f)(x)\Rightarrow
R(h)(x)\leqslant I\!I(f)(x)^{-1}$ for $x\in(0,D)\Rightarrow$
$$\inf_{x\in(0, D)}R(h)(x)\leqslant\inf_{x\in(0, D)}I\!I(f)(x)^{-1}
\leqslant\sup_{f\in\mathscr{F}_{I\!I}}\inf_{x\in(0,D)}I\!I(f)(x)^{-1}.$$
\end{frame}

%----------------------------------------------------------18
\begin{frame}
\frametitle{3(3). Main results: (I)}\small

\begin{block}{}
$$\sup_{h\in\mathscr{H}}\inf_{x\in(0,D)}R(h)(x)\geqslant\lambda_{0}.$$
\end{block}

(e) Since $a,\,b\in {\mathscr C}[0,D]$, by {\red the property of eigenfunction}({\blue Lemmas 1, 4, 5}) , there exists an eigenfunction $g$ such that $Lg=-\lambda_{0}g$,
$$g'(0)=0,\ g|_{(0,D)}>0,\ g'|_{(0,D)}<0\ \text{and}\ g\in{\mathscr C}^{1}(0,D)\cap{\mathscr C}[0,D].$$
Let $h={g'}/{g}$. Then $h\in{\mathscr C}^{1}(0, D)\cap {\mathscr C}[0,D]$,
$h(0)=0$, $h\in\mathscr{H}$, and $$R(h)(x)=-\frac{Lg(x)}{g(x)}=\lambda_{0}\qquad\text{for } x\in(0, D).$$

\end{frame}

%---------------------------------------------------------19
\begin{frame}
\frametitle{3(3). Main results: (II)}\small
\begin{block}{Proposition 2}
Suppose that $a, b\in {\mathscr C}^1 (0, D)\cap {\mathscr C} [0, D]$ and $a>0$ on $(0,D)$. Set
$$\overline{\mathscr{H}}=\big\{h:  h(0)=0,\,h\in {\mathscr C}^2(0,D)\cap\mathscr{C}[0,D],\text{ and } h|_{(0, D)}<0\big\}.$$
Then (1)  we have
 $\sup_{h\in\overline{\mathscr{H}}}\inf_{x\in(0,  D)}{\overline R}(h)(x)\geqslant\lambda_{0}$ and the equality sign holds once  $\mu(0, D)=\infty$.

 (2) In general, we have
$$\lambda_{0}=\sup\limits_{h\in\mathscr{H}_{\ast}}\inf\limits_{x\in(0,D)}\overline {R}(h)(x),$$
where
$\mathscr{H}_{\ast}=\big\{h: h(0)=0,\, h\in {\mathscr C}^2(0, D)\cap {\mathscr C}[0,D], h|_{(0, D)}<0, h'<-{a}^{-1}{b}h\text{ on } (0,D)\big\}.$
Moreover, the supremum can be attained.
\end{block}

This differential form can be used as a `bridge' between ND and DN.
\end{frame}
%---------------------------------------------------------------------------------20
\begin{frame}
\frametitle{3(3). Main results: (III)}\small

Applying Theorem 1(1) to the test function $\nu(x, D)^{\gamma}$ with $\gamma=1/2$, $ 1$:

\begin{block}{Theorem 2(Criterion and basic estimates)}
$$\lambda_{0}>0\Leftrightarrow \delta:=\sup_{x\in(0, D)} \mu (0, x)\,\nu (x, D)<\infty.$$
More precisely, we have
$$(4\delta)^{-1}\leqslant\lambda_{0}\leqslant\delta^{-1}.$$
In particular, when $D=\infty$, we have
$\lambda_{0}=0$ if $\nu(0, D)=\infty$, and $\lambda_{0}>0$ if
$\int_0^\infty\mu(0, x)\,\nu(\text{d} x)<\infty$.\end{block}
\end{frame}

%---------------------------------------------------------------------------------21
\begin{frame}
\frametitle{3(3). Main results: (III)}\small

Applying Theorem 1(2), repeatedly with $f=f_n$:

\begin{block}{Theorem 3(Approximating procedure)}
Assume that $\delta<\infty$. Set $\varphi(x)=\nu(x,D)$.

(1) Let $f_{1}=\sqrt{\varphi}$, $f_{n}=f_{n-1}I\!I(f_{n-1})$, and $\delta_{n}=\sup_{x\in(0, D)}I\!I(f_{n})(x)$. Then $\delta_{n}$ is decreasing in $n$ and $$ \lambda_{0}\geqslant\delta_{n}^{-1}\geqslant({4\delta})^{-1},\quad n\geqslant1.$$

(2) For fixed $x_{0},\ x_{1}\in [0, D)$ with $x_{0}<x_{1}$, define
$$f_{1}^{x_{0}, x_{1}}\!=\!
\nu (\cdot\vee x_{0},\, x_1)\, \mathbf{1}_{[0, x_{1})}, \ f_{n}^{x_{0},x_{1}}\!=\!\big(f_{n-1}^{x_{0}, x_{1}}I\!I\big(f_{n-1}^{x_{0},x_{1}}\big)\big)(\cdot\vee x_0)\,\mathbf{1}_{[0, x_{1})},\ n\geqslant1,$$
and let $\delta_{n}'=\sup_{x_{0}, x_{1}:x_{0}<x_{1}}\,\inf_{x< x_{1}}I\!I({f_{n}}^{x_{0}, x_{1}})(x)$. Then $\delta_{n}'$ is increasing in $n$ and
$$\delta^{-1}\geqslant
{\delta_{n}'}^{-1}\geqslant\lambda_{0},\quad  n\geqslant1.$$
\end{block}
\end{frame}

%---------------------------------------------------------------------------------22
\begin{frame}
\frametitle{3(3). Main results: (III)}\small

\begin{block}{Theorem 3(continued)}

(3) Define
$$\bar{\delta}_{n}=\sup\limits_{x_{0},x_{1}:x_{0}<x_{1}}\frac{\|{f_{n}}^{x_{0}, x_{1}}\|}{D({f_{n}}^{x_{0}, x_{1}})},\qquad n\geqslant 1. $$
Then $\bar{\delta}_{n}^{-1}\geqslant \lambda_{0},\ \bar{\delta}_{n+1}\geqslant\delta_{n}'\, (n\geqslant 1)$,  and $\bar{\delta}_{1}=\delta_{1}'$.
\end{block}

\begin{block}{Corollary (Improved estimates)}
$$\delta^{-1}\geqslant{\delta'_{1}}^{-1}\geqslant\lambda_{0}\geqslant\delta_{1}^{-1}\geqslant(4\delta)^{-1},$$
$$\aligned
\delta_{1}&=\sup\limits_{x\in(0,D)}
\bigg(\sqrt{\varphi(x)}\,\int_{0}^{x}\sqrt{\varphi}\,\text{d}\mu
+\frac{1}{\sqrt{\varphi(x)}}\int_{x}^{D}\varphi^{{3}/{2}}\text{d}\mu\bigg),\\
\delta_{1}'&=\sup\limits_{x\in(0,D)}\bigg(\mu(0,x)\,\varphi(x)
+\frac{1}{\varphi(x)}\int_{x}^{D}\varphi^2\,\text{d}\mu\bigg)\in[\delta,2\delta].\endaligned$$
\end{block}
\end{frame}

%-------------------------------------------------------------------------23
\frame{\frametitle{3(4). From ND to DN}\small For the DN case, the
similar results can be given by the parallel method.
{\hfill\red$\to$ `parallel results'}
\begin{block}{}
Duality relation: $\hat\nu:=\mu;\qquad \hat\mu:=\nu;$
$$L=\frac{\rm d}{{\rm d} \mu}\frac{\rm d}{{\rm d} \nu}; \quad\text{ dual with}\quad \hat L=\frac{\rm d}{{\rm d} \hat \mu}\frac{\rm d}{{\rm d} \hat \nu}.$$
Then {\red $\lambda_0(L)=\lambda_0(\hat L)$ and $\delta=\hat\delta$}, where $\lambda_0(\hat L)$ is the eigenvalue of the corresponding  diffusion process with DN boundaries.
\end{block}
The identity is a combination of the formula in differential form $\overline R$ (Proposition 2) in ND case and the one in differential form
$R$ in DN case. {\hfill\red$\to$ `dual results'}
}
%-------------------------------------------------------------------------24
\frame{\frametitle{3(4). From ND to DN }\small
{\red `parallel results' $\ne$ `dual results'}

\begin{block}{}
 The variational formulas and then the approximating procedure in DN case by parallel method are different from those deduced by the
dual approach. It is interesting that in the discrete situation, the approximating procedure given by the parallel method is often less powerful than those in terms of duality. Similar phenomenon happens in the continuous situation with $D<\infty$.
\end{block}
\begin{block}{M.F. Chen} General estimate of the first eigenvalue on manifolds, Front. Math. China 2011, 6(6): 1025-1043.
\end{block}
}
%--------------------------------------------------------------------25

\frame{\frametitle{3(5). The property of eigenfunction}\small
\begin{itemize}
\item The properties of eigenfunction in DN case were shown by  M.F. Chen in 1999.

\item The properties of eigenfunction in ND case are not so easy! The method for the DN case does not work now!
\end{itemize}
\begin{block}{}
 When $D<\infty$, one may simply reverse the variable to obtain one from the other of the ND- and DN- cases. But when $D=\infty$, these two cases are certainly different since the Dirichlet boundary at 0 is touchable but not the one at $\infty$.
 \end{block}

Six lemmas, except Lemma 2, are mainly devoted to describe the eigenfunction of $\lambda_0$ in ND case. The lemmas are essential in our study.
}
%----------------------------------------------------------------26
\frame{\frametitle{3(5). The property of eigenfunction}\small

Let $\mathscr{A}[\alpha, \beta]$ be the set of all absolutely continuous functions on $[\alpha, \beta]$.
\begin{block}{Lemma 1 (Existence) Zettl A. Sturm-Liouville Theory, AMS, 2005}
(1) $\exists$ uniquely a non-zero function $g\in {\mathscr C}^1[0, D]$ such that $$g'\in\{f\in\mathscr{A}[\alpha, \beta]: 0\leqslant\alpha<\beta<D\}\quad\text{ and }\quad L g=-\lambda g, \;a.e..$$

(2) Suppose additionally $a, b\in \mathscr{C}[0,D]$, then $g\in {\mathscr C}^2[0, D]$ and  $L g=-\lambda g$ on $[0, D]$.
\end{block}

We call the function $g$ given in Lemma 1(1) a.e. eigenfunction of $\lambda$.

\begin{block}{Lemma 2}

Define $\lambda_{\ast}=\inf\{D(f): f\in\mathscr{A}[0,D],\, \|f\|=1, \text{ and } f(D)=0\}$. Then $\tilde{\lambda}_{0}=\lambda_{\ast}$.

\end{block}
$\tilde{\lambda}_{0}:=\inf\Big\{D(f):\mu\big(f^{2}\big)=1,\,f\in \mathscr{C}^1(0, D)\cap \mathscr{C}[0,D]\text{ and }f(D)=0\Big\}.$
}
%--------------------------------------------------------27
\frame{\frametitle{3(5). The property of eigenfunction}\small

\begin{block}{Lemma 3}

$\lambda_0^{(0, p_n)}\downarrow\downarrow \lambda_0^{(0, D)}=\tilde{\lambda}_{0}$ as $p_n\uparrow\uparrow D$. Moreover, for $p\in (0, D)$, we have
$\lambda_1^{(0, p)}>\lambda_0^{(0, p)}$, where $\lambda_1^{(0, p)}$ is the principal NN-eigenvalue.
\end{block}
\begin{block}{Lemma 4(Monotonicity)}
Let $g$ be a non-zero a.e. eigenfunction of $\tilde\lambda_{0}>0$. Then $g$ is strictly monotone.
\end{block}

\small{Sketch of proof:

\rm 1. Since $\tilde\lambda_{0}>0$, $g\neq$ cons. It suffices to show that $g'|_{(0,D)}\ne 0$.

 2. If $\exists p\in (0,D)$ such that $g'(p)=0$, then $\tilde\lambda_{0}\geqslant \lambda_1^{(0, p)}$.

3. From the above arguments, we have  $$\tilde\lambda_{0}\geqslant \lambda_1^{(0, p)}>\lambda_0^{(0, p)}
> \lambda_0^{(0, D)}=\tilde{\lambda}_0.$$}
This is a contradiction.
}
%----------------------------------28
\frame{\frametitle{3(5). The property of eigenfunction}\small
\begin{block}{Lemma 5}

The a.e. eigenfunction $g$ of $\tilde\lambda_{0}$ is either positive or negative everywhere.
\end{block}
Sketch of proof:\rm

1. If $\tilde\lambda_0=0$, then $g=const.$ and so the assertion is obvious.

2. Let $\tilde\lambda_{0}>0$. Assume that $g'|_{(0,D)}<0$ and $g(0)>0$. We need only to prove that $g\ne 0$ on $(0, D)$.

3. If otherwise $g(p)=0$ for some $p\in (0, D)$, then, since $\lambda_0^{(0,p)}$ is the minimal ND-eigenvalue on $(0,p)$, the eigenequation restricted to $(0, p)$ shows that
$$\tilde\lambda_{0}\geqslant \lambda_0^{(0, p)}> \lambda_0^{(0, D)}=\tilde\lambda_0,$$
which is a contradiction.
}
%----------------------------------------------------29
\frame{\frametitle{3(5). The property of eigenfunction}\small
\begin{block}{Lemma 6 (Vanishing property)} Let $D=\infty$. If $\lambda_0>0$, then
its a.e. eigenfunction $g$ satisfies $g(\infty)=0$.
\end{block}
Sketch of proof:
 \rm
 Since $\lambda_0>0$ and $Lg=-\lambda_0g$, we have
 $$\int_{0}^{D}e^{-C(x)}{\rm d}x\int_{0}^{x}g(t)\mu({\rm d}t)=\frac{g(0)-g(\infty)}{\lambda_{0}}<\infty. $$
 Let $f=g-g(\infty)$. If $g(\infty)>0$, then $f\in\mathscr{F}_{I\!I}$. By the lower variational formula and Cauchy's mean value theorem,
 $$\aligned&\lambda_{0}^{-1}\leqslant\inf_{f\in\mathscr{F}_{I\!I}}\sup_{x\in(0,D)}I\!I(f)(x) \leqslant \sup_{x\in(0,D)}I\!I(f)(x) \\
 &=\sup_{x\in(0,D)}\int_{x}^{D}e^{-C(x)}{\rm d}x\int_{0}^{x}f(t)\mu({\rm d}t)\bigg/\lambda_0\int_{x}^{D}e^{-C(x)}{\rm d}x\int_{0}^{x}g(t)\mu({\rm d}t)\\
 &\leqslant\sup_{x\in(0,D)}\frac{f(x)}{\lambda_0g(x)}= \sup_{x\in(0,D)}\frac{1}{\lambda_0}\bigg(1-\frac{g(\infty)}{g(x)}\bigg)
 \leqslant \frac{1}{\lambda_0}\bigg(1-\frac{g(\infty)}{g(0)}\bigg)<\lambda_0^{-1}.
\endaligned$$
}
%----------------------------------------------------------------------------------------30
\frame{\frametitle{4. Other related research}
\begin{itemize}
\item \small{  J.H. Shao  and  Y.H. Mao studied principal eigenvalue of birth-death processes  {\red on trees} by  taking advantage of the related results on half line. (Variational formulas, DN)}
\item  \small{ Ma Y.T. discussed the regularity, recurrence and ergodic criterions, $\cdots$ for birth-death process {\red on trees} by constructing two corresponding birth-death process on $Z_+$. }
\item weighted Hardy inequality: $$\mu(|f|^q)\leqslant C \nu(|f|^p).$$
\end{itemize}
\begin{block}{A. Kufner, L.-E. Persson }\small
Weighted Inequalities of Hardy type, World Scientific, 2003.
\end{block}\small
  H.Y. Jin and Y.H. Mao obtained some estimates on the best constant of weighted {\red Hardy inequality} in $L^p$ space using the similar method for estimates on $\lambda_0$.(DD, continuous case)
}
%----------------------------------------------------------------------------------------31
\frame{\frametitle{4. Other related research}
\begin{itemize}
\item \small{ Especially, let $p=q=2, \mu(x)=e^{C(x)}/a(x)$, $\nu(x)=e^{C(x)}$. Then {\red $C=\lambda_0^{-1},$ }i.e., the estimates on mixed eigenvalue is just the estimates on the best constant of a special kind of Hardy inequality.}
\item\small {We now study the weighted Hardy inequality and mixed eigenvalues of $p$-Laplacian operator in discrete and continuous cases.}
\end{itemize}}
%---------------------------------------------32
\begin{frame}
\frametitle{5. Main references} \small \begin{thebibliography}{99}
\bibitem{r1}M.F. Chen, F.Y. Wang , {\it  Estimation of spectral gap for elliptic operators},\ Trans. Amer. Math. Soc. 349:3 1997,\ 1239-1267.
\bibitem{C1}M.F. Chen, {\it Anaylitic proof of dual variational formula for the first eigenvalue in dimension one},\  Sci. Sin. (A) 1999,\ 42:8,\ 805-815.
 \bibitem{r5}M.F. Chen,  {\it Variational formulas and approximation theorems for the first eigenvalue in dimension one},\  Sci. Sin. (A) 2001,\ 31:1,\ 28-36 (Chinese Edition); 44:4 (English
 Edition),\ 409-418.
 \bibitem{r6}M.F. Chen,  {\it Eigenvalues,\  Inequalities,\  and Ergodic Theory,\ }New York,\  Springer,\  2005.
 \bibitem{r2}M.F. Chen, {\it Speed of stability for birth-death process},\ Front. Math. China 5(3):379-515\ 2010      \end{thebibliography}
 \end{frame}
 %--------------------------------------------------------------------33
\begin{frame}\frametitle{5. Main references}
\begin{thebibliography}{99}\small
\bibitem{rn1}M.F. Chen, {\it Basic esimates and stability rate for one-dimentional diffusion}, Chapter 6 in "Probability Approximations and Beyond", Lecture notes in statistics 205, 2012, 75-99.\label{rn1}
\bibitem{r18}H.Y. Jin, Y.H. Mao, {\it Estimation of the optimal constants
in the $L^p$-Poincar\'e inequalities on the half line}, Acta Math.
Sin. (Chinese vertion)  2012, Vol.55, No.1.
\bibitem{rm}Y.T. Ma,  {\it Birth and death process on trees},\ Sci. Sin. (A), 2010,Vol.53, No.1:10.
\bibitem{rsm}J.H. Shao, Y.H. Mao, {\it Eiatimation of the Dirichlet Eigenvalue of birth-delth Process on trees.} Acta Math. Sin. (Chinese vertion) 2007, Vol.50, No.3.
\bibitem{rn2}M.F. Chen, L.D. Wang, Y.H. Zhang, {\it Mixed principal eigenvalues in dimension one}, Front. Math. China
2012+.
\end{thebibliography}\end{frame}
%----------------------------------------------------34
\begin{frame}

\centerline{ \emph{\textbf{\color{blue} Thank you for your attention!}}}

\end{frame}

\end{CJK*}

\end{document}
