# Long time asymptotics of mixed-type Kimura diffusions

Guillaume Bal Departments of Statistics and Mathematics and CCAM, University
of Chicago, Chicago, IL 60637<EMAIL_ADDRESS>Binglu Chen
Department of Mathematics, University of Chicago, Chicago, IL 60637;
<EMAIL_ADDRESS>Zhongjian Wang Departments of Statistics and CCAM,
University of Chicago, Chicago, IL 60637<EMAIL_ADDRESS>
###### Abstract

This paper concerns the long-time asymptotics of diffusions with degenerate
coefficients at the domain’s boundary. Degenerate diffusion operators with
mixed linear and quadratic degeneracies find applications in the analysis of
asymmetric transport at edges separating topological insulators. In one space
dimension, we characterize all possible invariant measures for such a class of
operators and in all cases show exponential convergence of the Green’s kernel
to such invariant measures. We generalize the results to a class of two-
dimensional operators including those used in the analysis of topological
insulators. Several numerical simulations illustrate our theoretical findings.
Key Words: Degenerate Diffusion Operators; Invariant Measure; Long Time
Asymptotics; Fredholm Index; Lyapunov Functions.
MSC: 35K65, 60J70, 47D06, 37C40.

###### Contents

  1. 1 Introduction
  2. 2 The $C^{0}$ Semigroup in one-space dimension
  3. 3 Invariant Measures in dimension one
    1. 3.1 Functional settings and index associated to $L$
    2. 3.2 Null Space of $\overline{L}^{*}$
    3. 3.3 Proof of Theorem 3.1
      1. 3.3.1 Outline of proof
      2. 3.3.2 Reduction to a first order system
      3. 3.3.3 Representation of solutions of the modeling operator
      4. 3.3.4 Proof of Theorem 3.1
  4. 4 Exponential convergence to invariant measures
    1. 4.1 Case of one/two tangent boundary points
      1. 4.1.1 Function Space $C(\alpha,\beta)$
      2. 4.1.2 Estimation of $\lambda_{0}$ and rate of convergence
    2. 4.2 Case with two transverse boundaries
      1. 4.2.1 (W)-Lyapunov-Poincaré inequality
      2. 4.2.2 Long time behaviors
  5. 5 Long time behaviour in two dimension
    1. 5.1 Lyapunov function
    2. 5.2 Exponential convergence to invariant measure
  6. 6 Numerical Experiments
    1. 6.1 One dimensional cases with mixed boundary conditions
    2. 6.2 2D with tangent diagonal
  7. A Appendix

##  1 Introduction

This paper analyzes the long time behavior of diffusion processes with
infinitesimal generator given by a mixed type Kimura operator $L$ on one-
dimensional and two-dimensional manifolds with corners.

In two-space dimensions, the mixed type Kimura operator $L$ acts on functions
defined on a manifold with corner $P$. A boundary point $p\in bP$ has a
relatively open neighborhood $V$ that is homomorphic to either
$\mathbb{R}_{+}^{2}$ or $\mathbb{R}_{+}\times\mathbb{R}$, where we assume that
$p$ maps to $(0,0)$. We say that $p$ is a corner in the first case and an edge
point in the second case. More details are presented in Section 5.

In this paper, $L$ is a degenerate second-order differential operator such
that for each edge, the coefficients of the normal part of the second-order
term vanish to order one or two. For an edge point $p$, when written in an
adapted system of local coordinates on $\mathbb{R}_{+}\times\mathbb{R}$, $L$
takes the form:

$\displaystyle
L=a(x,y)x^{m}\partial_{xx}+b(x,y)x^{m-1}\partial_{xy}+c(x,y)\partial_{yy}+d(x,y)x^{m-1}\partial_{x}+e(x,y)\partial_{y},$
(1)

where we assume that $a,b,c,d,e$ are smooth functions and that $a(x,y)>0$ and
$c(x,y)>0$. Also $m\in\\{1,2\\}$. When $m=1$, the edge $x=0$ is of Kimura type
in that the coefficients vanish linearly towards it. When $m=2$, the edge
$x=0$ is of quadratic type as the coefficients now vanish quadratically.

For a corner $p$, as an intersection point of two edges, $L$ has the following
normal form:

$\displaystyle
L=a(x,y)x^{m}\partial_{xx}+b(x,y)x^{m-1}y^{n-1}\partial_{xy}+c(x,y)y^{n}\partial_{yy}+d(x,y)x^{m-1}\partial_{x}+e(x,y)y^{n-1}\partial_{y},$
(2)

when written in an adapted system of local coordinates on
$\mathbb{R}_{+}^{2}$, where $a(x,y)>0$, $c(x,y)>0$, and $m,n\in\\{1,2\\}$.

Associated to the operator $L$ is a $C^{0}$ semigroup $\mathcal{Q}_{t}=e^{tL}$
solution operator of the Cauchy problem

$\displaystyle\partial_{t}u=Lu$

with initial conditions $u(x,0)=f(x)$ at $t=0$. The operator $e^{tL}$ and some
of its properties are presented in detail in [10]. The main objective of this
paper is to analyze the long time behavior of $e^{tL}$, and in particular
convergence to appropriately defined invariant measures.

It turns out that the number of possible invariant measures and their type
(absolutely continuous with respect to one-dimensional or two-dimensional
Lebesgue measures or not) strongly depend on the structure of the coefficients
$(a,b,c,d,e)$. We thus distinguish the different boundary types that influence
the long time asymptotics of transition probabilities.

######  Definition 1.1.

A Kimura edge $E$ is called a tangent (Kimura) edge when $d(0,y)=0$ and a
transverse (Kimura) edge when $d(0,y)>0$.

A quadratic edge $E$ is called a tangent (quadratic) edge when
$\frac{d(0,y)}{a(0,y)}<1$, a transverse (quadratic) edge when
$\frac{d(0,y)}{a(0,y)}>1$, and a neutral (quadratic) edge when
$\frac{d(0,y)}{a(0,y)}=1$.

We assume that:

######  Assumption 1.1.

Every edge is either tangent, transverse, or neutral.

Note that we do not consider the setting with $d(0,y)<0$ on a Kimura edge. In
such a situation, diffusive particles pushed by the drift term $d(0,y)<0$ have
a positive probability of escaping the domain $P$. We would then need to
augment the diffusion operator with appropriate boundary conditions.

The long-time analysis in two-dimensions for a class of operators including a
specific example of interest in the field of topological insulators [6] is
given in section 5. In the application in [6], the degenerate diffusion
equation describes reflection coefficients of wavefields propagating in
heterogeneous media, which model the separation between topological
insulators; see [5, 7]. Before addressing two-dimensional operators, we
consider the simpler one-dimensional setting, where we need to consider ten
different scenarios, listed in table 1 below, depending on the form of the
generator $L$ at the domain’s boundary. These cases are analyzed in detail in
sections 2-4.

To describe all invariant measure on the interval $[0,1]$ in one dimensional,
consider

$L=a(x)x^{m(0)}(1-x)^{m(1)}\frac{d^{2}}{dx^{2}}+b(x)x^{m(0)-1}(1-x)^{m(1)-1}\frac{d}{dx},$
(3)

with ${a}(x),{b}(x)\in C^{\infty}([0,1])$.

For $i=0,1$, we say that $x=i$ is of Kimura type when $m(i)=1$ and of
quadratic type when $m(i)=2$. When $x=i$ is of Kimura type, we assume that the
vector field $b(x)\frac{d}{dx}$ is inward pointing at $x=i$. For brevity, we
use $\widetilde{a}$ and $\widetilde{b}$ to denote

$\displaystyle\widetilde{a}(x)=a(x)x^{m(0)}(1-x)^{m(1)},\quad\widetilde{b}(x)=b(x)x^{m(0)-1}(1-x)^{m(1)-1}.$
(4)

######  Definition 1.2.

When $x=0\ (1\ \text{resp.})$ is a Kimura endpoint, we say it is a tangent
point if $b(0)=0\ (b(1)=0\ \text{resp.})$ and a transverse point if $b(0)>0\
(b(1)<0\ \text{resp.})$.

When $x=0\ (1\ \text{resp.})$ is a quadratic endpoint, we say it is a tangent
point if $\frac{{b}(0)}{{a}(0)}<1\ (\frac{{b}(1)}{{a}(1)}>-1\ \text{resp.})$,
a transverse point if $\frac{{b}(0)}{{a}(0)}>1\ (\frac{{b}(1)}{{a}(1)}<-1\
\text{resp.})$, and a neutral point if $\frac{{b}(0)}{{a}(0)}=1\
(\frac{{b}(1)}{{a}(1)}=-1\ \text{resp.})$.

The quadratic endpoint and the tangent Kimura endpoint are sticky boundary
points in the sense that the Dirac measure supported on them is an invariant
measure. When both endpoints are transverse, there is another invariant
measure $\mu$ with full support on the whole interval. By computing the index
of $L$ on an appropriate Hölder space, we characterize the kernel space of
$\overline{L}^{*}$ composed of invariant measures for the diffusion $L$.

In both cases, starting from a point in $P$, the corresponding transition
probability of the diffusion converges to the invariant measure at an
exponential rate. In cases with at least one tangent boundary point, we
consider a functional space of functions that vanish at the tangent boundary
points and show that $L$ has a spectral gap on such a space. In the absence of
tangent boundary points, we prove that the invariant measure $\mu$ satisfies
an appropriate Poincaré inequality so that $L$ also admits a spectral gap in
$L^{2}(\mu)$. Our main convergence results for $\mathcal{Q}_{t}=e^{tL}$, whose
properties are described in Theorem 2.1, are summarized in Theorems 4.1 and
4.2 below.

In two space dimensions, we do not consider all possible invariant measures as
a function of the nature of the drift terms $d$ and $e$ in the vicinity of
edges or corners. Instead, we restrict ourselves to the following case:

######  Assumption 1.2.

For $L$ on a 2 dimensional compact manifold with corners $P$, there is exactly
one tangent edge $H$, and when restricted to $H$, $L|_{H}$ is transverse to
both boundary points.

This case involves exactly one tangent edge with two transverse boundary
points so that, applying results from the one-dimensional case, we find that
$L$ has a unique invariant measure $\mu$ fully supported on (the one-
dimensional edge) $H$. Starting from any point $p$ not on the quadratic edge,
we show in Theorem 5.2 that the transition probability converges to $\mu$ at
an exponential rate in the Wasserstein distance sense. The main tool used in
the convergence is the construction of a Lyapunov function in Theorem 5.1.

The setting of $P$ a triangle with two transverse Kimura edges while the third
edge is quadratic with transverse endpoints as described in Assumption 1.2
finds applications in the analysis of the asymmetric transport observed at an
edge separating topological insulators [6]; see Remark 5.2 below. The
corresponding one-dimensional version with one transverse Kimura point and one
tangent quadratic point (see entry $(1,4)$ in Table 1 below) also appears in
the analysis of reflection coefficients for one-dimensional wave equations
with random coefficients [20].

There is a large literature on the analysis of the long-time behavior of
$e^{tL}$ when $L$ is non-degenerate and when $L$ is of Kimura type. In the
latter case, $L$ is the generalized Kimura operator studied in [16]. In that
work, $L$ is analyzed on a weighted Hölder space, denoted by
$L_{\gamma}:C^{0,2+\gamma}(P)\to C^{0,\gamma}(P)$. $L_{\gamma}$ is then
Fredholm of index zero, which is used to characterize the nullspace of an
adjoint operator and show that the non-zero spectrum lies in a half plane
$\text{Re}\ \mu<\eta<0$ so that for $f\in C^{0,\gamma}(P)$, $e^{tL}f$
converges to a stable limiting solution at an exponential rate.

However, in the presence of quadratic edge/point, $L_{\gamma}$ is not
Fredholm. The reason is that near such quadratic edges or points, the operator
may be modeled by an elliptic (non-degenerate) operator on an infinite domain
(with thus continuous spectrum in the vicinity of the origin). We thus need
another approach that builds on the following previous works. In [1], the
growth bound of a strongly continuous positive semigroup is associated with
the Lyapunov function. In [4], for a time continuous Markov process admitting
a (unique) ergodic invariant measure, the rate of convergence to equilibrium
is studied using a weaker version of a Lyapunov function called a
$\phi$-Lyapunov function and an appropriate Poincaré inequality.

An outline of the rest of this paper is as follows. The semigroup $e^{tL}$ is
analyzed in section 2 in the one-dimensional case. The space of invariant
measures associated with a one-dimensional diffusion $L$, which depends on the
structure of the drift term at the two boundary points, is constructed in
section 3; see Table 1 for a summary. The exponential convergence of the
kernel of $e^{tL}$ (the Green’s function) to an appropriate invariant measure
over long times is demonstrated in section 4.

The operator $e^{tL}$ in the two-dimensional setting is analyzed in [10]. For
the class of operators satisfying Assumption 1.2, the construction of the
(unique normalized) invariant measure and the exponential convergence of the
kernel of $e^{tL}$ to this measure in the Wasserstein sense are given in
section 5. Numerical simulations of stochastic differential equations
presented in section 6 illustrate the theoretical convergence results obtained
in dimensions one and two.

##  2 The $C^{0}$ Semigroup in one-space dimension

Let $L$ be the one-dimensional mixed-type Kimura operator on $[0,1]$ given in
(3). Let $\mathcal{Q}_{t}=e^{tL}$ be the solution operator of the Cauchy
problem for the generator $L$ and denote by $q_{t}(x,y)$ its kernel. Its main
properties are summarized in the following result:

######  Theorem 2.1.

The operator $\mathcal{Q}_{t}$ defines a positivity preserving semigroup on
$C^{0}([0,1])$. For $f\in C^{0}([0,1])$, the function
$u(x,t)=\mathcal{Q}_{t}f(x)$ solves the Cauchy Problem for $L$ with initial
condition $f(x)$ in the sense that

$\displaystyle\underset{t\to
0+}{\lim}||\mathcal{Q}_{t}f-f||_{\mathcal{C}^{0}}=0.$ (5)

###### Proof.

If both endpoints are of Kimura type, $L$ is the 1D Kimura operator. By [15,
Sec 9, Theorem 1], $\mathcal{Q}_{t}$ defines a positive and strongly
continuous semi-group on $C^{m}([0,1])$ for $m\in\mathbb{N}$. If both
endpoints are of quadratic type, we can regard $L$ as a uniform parabolic
operator on $\mathbb{R}$ (using, e.g., a change of variables $x=e^{z}$ in the
vicinity of $x=0$ as we do below), which is well studied (see, e.g., [21]).

For the remaining case, we might as well assume that $x=0$ is a Kimura
endpoint and $x=1$ a quadratic endpoint. We intend to build the global
solution out of local solution near the boundary. Let
$([0,1-\eta],\phi_{0}),([\eta,1],\phi_{1})$ for some $0<\eta<\frac{1}{4}$
small be the coordinate charts so that pulling back $L$ to these coordinate
charts gives two local operators

$\displaystyle
L_{0}=x\partial^{2}_{x}+b_{0}\partial_{x}+xc(x)\partial_{x},\quad
x\in[0,\phi_{0}(1-\eta)),$ $\displaystyle
L_{1}=\partial_{z}^{2}+d(z)\partial_{z},\quad z\in(\phi_{1}(\eta),\infty).$

We extend these two local operators to the whole sample space

$\displaystyle\widetilde{L}_{0}=x\partial^{2}_{x}+b_{0}\partial_{x}+xc(x)\varphi_{0}(x)\partial_{x},\quad\widetilde{L}_{1}=\partial_{z}^{2}+d(z)\varphi_{1}(z)\partial_{z}$

where $\varphi_{0}(x)$ is a smooth cutoff function so that

$\displaystyle\varphi_{0}(x)=\begin{cases}\begin{array}[]{cc}1&\text{for}\
x\in[0,\phi_{0}(1-2\eta)]\\\ 0&\text{for}\
x>\phi_{0}(1-\eta),\end{array}\end{cases}\varphi_{1}(z)=\begin{cases}\begin{array}[]{cc}1&\text{for}\
z\in[\phi_{1}(2\eta),\infty)\\\ 0&\text{for}\
z<\phi_{1}(\eta).\end{array}\end{cases}$

Let $\widetilde{Q}^{0}_{t},\widetilde{Q}^{1}_{t}$ be the solution operators of
$\widetilde{L}_{0},\widetilde{L}_{1}$ and denote their kernels by
$\widetilde{q}^{0}_{t},\widetilde{q}^{1}_{t}$ respectively. Define smooth
cutoff functions $0\leq\chi,\psi_{0},\psi_{1}\leq 1$ so that

$\displaystyle\text{supp}\psi_{0}\subset[0,1-2\eta],\
\text{supp}\psi_{1}\subset[2\eta,1],\ \psi_{0}|_{\text{supp}\chi}\equiv 1,\
\psi_{1}|_{\text{supp}(1-\chi)}\equiv 1.$ (6)

Given $f\in C^{0}([0,1])$ and $g\in C^{0}([0,1]\times[0,T])$, set the
homogeneous and inhomogeneous solution operator as

$\displaystyle\widetilde{Q}_{t}f=\psi_{0}\widetilde{Q}^{0}_{t}[\chi
f]+\psi_{1}\widetilde{Q}^{1}_{t}[(1-\chi)f],$ $\displaystyle
A_{t}g=\int_{0}^{t}\widetilde{Q}_{t-s}g(s)ds.$

Then

$\displaystyle(\partial_{t}-L)\widetilde{Q}_{t}f=E^{0}_{t}f:=[\psi_{0},L]\widetilde{Q}^{0}_{t}[\chi
f]+[\psi_{1},L]\widetilde{Q}^{1}_{t}[(1-\chi)f],$
$\displaystyle(\partial_{t}-L)A_{t}g=(Id-E_{t})g:=g-[\psi_{0},L]A^{0}_{t}[\chi
g]-[\psi_{1},L]A^{1}_{t}[(1-\chi)g].$

Our choice of $\chi,\psi_{0},\psi_{1}$ (6) ensures that
$\text{dist}(\text{supp}[\psi_{0},L],\text{supp}\chi)>0$,
$\text{dist}(\text{supp}[\psi_{1},L],\text{supp}(1-\chi))>0$, which ensures
that $E^{0}_{t},E_{t}$ are bounded operators with operator norms bounded by
$O(e^{-\frac{c}{t}})$ for some constant $c>0$ as $t\to 0^{+}$. Hence for $T>0$
small enough, there exists an inverse $(Id-E_{t})^{-1}$, which can be
expressed as a convergent Neumann series in the operator norm topology of
$C^{0}([0,1]\times[0,T])$. Finally we can express the solution operator by

$\displaystyle\mathcal{Q}_{t}f=\widetilde{Q}_{t}f-A_{t}(Id-
E_{t})^{-1}E^{0}_{t}f.$

Since both $\widetilde{Q}^{0}_{t},\widetilde{Q}^{1}_{t}$ are strongly
continuous, then so is $\widetilde{Q}_{t}$:

$\underset{t\to 0}{\lim}\ ||\widetilde{Q}_{t}f-f||_{C^{0}}=0.$

As $(Id-E_{t})^{-1}E^{0}_{t}$ is a bounded map from $C^{0}([0,1]$ to
$C^{0}([0,1]\times[0,T])$, we have

$||A_{t}(Id-E_{t})^{-1}E^{0}_{t}||_{C^{0}([0,1])\to(C^{0}([0,1]),t)}=o(t).$

Therefore (5) holds. Let $\widetilde{q}_{t},h_{t}$ be the heat kernels of
$\widetilde{Q}_{t},(Id-E_{t})^{-1}E^{0}_{t}$. We express the heat kernel of
$\mathcal{Q}_{t}$ as

$\displaystyle
q_{t}(x,y)=\widetilde{q}_{t}(x,y)-\int_{0}^{t}\int_{0}^{1}\widetilde{q}_{t-s}(x,z)h_{s}(z,y)dzds.$

∎

##  3 Invariant Measures in dimension one

In this section, we aim to find all invariant measures of $L$ in spatial
dimension one. For convenience of computation, in this section we first choose
a global coordinate $\phi$ so that $(W_{0},\phi),(W_{1},\phi)$ is a cover of
$[0,1/3],[2/3,1]$ under which $L$ takes the following normal form:

  1. 1.

In $([0,1/3],\phi)$, $L_{0}$ has two possible forms:

$\displaystyle L_{0}=x\partial_{x}^{2}+b(x)\partial_{x},\ \text{if 0 is a
Kimura endpoint}$ $\displaystyle L_{0}=\partial_{z}^{2}+b(z)\partial_{z},\
\text{if 0 is a quadratic endpoint}$

  2. 2.

In $([2/3,1],\phi)$, $L_{1}$ has two possible forms:

$\displaystyle L_{1}=(1-x)\partial_{x}^{2}+b(x)\partial_{x},\ \text{if 1 is a
Kimura endpoint}$ $\displaystyle L_{1}=\partial_{z}^{2}+b(z)\partial_{z},\
\text{if 1 is a quadratic endpoint}$

where we use $b$ to denote the first-order term in all cases.

######  Notation 1.

We call the global coordinate $\phi$ on $[0,1]$ heat coordinates if
$L_{0},L_{1}$ have forms above under $\phi$.

Let

$\displaystyle b_{\pm}=\underset{x\to 1,0}{\lim}b(x)\qquad\text{ or }\qquad
b_{\pm}=\underset{z\to\pm\infty}{\lim}b(z).$ (7)

A straightforward derivation shows that, if in the original coordinate,
$L_{0}=x^{2}\partial_{x}^{2}+(b_{-}+1)x\partial_{x}$, then after turning to
heat coordinates $x=e^{z}$, $L_{0}$ takes the form
$\partial_{z}^{2}+b_{-}\partial_{z}$.

###  3.1 Functional settings and index associated to $L$

Our functional setting involves local Hölder spaces, which differ from the
usual Hölder space (with $|\cdot|_{k+\gamma}$ to denote its norm) near the
boundaries and are variations of those used in [16]. In the following
definition, we assume $c$ is a point away from the interval’s boundaries.

  1. 1.

For quadratic type boundaries, with $U=(-\infty,c]$,

    1. (a)

$f\in C^{0}(U)$ belongs to $\mathcal{D}^{\gamma}(U)$ if the function $f$ can
be continuously extend to $z=-\infty$ and the following norm is finite:

$\displaystyle{||f||_{\gamma,U}=|f|_{\gamma,U}}+\underset{z_{1}\leq z_{2}\leq
c}{\sup}\big{|}\int_{z_{1}}^{z_{2}}fdz\big{|},\text{when it is not neutral};$
(8) $\displaystyle{||f||_{\gamma,U}=|f|_{\gamma,U}}+\underset{z_{1}\leq
z_{2}\leq c}{\sup}|\int_{z_{1}}^{z_{2}}fdz|+\underset{z_{1}\leq z_{2}\leq
c}{\sup}|\int_{z_{1}}^{z_{2}}\int_{-\infty}^{z}fdxdz|$ (9)
$\displaystyle,\text{when it is neutral};$

    2. (b)

$f\in C^{1}(U)$ belongs to $C^{1+\gamma}(U)$ if the functions
$\partial_{z}f,f$ can be continuously extend to $z=-\infty$ and local
$C^{1+\gamma}$ norm is finite:

$\displaystyle||f||_{1+\gamma,U}=|f|_{1+\gamma,U}+||\partial_{z}f||_{\gamma,U};$
(10)

    3. (c)

$f\in C^{2}(U)$ belongs to $C^{2+\gamma}(U)$ if the functions
$\partial_{z}^{2}f,\partial_{z}f,f$ extend continuously to $z=-\infty$ and the
local $C^{2+\gamma}$ norm is finite:

$\displaystyle{||f||_{2+\gamma,U}=|f|_{2+\gamma,U}}+||\partial_{z}^{2}f||_{\gamma,U}.$
(11)

  2. 2.

For Kimura type boundaries, with $U=(0,c]$,

    1. (a)

$f\in C^{0}(U)$ belongs to $\mathcal{D}^{\gamma}(U)$ if the function $xf$ can
be continuously extend to $x=0$, and the local $C^{\gamma}$ norm is finite:

$\displaystyle{||f||_{\gamma,U}}=|xf|_{\gamma,U};$ (12)

    2. (b)

$f\in C^{0}(\overline{U})\cap C^{1}(U)$ belongs to $C^{1+\gamma}(U)$ if the
function $x\partial_{x}f$ can be continuously extend to $x=0$ and vanish, and
the local $\mathcal{C}^{1+\gamma}$ norm is finite:

$\displaystyle{||f||_{1+\gamma,U}}=|f|_{\gamma,U}+|x\partial_{x}f|_{\gamma,U};$
(13)

    3. (c)

$f\in C^{1}(\overline{U})\cap C^{2}(U)$ belongs to $C^{2+\gamma}(U)$ if the
function $x\partial_{x}^{2}$ can be continuously extend to $x=0$ and vanish,
and the local $\mathcal{C}^{2+\gamma}$ norm is finite:

$\displaystyle||f||_{2+\gamma,U}=|f|_{\gamma,U}+|\partial_{x}f|_{\gamma,U}+|x\partial_{x}^{2}f|_{\gamma,U}.$
(14)

######  Remark 3.1.

At neutral quadratic endpoint, interchanging the two integral signs, we have

$\displaystyle\int_{z_{1}}^{z_{2}}\int_{-\infty}^{z}f(s)sdsdz=\int_{-\infty}^{z_{2}}(z_{2}-s)f(s)ds,$

so $\int_{-\infty}^{z_{2}}sf(s)ds<\infty$.

We now build global norms on spaces of functions on $[0,1]$ out of the above
local norms.

######  Definition 3.1.

Let $W_{2}\subset\subset(0,1)$ covering $[0,1]\setminus(W_{0}\cup W_{1})$ and
$\varphi_{0},\varphi_{1},\varphi_{2}$ be a partition of unity subordinate to
this cover. A function $f\in C^{2+\gamma}([0,1])$ if
$(\varphi_{i}f)\circ\phi\in C^{2+\gamma}(W_{i})$ for each $i$ and the global
norm is

$\displaystyle||f||_{2+\gamma}=\underset{i}{\sum}||(\varphi_{i}f)\circ\phi||_{2+\gamma,W_{i}}.$

Motivated by (21) below, we define $f\in C^{1+\gamma}([0,1])$ if
$\phi^{\prime}_{i}\cdot(\varphi_{i}f)\circ\phi\in C^{1+\gamma}(W_{i})$ for
each $i$ and the global norm is

$\displaystyle||f||_{1+\gamma}=\underset{i}{\sum}||\phi^{\prime}_{i}(\varphi_{i}f)\circ\phi||_{1+\gamma,W_{i}}.$

Finally $f\in\mathcal{D}^{\gamma}([0,1])$ if
$(\phi^{\prime}_{i})^{2}\cdot(\varphi_{i}f)\circ\phi\in C^{\gamma}(W_{i})$ for
each $i$ and the global norm is

$\displaystyle||f||_{\gamma}=\underset{i}{\sum}||(\phi^{\prime}_{i})^{2}(\varphi_{i}f)\circ\phi||_{\gamma,W_{i}}.$

Different choices of coverings give rise to equivalent norms. Endowed with
these norms, we now show that the domains and target spaces of $M$ are all
Banach spaces.

######  Proposition 3.1.

For $0<\gamma<1$, all the spaces defined in Definition 3.1 are all Banach
spaces.

###### Proof.

Take $C^{2+\gamma}([0,1])$ as an example. Let

$\displaystyle i:C^{2+\gamma}([0,1])\to C^{2+\gamma}(W_{0})\times
C^{2+\gamma}(W_{1})\times C^{2+\gamma}(W_{2})$

be the inclusion by mapping $f$ to
$((\varphi_{0}f)\circ\phi,(\varphi_{1}f)\circ\phi,(\varphi_{2}f)\circ\phi)$.
This inclusion is closed since $\varphi_{0},\varphi_{1},\varphi_{2}$ is a
partition of unity. Thus we need to show that each local $(0,\gamma)$ space
defined at the beginning of this subsection is Banach.

The cases when $U=(0,c]$ are verified in [16]. When $U=(-\infty,c]$, we first
show that $\mathcal{D}^{\gamma}(U)$ is a Banach space. As usual we take a
Cauchy sequence $\\{u_{i}\\}_{i\geq 1}$ in $\mathcal{D}^{\gamma}(U)$. The
standard argument for the usual Hölder space (see [25, Remark 3.1.3]) applies
to show that there exists a limit $u\in C^{\gamma}(U)$. To show that $u$ is in
$\mathcal{D}^{\gamma}(U)$, we only need to check that $u$ can continuously
extend to $-\infty$ and $\underset{z_{1}\leq z_{2}\leq
c}{\sup}|\int_{z_{1}}^{z_{2}}udz|$ exists. For the first part,
$\\{u_{i}(-\infty)\\}_{i\geq 1}$ is convergent and the limit coincides with
the limit of $u$ at $-\infty$. For the second part, for any fixed $z_{1}\leq
z_{2}\leq c$, $|\int_{z_{1}}^{z_{2}}u_{i}dz|$ converges to
$|\int_{z_{1}}^{z_{2}}udz|$. And since $|\int_{z_{1}}^{z_{2}}u_{i}dz|$ are
uniformly bounded for $i,z_{1},z_{2}$, $\underset{z_{1}\leq z_{2}\leq
c}{\sup}|\int_{z_{1}}^{z_{2}}udz|$ exists. Therefore we proved that
$u\in\mathcal{D}^{\gamma}(U)$, which implies that $\mathcal{D}^{\gamma}(U)$ is
a Banach space.

For $C^{1+\gamma}(U),C^{2+\gamma}(U)$, the above proof applies to show that
there exists a limit $u\in C^{1+\gamma}(U),C^{2+\gamma}(U)$ respectively, and
$\partial_{z}f,\partial_{zz}f\in\mathcal{D}^{\gamma}(U)$, so that these two
spaces are also Banach spaces. ∎

$\displaystyle
L_{\gamma}:C^{2+\gamma}([0,1])\longrightarrow\alpha\cdot\mathcal{D}^{\gamma}([0,1]),$
(15)

where $\alpha(x)=x^{m(0)}(1-x)^{m(1)}$ with $m(0)=1$ if $x=0$ is Kimura and
$m(0)=0$ otherwise, while $m(1)$ is defined similarly. We first state the main
theorem of this section and leave the proof to Section 3.3.

######  Theorem 3.1.

$L_{\gamma}$ is a Fredholm operator and its index is

$\displaystyle\text{ind}\ (L_{\gamma})=\kappa^{+}+\kappa^{-}$

where $\kappa^{+},\ \kappa^{-}$ are the number of positive $b_{+}$ and
negative $b_{-}$, respectively, where $b_{\pm}$ is defined in (7).

###  3.2 Null Space of $\overline{L}^{*}$

Let $\overline{L}$ denote the $C^{0}([0,1])$-graph closure of $L$ with domain
$C^{2+\gamma}([0,1])$. Having obtained the index of $L_{\gamma}$, we are now
able to find the null space of the adjoint operator $\overline{L}^{*}$. We
first use the maximum principle below to find the kernel space of
$L_{\gamma}$.

######  Lemma 3.1 (Maximum Principle).

Suppose that $w\in C^{2+\gamma}([0,1])$ is a subsolution of $L$, $Lw\geq 0$ in
a neighborhood, $U$ of a transverse boundary point $p$. If $w$ attains a local
maximum at $p$, then $w$ is a constant on $U$.

###### Proof.

The case when $L$ is of generalized Kimura type is studied in [16], so we only
need to consider quadratic boundaries. Let $x$ denote normalized local
coordinates of $0$ so that

$\displaystyle L=x^{2}\partial_{x}^{2}+(b+1)x\partial_{x},b\geq 0$

By subtracting $v(-\infty)$ from $v$, we may assume that $v(-\infty)=0$.
Integrating $(\partial_{z}^{2}+b\partial_{z})v$ we see that
$\partial_{z}v+bv\geq 0$ in a neighborhood $U$ of $-\infty$. Thus
$\partial_{z}v\geq-bv\geq 0$ in $U$. As $w$ is not a constant, we can expand
this neighbourhood until some $z_{0}$ such that $\partial_{z}v>0$. Then
$v(z_{0})=\int_{-\infty}^{z_{0}}\partial_{z}vdz>0$, which contradicts the fact
that $-\infty$ is a local maximum. ∎

######  Theorem 3.2.

$\text{dim ker}\ L_{\gamma}=2$ if and only if $L_{\gamma}$ is tangent to both
endpoints; otherwise $\text{dim ker}\ L_{\gamma}=1$.

###### Proof.

The kernel of $L_{\gamma}$ is in the linear space of $\\{1,S(x)\\}$, where
$S(x)$ is the scale function of the process. For

$L=\widetilde{a}(x)\partial_{xx}+\widetilde{b}(x)\partial_{x},$ (16)

the scale function is defined as

$\displaystyle
S(x)=C\int^{x}\text{exp}\left[-\int_{\frac{1}{2}}^{\eta}\frac{\widetilde{b}(\xi)}{\widetilde{a}(\xi)}d\xi\right]d\eta$
(17)

and has derivatives

$\displaystyle
S^{\prime}(x)=\text{exp}\left[-\int^{x}\frac{\widetilde{b}(\xi)}{\widetilde{a}(\xi)}d\xi\right],\qquad
S^{\prime\prime}(x)=-S^{\prime}(x)\frac{\widetilde{b}(x)}{\widetilde{a}(x)}.$

Denote $\frac{\widetilde{b}(x)}{\widetilde{a}(x)}=\frac{c(x)}{x(1-x)}$ and
$c_{0}=c(0)$, $c_{1}=c(1)$. The integrand
$\text{exp}\left[-\int_{\frac{1}{2}}^{\eta}\frac{\widetilde{b}(\xi)}{\widetilde{a}(\xi)}d\xi\right]\sim\eta^{-c_{0}},(1-\eta)^{c_{1}}$
as $\eta$ approaches $0+,1-$ respectively, so $S$ is integrable when
$c_{0}<1,c_{1}>-1$.

For Kimura point $x=0$, if $0<c_{0}<1$, then $xS^{\prime\prime}(x)\sim
x^{-c_{0}}$ as $x\to 0$, which is not finite and thus not in
$\mathcal{D}^{\gamma}$. If $c_{0}=0$, $S^{\prime\prime}$ is smooth at $x=0$.
For a quadratic point $x=0$, if $c_{0}<1$, turning into heat coordinates
$z=lnx$, $S(z)\sim e^{(1-c_{0})z}$, it is in local $C^{2+\gamma}$ space. We
conclude that $S$ is in $C^{2+\gamma}([0,1])$ if and only if $L$ is tangent to
both endpoints. ∎

######  Definition 3.2.

We define

$\displaystyle bP_{\text{ter}}(L)=\\{\text{quadratic endpoint},P\\}$

if $L$ is transverse to both endpoints, otherwise we define

$\displaystyle bP_{\text{ter}}(L)=\\{\text{quadratic endpoint, Kimura tangent
endpoint}\\}.$

Here we use the notation $bP_{\text{ter}}$ from [16] to denote terminal
boundary.

We first explain that to each element of $bP_{\text{ter}}([0,1])$ there is an
element of the nullspace of $\overline{L}^{*}$. For any quadratic endpoint or
Kimura tangent endpoint $p$, $p$ is a sticky boundary point so $\delta(p)$ is
in $\text{ker}\ \overline{L}^{*}$. For $w\in C^{2+\gamma}([0,1])$, we have

$\displaystyle\langle L_{\gamma}w,\delta(p)\rangle=0,$

that is to say $\delta(p)\in\text{Ker}\ L_{\gamma}^{*}$. This equality still
holds for $w\in\text{Dom}(\overline{L})$, where $\overline{L}$ is the
$\mathcal{C}^{0}$ graph closure of $L_{\gamma}$. Hence
$\delta(p)\in\text{Dom}(\overline{L}^{*})$, and $\overline{L}^{*}\delta(p)=0$.
If $L$ is transverse to both endpoints, we can explicitly construct $\mu$,
which is supported on the whole interval, as follows:

##### Construction of $\mu$

We first reduce $L$ to the standard form

$\displaystyle L_{z}=\frac{1}{2}\partial_{zz}-\nabla
U(z)\partial_{z}=-\frac{1}{2}\partial_{z}^{*}\partial_{z}$

with $\partial^{*}_{z}=\partial_{z}+2\nabla U(z)$ on a probability space
$(X,\mu)$. It is known that $L_{z}$ has an invariant measure $\mu$

$\displaystyle\mu(dz)=\frac{e^{-2U(z)}}{Z}dz$

where $Z$ is a normalizing constant. We now check that $\mu$ is a probability
measure in the different boundary cases.

  1. 1.

Two transverse quadratic boundary. For
$L=x^{2}(1-x)^{2}\partial_{xx}+x(1-x)b(x)\partial_{x}$, we first do a
coordinate change:

$\displaystyle z=\frac{1}{\sqrt{2}}\text{ln}\frac{x}{1-x},\
x=\frac{e^{\sqrt{2}z}}{1+e^{\sqrt{2}z}},$

so that

$\displaystyle L_{z}=\frac{1}{2}\partial_{zz}-\nabla U(z)\partial_{z},\
z\in(-\infty,\infty)$

with $\nabla U=-\frac{1}{\sqrt{2}}\left(b(x)+2x-1\right)$. Then

$\displaystyle\mathit{e^{-2U(z)}=\begin{cases}\begin{array}[]{cc}\Theta(e^{\sqrt{2}(b(0)-1)z}),\
z\to-\infty\\\ \Theta(e^{\sqrt{2}(b(1)+1)z}),\
z\to+\infty\end{array}\end{cases}}.$

Thus, $Z<\infty$ exactly when $b(0)>1,b(1)<-1$, i.e. both quadratic endpoints
are transverse.

  2. 2.

One transverse Kimura and one quadratic boundary. For
$L=x^{2}(1-x)\partial_{xx}+xb(x)\partial_{x}$, we do a coordinate change by
letting

$\displaystyle\frac{\partial z}{\partial x}=\frac{1}{\sqrt{2}x\sqrt{1-x}}$

so that

$\displaystyle L_{z}=\frac{1}{2}\partial_{zz}-\nabla U(z)\partial_{z},\
z\in(-\infty,0]$

with $\nabla U=-\frac{1}{\sqrt{2}}(\frac{3x-2+2b(x)}{2\sqrt{1-x}})$. Then

$\displaystyle\mathit{e^{-2U(z)}=\begin{cases}\begin{array}[]{cc}\Theta(e^{\sqrt{2}(b(0)-1)z}),\
z\to-\infty\\\ \Theta(z^{-1-2b(1)}),\ z\to 0\end{array}\end{cases}}.$

Thus, $Z<\infty$ exactly when $b(0)>1,b(1)<0$, i.e. both endpoints are
transverse.

  3. 3.

Two transverse Kimura boundary. For $L=x(1-x)\partial_{xx}+b(x)\partial_{x}$,
we do a coordinate change

$\displaystyle\frac{\partial z}{\partial x}=\frac{1}{\sqrt{2x(1-x)}}$

so that

$\displaystyle L_{z}=\frac{1}{2}\partial_{zz}-\nabla U(z)\partial_{z},\
z\in[0,\frac{\pi}{\sqrt{2}}]$

with $\nabla U=-\frac{x-1/2+b(x)}{\sqrt{2x(1-x)}}$. Then

$\displaystyle\mathit{e^{-2U(z)}=\begin{cases}\begin{array}[]{cc}\Theta(z^{2b(0)-1}),\
z\to 0\\\ \Theta(z^{-1-2b(1)}),\
z\to\frac{\pi}{\sqrt{2}}\end{array}\end{cases}}.$ (18)

Again, $Z<\infty$ exactly when $b(0)>0,b(1)<0$, i.e. both endpoints are
transverse.

######  Proposition 3.2.

$\displaystyle\text{dim ker}\ \overline{L}^{*}=|bP_{\text{ter}}(L)|.$

###### Proof.

We know that $\text{dim\ ker}\ L_{\gamma}=2$ if $L$ is tangent to both
endpoints and otherwise $\text{dim\ ker}\ L_{\gamma}=1$. This is equivalent
to:

$\displaystyle\text{dim\ ker}\ L_{\gamma}=\text{max}\\{1,|\text{tangent\
points}|\\}.$

By Theorem 3.1, $\text{ind}(L_{\gamma})=\kappa^{+}+\kappa^{-}=|\text{tangent\
quadratic endpoints}|$, so $\text{dim ker}\
L_{\gamma}^{*}=|bP_{ter}(L_{\gamma})|$, where

$\displaystyle
bP_{ter}(L_{\gamma})=\begin{cases}\begin{array}[]{cc}P&\text{$L$ is
transverse/neutral to both endpoints,}\\\ \text{Kimura tangent
endpoint}&\text{otherwise}.\end{array}\end{cases}$

Write

$\displaystyle C^{0}([0,1])^{*}=C^{0}_{0}([0,1])^{*}\oplus A^{*}$

where $A^{*}=\\{\delta(p)|\ p\ \text{quadratic}\\}$. $A^{*}\subset\text{ker}\
\overline{L}^{*}$. Since $\widetilde{a}\cdot\mathcal{D}^{\gamma}([0,1])$ is
dense in the subspace $C_{0}^{0}([0,1])$, every $\mu\in\text{ker}\
{L_{\gamma}}^{*}$ can be uniquely extended to a measure in $\text{ker}\
\overline{L}^{*}\cap C^{0}_{0}([0,1])^{*}$, so there is an inclusion map

$\displaystyle i:\text{ker}\ \overline{L}^{*}\cap
C^{0}_{0}([0,1])^{*}\longrightarrow\text{ker}\ L^{*}_{\gamma}.$

And $\text{codim}(i)=1$ if there are one or more neutral point(s) and no
tangent point, while $\text{codim}(i)=0$. This is equivalent to

$\displaystyle\text{codim}(i)=|bP_{\text{ter}}(L_{\gamma})|-|bP_{ter}(L)|+|\text{quadratic
endpoint}|.$

In conclusion

$\displaystyle\text{dim ker}\ \overline{L}^{*}=\text{dim ker}\
L_{\gamma}^{*}-\text{codim}(i)+|\text{quadratic\ point}|=|bP_{ter}(L)|.$

∎

The following table summarizes the invariant measures found in the ten
different cases of interest:

| K Trans | K Tan | Q Trans | Q Tan **
---|---|---|---|---
Kimura Transverse (K Trans) |  $\mu$* | $\delta_{1}$ |  $\mu$, $\delta_{1}$ | $\delta_{1}$
Kimura Tangent (K Tan) | $\delta_{0}$ |  $\delta_{0}$, $\delta_{1}$ | $\delta_{0}$ |  $\delta_{0}$, $\delta_{1}$
Quadratic Transverse (Q Trans) |  $\mu$, $\delta_{0}$ |  $\delta_{0}$, $\delta_{1}$ |  $\mu$, $\delta_{0}$, $\delta_{1}$ |  $\delta_{0}$, $\delta_{1}$
Quadratic Tangent (Q Tan)** | $\delta_{0}$ |  $\delta_{0}$, $\delta_{1}$ | $\delta_{1}$ |  $\delta_{0}$, $\delta_{1}$

  * *

$\mu$ refers to an invariant measure supported on (0,1)

  * **

include neutral case

Table 1: Invariant measures for all cases of boundary types at $x=0$ (rows)
and at $x=1$ (columns).

###  3.3 Proof of Theorem 3.1

For convenience of computation, we fix the spatial domain as the interval
$[0,3]$ instead of $[0,1]$ whenever convenient in this subsection.

####  3.3.1 Outline of proof

It remains to characterize the range of $L_{\gamma}$. First, we turn the
second-order operator $L_{\gamma}$ to a first-order system $M$ in (20), with
an index equal to the index of $L_{\gamma}$ (see Lemma 3.2 below). Next we
continuously deform $M$ to another first-order system $\widetilde{M}$ with
constant coefficients in the vicinity of the two endpoints. Such a deformation
does not change the index (see Proposition 3.3 below), so the original problem
now is equivalent to the easier linear system $\widetilde{M}$. These
constructions are presented in subsection 3.3.2.

The problem $\widetilde{M}u=f=\left(\begin{array}[]{c}f_{1}\\\
f_{2}\end{array}\right)$, given $u(1)\in\mathbb{R}^{2}$, has a uniquely
defined solution $u$. In order for $u$ to belong to the domain space, we
observe that $\alpha f_{1}$ has to vanish at tangent Kimura endpoints and
$u(1),f$ have to be related by

$\displaystyle E_{-}u(1)=I_{1}f,\quad E_{+}\Lambda u(1)=-I_{2}f;$ (19)

see the definition of $I_{1},I_{2},E_{\pm}$, and $\Lambda$ in Section 3.4.
Conversely if $\alpha f_{1}$ vanishes at tangent Kimura endpoints and if we
can find $u(1)$ satisfying the relation (19), then the uniquely defined $u$ is
the solution of $\widetilde{M}u=f$ (see Lemma 3.3, Lemma 3.4). Thus $f$ in the
range of $\widetilde{M}$ is equivalent to the existence of
$u(1)\in\mathbb{R}^{2}$ satisfying (19). This construction is presented in
subsection 3.3.3.

We next show in Lemma 3.6 that the above constraint is equivalent to

$\displaystyle\Lambda
I_{1}f+I_{2}f\in\text{ker}(E_{+})+\Lambda\cdot\text{ker}(E_{-}).$

We thus define the linear map

$\displaystyle\Phi:C^{1+\gamma}([0,3])\times{\mathcal{D}^{\gamma}([0,3])}\longrightarrow\frac{\mathbb{R}^{2}}{\text{ker}(E_{+})+\Lambda\text{ker}(E_{-})},$

by assigning $f$ to the coset $[\Lambda I_{1}f+I_{2}f]$. Clearly the range of
$\widetilde{M}$ lies in the kernel of $\Phi$. A computation in Lemma 3.7 shows
that $\Phi$ is surjective and $\text{dim\
ker}(\Phi)/R(\widetilde{M})=|\text{kimura tangent points}|$. We now have all
the ingredients to compute the codimension of $\widetilde{M}$. The final
results and proofs are given in subsection 3.3.4.

####  3.3.2 Reduction to a first order system

To simplify notation, we consider $L_{\gamma}$ defined on the interval
$[0,3]$. We rewrite the system $L_{\gamma}u=f$ as a first-order system $M$:

$\displaystyle M:C^{2+\gamma}([0,3])\times C^{1+\gamma}([0,3])\longrightarrow
C^{1+\gamma}([0,3])\times\mathcal{D}^{\gamma}([0,3]).$ (20)

associated to the expression

$\displaystyle\begin{pmatrix}u_{0}\\\
u_{1}\end{pmatrix}\mapsto\begin{pmatrix}u_{0}^{\prime}\\\
u_{1}^{\prime}\end{pmatrix}+A(L_{\gamma})\begin{pmatrix}u_{0}\\\
u_{1}\end{pmatrix}=\begin{pmatrix}u_{0}^{\prime}-u_{1}\\\
u_{1}^{\prime}+\frac{b}{a}u_{1}\end{pmatrix},\qquad
A(L_{\gamma}):=\left(\begin{array}[]{cc}0&-1\\\
0&\frac{b}{a}\end{array}\right).$

where $a,b$ are the coefficients of second-order and first-order term of $L$
under heat coordinate $\phi$, i.e.
$L_{\phi}=a(x)\partial_{xx}+b(x)\partial_{x}$.

For a smooth coordinate change $z=z(x)$, we let $L_{z}$ be the operator
corresponding to $L$ under coordinate change and define $A(L_{z})$ and $M_{z}$
as above. If $M\begin{pmatrix}u_{0}\\\
u_{1}\end{pmatrix}=\begin{pmatrix}f_{0}\\\ f_{1}\end{pmatrix}$, then

$\displaystyle M_{z}\begin{pmatrix}u_{0}[x(z)]\\\
u_{1}[x(z)]x^{\prime}(z)\end{pmatrix}=\begin{pmatrix}f_{0}[x(z)]x^{\prime}(z)\\\
f_{1}[x(z)](x^{\prime}(z))^{2}\end{pmatrix}.$ (21)

In the local charts $([0,1],\phi^{-1}),([2,3],\phi^{-1})$,
$A(t)=\left(\begin{array}[]{cc}0&-1\\\ 0&b(z)\end{array}\right)$ when 0 is a
quadratic point with heat coordinate, while
$A(t)=\left(\begin{array}[]{cc}0&-1\\\ 0&\frac{{b}(x)}{x}\end{array}\right)$
when $0$ is a Kimura point.

######  Lemma 3.2.

Assume that $M$ is Fredholm, then $L_{\gamma}$ is also Fredholm and

$\displaystyle\text{ind}(L_{\gamma})=\text{ind}(M).$ (22)

###### Proof.

There is an isomorphism between
$\text{ker}(M)\longrightarrow\text{ker}(L_{\gamma})$ by mapping
$(u,u^{\prime})$ to $u$, so the dimensions of the kernel spaces are the same.

If $(f_{0},f_{1})\in R(M)$ and $M\begin{pmatrix}u_{0}\\\
u_{1}\end{pmatrix}=\begin{pmatrix}f_{0}\\\ f_{1}\end{pmatrix}$, then
$L_{\gamma}u_{0}=\widetilde{a}f_{0}^{\prime}+\widetilde{b}f_{0}+\widetilde{a}f_{1}$.
Based on this observation we define the map

$\displaystyle\pi:C^{1+\gamma}([0,3])\times\mathcal{D}^{\gamma}([0,3])\longrightarrow\alpha\cdot\mathcal{D}^{\gamma}([0,3])$
$\displaystyle(f_{0},f_{1})\mapsto\widetilde{a}f_{0}^{\prime}+\widetilde{b}f_{0}+\widetilde{a}f_{1}.$

We claim that

$\displaystyle\text{codim ran}(\pi)=\text{codim\ ran}(L_{\gamma})-\text{codim
ran}(M).$ (23)

We first show that

$\displaystyle f\in\text{ran}(M)\Leftrightarrow\pi
f\in\text{ran}(L_{\gamma}).$ (24)

The inclusion part is straightforward. For the converse part, if
$\pi(f_{0},f_{1})=L_{\gamma}u\in\text{ran}(L_{\gamma})$, we verify that
$M\begin{pmatrix}u\\\ u^{\prime}-f_{0}\end{pmatrix}=\begin{pmatrix}f_{0}\\\
f_{1}\end{pmatrix}\in\text{ran}(M)$. Next we show that

$\displaystyle\pi(\text{ran}(M))=\text{ran}(L_{\gamma}).$ (25)

Again the inclusion part is straightforward. For the converse part, if
$g=L_{\gamma}u\in\text{ran}(L_{\gamma})$, then
$\pi(0,\frac{g}{\widetilde{a}})=g\in\text{ran}(L_{\gamma})$. Note that by
(24), this implies that $(0,\frac{g}{\widetilde{a}})\in\text{ran}(M)$. (24)
and (25) imply that the quotient map defined by

$\displaystyle\pi^{\prime}:[C^{1+\gamma}([0,3])\times\mathcal{D}^{\gamma}([0,3])]/\text{ran}(M)\longrightarrow\alpha\cdot\mathcal{D}^{\gamma}([0,3])/\text{ran}(L_{\gamma})$

is injective. So $\text{codim}\ \text{ran}(\pi^{\prime})=\text{codim}\
\text{ran}(L_{\gamma})-\text{codim}\ \text{ran}(M)$. At the same time,
$\pi(\text{ran}(M))=\text{ran}(L_{\gamma})$ indicates $\text{codim}\
\text{ran}(\pi^{\prime})=\text{codim}\ \text{ran}(\pi)$ so that (23) holds.
Clearly, $\pi$ is surjective since $\pi(0,\frac{h}{\widetilde{a}})=h$.
Combined with (23), this proves (22). ∎

##### Index of the model operator $\widetilde{M}$

Alongside $M$, we consider the operator $\widetilde{M}$ associated with a
continuous function $A(t)$ such that

$\displaystyle A(t)=A_{-}\ \text{for}\ t\leq 1,A(t)=A_{+}\ \text{for}\ t\geq
2$

where $A_{-}=\left(\begin{array}[]{cc}0&-1\\\ 0&b_{-}\end{array}\right)$ if 0
is a quadratic point, $A_{-}=\left(\begin{array}[]{cc}0&-1\\\
0&\frac{b_{-}}{x}\end{array}\right)$ if 0 is a kimura point, and $A_{+}$ is
similarly defined. Since $M$ and $\widetilde{M}$ can be reduced to each other
by a continuous deformation in the class of Fredholm operators, we have

######  Proposition 3.3.

The operator $M$ is Fredholm if and only if the operator $\widetilde{M}$ is
Fredholm and

$\displaystyle\text{ind}(M)=\text{ind}(\widetilde{M}).$ (26)

###### Proof.

We need to show that $M-\widetilde{M}$ is a compact map since then by [26,
§27.1 Theorem 3], $\widetilde{M}$ has the same index as $M$.

Given a bounded sequence $u_{k}$ in $C^{2+\gamma}([0,3])\times
C^{1+\gamma}([0,3])$, we need to show that there exists a convergent
subsequence $v_{k}$ of $(M-\widetilde{M})u_{k}$. $v_{k}$ convergent in
$C^{1+\gamma}([0,3])\times\mathcal{D}^{\gamma}([0,3])$ is equivalent to
$v_{k}|_{[0,1]},v_{k}|_{[1,2]},v_{k}|_{[2,3]}$ convergent in the local spaces,
respectively. Thus it suffices to show that
$(M-\widetilde{M})|_{[0,1]},(M-\widetilde{M})|_{[2,3]}$ are compact.

To check both Kimura and quadratic cases, we assume that $L_{0}$ is Kimura
type and $L_{1}$ is of quadratic type.

At $x=0$, given a bounded sequence $(u_{n},q_{n})$ in
$C^{2+\gamma}([0,1])\times C^{1+\gamma}([0,1])$ , we have

$\displaystyle(M-\widetilde{M})(u_{n},q_{n})=(0,\frac{b(x)-b_{-}}{x}q_{n}).$

Since $\frac{b(x)-b_{-}}{x}$ is smooth and bounded, it remains to show by
definition of $\mathcal{D}^{\gamma}([0,1])$ that $xq_{n}$ has a convergent
subsequence in $C^{\gamma}([0,1])$.

Since $\\{q_{n},xq_{n}^{\prime}\\}$ are bounded in $C^{\gamma}([0,1])$, then
$\\{xq_{n}\\}$ are uniformly bounded and uniformly equicontinuous. Therefore
by the Arzela-Ascoli theorem, there exists a convergent subsequence $v_{n}$ of
$xq_{n}$ in $C^{0}([0,1])$. For the $[\cdot]_{\gamma}$ part, notice that for
$g_{n}(y)=v_{n}(x)$ under the coordinate change $y=x^{\gamma}$,

$\displaystyle[v_{n}]_{\gamma}=O(||g_{n}^{\prime}||_{\infty}).$ (27)

Since $g_{n}^{\prime}=\frac{1}{\gamma}v_{n}^{\prime}x^{1-\gamma}$ and
$\\{v^{\prime}_{n}\\}$ are uniformly bounded in $C^{\gamma}([0,1])$,
$\\{g^{\prime}_{n}\\}$ are uniformly bounded and uniformly equicontinuous.
Thus, there exists a subsequence of $g^{\prime}_{n}$ convergent in
$C^{0}([0,1])$. By (27), the corresponding subsequence of $v_{n}$ converges in
$C^{\gamma}([0,1])$. Therefore $(M-M^{\prime})|_{[0,1]}$ is a compact
operator.

At $x=3$, we prove the result using heat coordinates in $[-\infty,0]$. Given a
bounded sequence $(u_{n},q_{n})$ in $C^{2+\gamma}((-\infty,0])\times
C^{1+\gamma}((-\infty,0])$, write

$\displaystyle(M-\widetilde{M})(u_{n},q_{n})=(0,(b(z)-b_{+})q_{n}).$

Since $\\{q^{\prime}_{n}\\}$ is bounded in $C^{\gamma}(T)$for any compact set
$T$, there exists a convergent subsequence of $\\{q_{n}\\}$ in
$C^{\gamma}(T)$. We choose a convergent subsequence $\\{q^{k}_{1}\\}_{k\geq
1}$ on $[-1,0]$, and $\\{q^{k}_{2}\\}_{k\geq 1}$ a convergent subsequence of
$\\{q^{k}_{1}\\}$ on $[-2,0]$. Iteratively, we choose $\\{q^{k}_{n}\\}_{k\geq
1}$ to be a convergent subsequence of $\\{q^{k}_{n-1}\\}$ on $[-n,0]$.

We show below that the diagonal sequence
$\\{\left(b(z)-b_{+}\right)q^{k}_{k}\\}_{k\geq 1}$ converges in
$\mathcal{D}^{\gamma}((-\infty,0])$. Since $||q_{k}^{k}||_{\gamma}$ is bounded
and $b(z)-b_{-}$ tends to 0 as $z$ tends to $-\infty$, there exist $N>0,K>0$
such that

$\displaystyle\max\Big{\\{}|(b(z)-b_{-})q_{k}^{k}|,\quad[(b(z)-b_{-})q_{k}^{k}]_{\gamma},\quad\underset{z_{1}\leq
z_{2}\leq-N}{\sup}\left|\int_{z_{1}}^{z_{2}}(b(z)-b_{-})q_{k}^{k}dz\right|\Big{\\}}\leq\epsilon/4,\text{
on }\ (-\infty,-N]$ $\displaystyle\\{q^{k}_{k}\\}_{k\geq K}\ \text{converge
in}\ C^{\gamma}([-N,0]).$

So we can pick $N^{\prime}$ such that
$||(b(z)-b_{-})q^{m}_{m}-(b(z)-b_{-})q^{n}_{n}||_{\gamma,[-N,0]}<\frac{\epsilon}{2}$
when $m,n>N^{\prime}$. Combining the above, we have
$||(b(z)-b_{-})q^{m}_{m}-(b(z)-b_{-})q^{n}_{n}||_{\gamma,(-\infty,0]}<\epsilon$
when $m,n>N^{\prime}$. We proved that the diagonal sequence
$\\{\left(b(z)-b_{+}\right)q^{k}_{k}\\}_{k\geq 1}$ converged in
$\mathcal{D}^{\gamma}((-\infty,0])$. Therefore $(M-M^{\prime})|_{[2,3]}$ is a
compact operator. It is not hard to show that $(M-M^{\prime})|_{[1,2]}$ is
also a compact operator so that $M-\widetilde{M}$ is compact. ∎

####  3.3.3 Representation of solutions of the modeling operator

The results in the previous subsection show that
$\text{ind}(L_{\gamma})=\text{ind}(\widetilde{M})$ if $\widetilde{M}$ is
Fredholm. Now we turn to proving that $\widetilde{M}$ is Fredholm and
computing the index of $\widetilde{M}$. Recall that

$\displaystyle\widetilde{M}:C^{2+\gamma}([0,3])\times
C^{1+\gamma}([0,3])\longrightarrow
C^{1+\gamma}([0,3])\times\mathcal{D}^{\gamma}([0,3])$ (28)

is associated to the expression

$\displaystyle\begin{pmatrix}u_{0}\\\
u_{1}\end{pmatrix}\mapsto\begin{pmatrix}u_{0}^{\prime}\\\
u_{1}^{\prime}\end{pmatrix}+A(t)\begin{pmatrix}u_{0}\\\
u_{1}\end{pmatrix},\qquad A(t)=A_{-}\ \text{for}\ t\leq 1,\ \ A(t)=A_{+}\
\text{for}\ t\geq 2.$

Let $f\in C^{1+\gamma}([0,3])\times\mathcal{D}^{\gamma}([0,3])$ be given, and
suppose $u\in C^{2+\gamma}([0,3])\times C^{1+\gamma}([0,3])$ solves

$\displaystyle\frac{du}{dt}+Au=f.$

Let $Y=Y(t)$ denote the fundamental matrix associated to $\widetilde{M}$,
defined in different intervals by

$\displaystyle\frac{dY}{dt}+AY(t)=0,\ Y(1)=I,\text{on}\ [0,2)$
$\displaystyle\frac{dY}{dt}+AY(t)=0,\ Y(2+)=I,\text{on}\ (2,3].$

Then $\Lambda:=Y(2-)$ is an invertible matrix and

$\displaystyle u(2)=\Lambda u(1)+\Lambda\int_{1}^{2}Y(s)^{-1}f(s)ds.$

Given $u(1)$, we can uniquely determine $u(t)$ on $[0,3]$:

$\displaystyle u(t)=Y(t)u(1)-\int_{t}^{1}Y(t)Y^{-1}(s)f(s)ds,\ t\leq 1$ (29)
$\displaystyle u(t)=Y(t)\left(\Lambda
u(1)+\Lambda\int_{1}^{2}Y(s)^{-1}f(s)ds\right)+\int_{2}^{t}Y(t)Y^{-1}(s)f(s)ds,\
t\geq 2.$ (30)

We analyze the above solutions for different boundary points. For quadratic
endpoints, we directly assume $t$ is in heat coordinates.

##### 1\. Quadratic endpoints with $b_{\pm}\neq 0$,

$A_{\pm}=\left(\begin{array}[]{cc}0&-1\\\ 0&b_{\pm}\end{array}\right)$
constant matrix. Then

$\displaystyle
Y(t)=\left\\{\begin{array}[]{cc}e^{-(t-1)A_{-}},&t\in(-\infty,1]\\\
e^{-(t-2)A_{+}},&t\in[2,+\infty).\end{array}\right.$ (33)

$A_{\pm}$ has two different eigenvalues $0,b_{\pm}$. We let $E_{-}$ be the
spectral projector associated to the set of all positive eigenvalues of
$A_{-}$, and $E_{+}$ be the spectral projector associated to the set of all
negative eigenvalues of $A_{+}$. Apply the operator $E_{-},E_{+}$ on
$\eqref{solution_u1},\eqref{solution_u2}$, and let $t$ go to $-\infty,\infty$
to get:

$\displaystyle E_{-}u(1)=\int_{-\infty}^{1}e^{{(s-1)}A_{-}}E_{-}f(s)ds,$ (34)
$\displaystyle E_{+}\Lambda
u(1)=-E_{+}\Lambda\int_{1}^{2}Y(s)^{-1}f(s)ds-\int_{2}^{\infty}E_{+}e^{(s-2)A_{+}}f(s)ds.$
(35)

Using this relation, we recast $u(t)$ for $t<1$ as:

$\displaystyle
u(t)=e^{-(t-1)A_{-}}(I-E_{-})u(1)-\int_{t}^{1}e^{-(t-s)A_{-}}(I-E_{-})f(s)ds+\int_{-\infty}^{t}e^{-(t-s)A_{-}}E_{-}f(s)ds$
(36)

and for $t>2$ as:

$\displaystyle
u(t)=e^{-(t-1)A_{+}}(I-E_{+})\Lambda\left[u(1)+\int_{1}^{2}Y(s)^{-1}f(s)ds\right]$
(37)
$\displaystyle+\int_{2}^{t}e^{-(t-s)A_{+}}(I-E_{+})f(s)ds-\int_{t}^{\infty}e^{-(t-s)A_{+}}E_{+}f(s)ds.$

##### 2\. Kimura transverse point

In this case

$\displaystyle Y(t)=\left(\begin{array}[]{cc}1&F(t^{-b_{-}})-F(1)\\\
0&t^{-b_{-}}\end{array}\right),\ t\in[0,1],$ $\displaystyle
Y(t)=\left(\begin{array}[]{cc}1&F((\frac{t}{2})^{-b_{+}})-F(1)\\\
0&(\frac{t}{2})^{-b_{+}}.\end{array}\right),\ t\in[2,3],$

where $F$ denotes the antiderivative. We let
$E_{\pm}=\left(\begin{array}[]{cc}0&0\\\ 0&1\end{array}\right)$. Multiplying
by $E_{-}Y(t)^{-1}$ both sides of (29), (30), and letting $t\to 0,t\to 3$,
respectively, we obtain

$\displaystyle E_{-}u(1)=\int_{0}^{1}E_{-}Y(s)^{-1}f(s)ds,$ (38)
$\displaystyle E_{+}\Lambda
u(1)=-E_{+}\Lambda\int_{1}^{2}Y(s)^{-1}f(s)d-\int_{2}^{3}E_{+}Y^{-1}(s)f(s)ds.$
(39)

Using this relation we recast $u(t)$ for $t<1$ as:

$\displaystyle
u(t)=Y(t)(I-E_{-})u(1)+Y(t)\int_{0}^{t}E_{-}Y^{-1}(s)f(s)ds-Y(t)\int_{t}^{1}(I-E_{-})Y^{-1}(s)f(s)ds$
(40)

and for $t>2$ as:

$\displaystyle
u(t)=Y(t)(I-E_{+})\Lambda\left(u(1)+\int_{1}^{2}Y(s)^{-1}f(s)ds\right)$ (41)
$\displaystyle+\int_{2}^{t}Y(t)(I-E_{+})Y^{-1}(s)f(s)ds-\int_{t}^{3}Y(t)E_{+}Y^{-1}(s)f(s)ds.$

##### 3\. Quadratic endpoints with $b_{\pm}=0$

In this case

$\displaystyle Y(t)=\left(\begin{array}[]{cc}1&t-1\\\ 0&1\end{array}\right),\
t\in[0,1],\qquad\quad Y(t)=\left(\begin{array}[]{cc}1&t-2\\\
0&1.\end{array}\right),\ t\in[2,3].$

We let $E_{\pm}=\left(\begin{array}[]{cc}0&0\\\ 0&1\end{array}\right)$.
Multiply by $e^{t}$ on both sides of (29), (30) and let $t\to\pm\infty$. We
again obtain (38), (39) with the integral lower/upper points $0,3$ replaced by
$-\infty,\infty$. So we can re-express $u(t)$ as (40), (41) again with
integral lower/upper points $0,3$ replaced with $-\infty,\infty$.

##### 4\. Kimura tangent point

In this case, $A_{\pm}=\left(\begin{array}[]{cc}0&-1\\\
0&0\end{array}\right)$. We observe that if
$f=(f_{0},f_{1})=\widetilde{M}(u_{0},u_{1})\in\text{ran}(\widetilde{M})$, then
$f_{1}=u_{1}^{\prime}$, so that $s(s-3)f_{1}(s)$ vanishes at Kimura tangent
endpoints by definition of $C^{1+\gamma}([0,3])$. We next let
$E_{\pm}=\left(\begin{array}[]{cc}0&0\\\ 0&0\end{array}\right)$. Obviously
(38), (39) are true then. So we can also recast $u(t)$ as (40), (41).

Summarizing these boundary cases, we make the following definition.

######  Definition 3.3.

We define

$\displaystyle I_{1}f=\int_{l}^{1}E_{-}Y(s)^{-1}f(s)ds,$ $\displaystyle
I_{2}f=E_{+}\Lambda{\int_{1}^{2}}Y(s)^{-1}f(s)ds+\int_{2}^{r}E_{+}Y^{-1}(s)f(s)ds,$

with $l,r=-\infty,\infty$ at a quadratic endpoint and $l,r=0,3$ at a Kimura
endpoint.

In all cases, when $f$ satisfying the corresponding boundary conditions is in
the range of $\widetilde{M}$, then $I_{1}f,I_{2}f$ are related by

$\displaystyle E_{-}u(1)=I_{1}f,\quad E_{+}\Lambda u(1)=-I_{2}f.$ (42)

For $f$ satisfying the relation (42), we fix a constant $u(1)$ as in (42), and
then define $u(t)$ on the whole domain by (40) and (41) (or (36) and (37) for
quadratic endpoints). We now show that such a $u$ is indeed the solution.

######  Lemma 3.3.

For $U=(-\infty,c]$ or $U=[c,\infty)$, if $f=(f_{0},f_{1})\in
C^{1+\gamma}(U)\times\mathcal{D}^{\gamma}(U)$ together with some
$u(1)\in\mathbb{R}^{2}$ satisfy (42), then $u(t)$ defined by (36),(37) is in
$C^{2+\gamma}(U)\times C^{1+\gamma}(U)$ and solves

$\displaystyle\frac{du}{dt}+Au=f.$

###### Proof.

When $b_{\pm}\neq 0$, $A_{\pm}$ can be diagonalized as

$\displaystyle A_{\pm}=\left(\begin{array}[]{cc}1&-1\\\
0&b_{\pm}\end{array}\right)\left(\begin{array}[]{cc}0&0\\\
0&b_{\pm}\end{array}\right)\left(\begin{array}[]{cc}1&-1\\\
0&b_{\pm}\end{array}\right)^{-1}=CB_{\pm}C^{-1}.$

By changing a basis $u\mapsto C^{-1}u$, we may assume that $A_{\pm}$ has the
diagonal form $B_{\pm}$ and $E_{\pm}$ is the corresponding projector of
$A_{\pm}$. For convenience we only prove (36) as (37) is obtained similarly.

For the solution $u=(u_{0},u_{1})$, it is straightforward to see that
$u_{0}\in C^{2}((-\infty,c])$,
$\underset{z\to-\infty}{\lim}\partial_{z}u=\underset{z\to\
-\infty}{\lim}\partial_{z}^{2}u=0$ and satisfies
$\partial_{z}^{2}u+b_{-}\partial_{z}u=f_{1}^{\prime}+f_{2}+bf_{1}$. Hence the
classical result of elliptic operator in Hölder spaces gives that $u_{0}\in
C^{2+\gamma}((-\infty,c])$. Then $u_{1}=u_{0}^{\prime}-f_{1}\in
C^{1+\gamma}((-\infty,c])$.

When $b_{\pm}=0$, we explicitly express $u(t)$ for $t<1$ as follows:

$\displaystyle
u_{0}(t)=u_{0}(1)-u_{1}(1)+tu_{1}(t)-\int_{t}^{1}[f_{0}(s)-sf_{1}(s)]ds,$ (43)
$\displaystyle u_{1}(t)=u_{1}(1)-\int_{t}^{1}f_{1}(s)ds.$ (44)

Since $f$ satisfies (34), by Remark 3.1, $u_{0},u_{1}$ are integrable up to
$-\infty$. $u_{1}^{\prime}(t)=f_{1}(t)\in C^{\gamma}((-\infty,c])$ so
$u_{1}\in C^{1+\gamma}((-\infty,c])$. And

$u_{0}^{\prime}(t)=u_{1}(1)+f_{0}(t)-\int^{0}_{t}f_{1}(s)ds,\ \
u_{0}^{\prime\prime}(t)=f_{0}^{\prime}(t)+f_{1}(t).$

$u_{0}^{\prime\prime}\in C^{\gamma}((-\infty,c])$ and $u_{0}^{\prime}$ is
integrable up to $-\infty$, so $u_{0}^{\prime}\in C^{1+\gamma}((-\infty,c])$.
Thus, $u\in C^{2+\gamma}((-\infty,c])\times C^{1+\gamma}((-\infty,c])$ and
solves $\frac{du}{dt}+Au=f$. ∎

######  Lemma 3.4.

For $U=[0,1]$ or $U=[2,3]$, if $f=(f_{0},f_{1})\in
C^{1+\gamma}(U)\times\mathcal{D}^{\gamma}(U)$ together with some
$u(1)\in\mathbb{R}^{2}$satisfies (42) and $s(s-3)f_{1}(s)$ vanishes at Kimura
tangent endpoints, then $u(t)$ defined by (40),(41) is in
$C^{2+\gamma}(U)\times C^{1+\gamma}(U)$ and solves

$\displaystyle\frac{du}{dt}+Au=f.$

###### Proof.

At a transverse Kimura point, we write $u(t)$ for $t<1$ as

$\displaystyle u(t)=\left(\begin{array}[]{cc}1&0\\\
0&0\end{array}\right)u(1)+\int_{0}^{t}\left(\begin{array}[]{cc}0&s^{b_{-}}G(t)\\\
0&t^{-{b_{-}}}s^{b_{-}}\end{array}\right)f(s)ds-\int_{t}^{1}\left(\begin{array}[]{cc}1&s^{b_{-}}G(s)\\\
0&0\end{array}\right)f(s)ds,$ (51)

where $G(t)=F(t^{-{b_{-}}})-F(1)$. By direct computation we have

$\displaystyle
u_{1}^{\prime}(t)=f_{1}(t)-at^{-({b_{-}}+1)}\int_{0}^{t}s^{b_{-}}f_{1}(s)ds,$
$\displaystyle
u_{0}^{\prime}(t)=t^{-{b_{-}}}\int_{0}^{t}s^{{b_{-}}}f_{1}(s)ds+f_{0}(t),u_{0}^{\prime\prime}(t)=f_{1}(t)-{b_{-}}t^{-({b_{-}}+1)}\int_{0}^{t}s^{b_{-}}f_{1}(s)ds+f_{0}^{\prime}(t).$

Thus $u_{0}^{\prime\prime},u_{1}^{\prime}\in\mathcal{D}^{\gamma}([0,1])$ and
by definition $u\in C^{2+\gamma}(U)\times C^{1+\gamma}(U)$ and solves
$\frac{du}{dt}+Au=f$.

At a tangent Kimura point, for $t\in[0,1]$, we write $u(t)$ for $t<1$ as:

$\displaystyle
u_{0}(t)=u_{0}(1)+(t-1)u_{1}(1)-\int_{t}^{1}[f_{0}(s)+(t-s)f_{1}(s)]ds,$ (52)
$\displaystyle u_{1}(t)=u_{1}(1)-\int_{t}^{1}f_{1}(s)ds.$ (53)

$u_{1}$ is integrable up to $t=0$ iff $\underset{s\to 0}{\lim}\ sf_{1}(s)=0$.
In such a case, $u_{1}^{\prime}(t)=f_{1}(t)\in\mathcal{D}^{\gamma}([0,1])$ and
so $u_{1}\in C^{1+\gamma}([0,1])$. Since

$\displaystyle u_{0}^{\prime}(t)=u_{1}(1)+f_{0}(t)-\int^{1}_{t}f_{1}(s)ds,\
u_{0}^{\prime\prime}(t)=f_{0}^{\prime}(t)+f_{1}(t),$

$u\in C^{2+\gamma}(U)\times C^{1+\gamma}(U)$ and solves $\frac{du}{dt}+Au=f$.
For $t\in[2,3]$, the proof is essentially the same. ∎

####  3.3.4 Proof of Theorem 3.1

######  Lemma 3.5.

$dim\ \text{ker}(\widetilde{M})=dim\
[\text{ker}(E_{-})\cap\Lambda^{-1}\text{ker}(E_{+})]$.

###### Proof.

If $\widetilde{M}u=0$, then $\Lambda u(1)\in\text{ker}(E_{+})$ and similarly
$u(1)\in\text{ker}(E_{-})$. On the other hand, we set
$u(1)\in\text{ker}(E_{-})\cap\Lambda^{-1}\text{ker}(E_{+})$, and define $u$ at
the quadratic boundary point by

$\displaystyle\text{for}\ t\leq 1:u(t)=e^{-(t-1)A_{-}}u(1),\quad\text{for}\
t\geq 2:u(t)=e^{-(t-2)A_{+}}\Lambda u(1)$

and at the Kimura endpoint by

$\displaystyle\text{for}\ t\leq 1:u(t)=\Phi(t)u(1),\quad\text{for}\ t\geq
2:u(t)=\Phi(t)\Lambda u(1)$

and $u(t)$ is uniquely characterized on $[1,2]$ by $u(1)$. Then, $u\in
C^{2+\gamma}([0,3]\times C^{1+\gamma}([0,3])$ and solves $\widetilde{M}u=0$.
When such $u$ exists, it is uniquely determined by $u(1)$. We thus proved that
the map

$\displaystyle
N:\text{ker}(\widetilde{M})\longrightarrow\text{ker}(E_{-})\cap\Lambda^{-1}\text{ker}(E_{+})$

by assigning $u$ to $u(1)$ is a bijection. Therefore $\text{dim}\
\text{ker}(\widetilde{M})=\text{dim}\
[\text{ker}(E_{-})\cap\Lambda^{-1}\text{ker}(E_{+})]$. ∎

######  Lemma 3.6.

The condition (42) is equivalent to

$\displaystyle\Lambda
I_{1}f+I_{2}f\in\text{ker}(E_{+})+\Lambda\cdot\text{ker}(E_{-}).$ (54)

###### Proof.

Suppose (42) is satisfied, then for some $v_{1},v_{2}\in\mathbb{R}^{2}$ we
have

$\displaystyle u(1)=I_{1}f+(I-E_{-})v_{2},\quad\Lambda
u(1)=-I_{2}(f)+(I-E_{+})v_{1}.$

Multiplying the first equality by $\Lambda$ and subtracting the second
equality, we obtain (54).

Conversely, suppose that (54) is true. Then, for some
$v_{1},v_{2}\in\mathbb{R}^{2}$,

$\displaystyle I_{2}f+\Lambda I_{1}f=-\Lambda(I-E_{-})v_{2}-(I-E_{+})v_{1}.$

Define $u(1)=I_{1}f+(I-E_{-})v_{2}$. Using that $E^{2}_{\pm}=E_{\pm},\
E_{\pm}(I-E_{\pm})=0$, we verify that (42) is satisfied. ∎

We define the linear map

$\displaystyle\Phi:C^{1+\gamma}([0,3])\times{\mathcal{D}^{\gamma}([0,3])}\longrightarrow\frac{\mathbb{R}^{2}}{\text{ker}(E_{+})+\Lambda\cdot\text{ker}(E_{-})}.$

by assigning $f$ to the coset $[\Lambda I_{1}f+I_{2}f]$.

$\text{ran}(\widetilde{M})\subset\text{ker}(\Phi)$ is already known and
moreover

$\text{ran}(\widetilde{M})\subset\text{ker}(\Phi)\cap\\{(f_{1},f_{2}):s(s-3)f_{2}(s)\
\text{vanishes at Kimura tangent points.}\\}$

Indeed we show below that they are equal and thus obtain

######  Lemma 3.7.

We have

  1. 1.

$\text{dim}\ \text{ker}(\Phi)/\text{ran}(\widetilde{M})=|\text{Kimura tangent
points}|$,

  2. 2.

$\Phi$ is surjective.

###### Proof.

We show that

$\displaystyle\text{ker}(\Phi)\cap\\{(f_{1},f_{2}):s(s-3)f_{2}(s)\
\text{vanishes at Kimura tangent points}\\}=\text{ran}(\widetilde{M}).$ (55)

First we assume that there is no Kimura tangent endpoint. Given
$f\in\text{ker}(\Phi)$, then for some $u(1)\in\mathbb{R}^{2}$, $E_{+}\Lambda
u(1)=-I_{1}f,\ E_{-}u(1)=I_{2}f$. Define $u$ by (40), (41). Then by Lemmas
3.3, 3.4, $u\in C^{2+\gamma}([0,3])\times C^{1+\gamma}([0,3])$ solves
$\widetilde{M}u=f$, so that $f\in\text{ran}(\widetilde{M})$, and hence
$\text{ran}(\widetilde{M})=\text{ker}(\Phi)$.

If there is at least one Kimura tangent point, then
$\text{ker}(E_{+})+\Lambda\text{ker}(E_{-})=\mathbb{R}^{2}$, so
$\text{ker}(\Phi)$ is the whole space
$C^{1+\gamma}([0,3])\times{\mathcal{D}^{\gamma}([0,3])}$. For $f$ in the left
hand space of (55), we define $u$ by (40), (41). Again by Lemmas 3.3, 3.4,
such $u$ is the solution of $\widetilde{M}u=f$, and hence we proved (55) in
this case.

Next we prove that $\Phi$ is surjective. If there is at least one Kimura
tangent endpoint, then
$\text{ker}(E_{+})+\Lambda\text{ker}(E_{-})=\mathbb{R}^{2}$, so the target
space of $\Phi$ is $\\{0\\}$ and $\Phi$ is surjective.

Now we assume that there is no Kimura tangent point. Pick
$x\in\mathbb{R}^{2}$. We need to find $f\in
C^{1+\gamma}([0,3])\times\mathcal{D}^{\gamma}([0,3])$ such that

$\displaystyle\Lambda
I_{1}f+I_{2}f-x\in\text{ker}(E_{+})+\Lambda\cdot\text{ker}(E_{-}).$

Recall that

$\displaystyle
I_{1}f=E_{+}\Lambda\int_{-1}^{1}Y(s)^{-1}f(s)ds+\int_{1}^{\infty}e^{(s-1)A_{+}}E_{+}f(s)ds,$
$\displaystyle I_{2}f=\int_{-\infty}^{-1}e^{(s+1)A_{-}}E_{-}f(s)ds.$

We first take

$\displaystyle f(t)=0\ \forall\ t<1,\qquad f(t)=-gA_{+}x\ \forall t>1.$

We could choose an appropriate function $g\in
L^{1}([1,\infty)\cap\mathcal{C}_{b}^{0}(\mathbb{R})$ so that

$\displaystyle I_{1}(f)+\Lambda I_{2}f-x=-(I-E_{+})x$

and therefore $\Phi(f)=[x]$. However $f$ would not be continuous. Thus
modifying this idea we set

$\displaystyle f_{k}(t)=0\ \text{for}\ t\leq 1,$ $\displaystyle
f_{k}(t)=k(1-t)gA_{+}x\ \text{for}\ 1\leq t\leq 1+\frac{1}{k},$ $\displaystyle
f_{k}(t)=-gA_{+}x\ \text{for}\ t\geq 1+\frac{1}{k}.$

We compute

$\displaystyle I_{1}(f)+\Lambda I_{2}f-x=-(I-E_{+})x+x_{k}$

where $x_{k}\to 0\ \text{as}\ k\to\infty$. Since the projection map is
continuous we have $\Phi(f_{k})\to[x]$ and since the image of $\Phi$ is closed
we conclude that $[x]\in\text{ran}(\Phi)$. We construct $f$ similarly in all
other cases. ∎

From Lemma 3.7 we have

$\displaystyle\text{codim}\ \text{ran}(\widetilde{M})=\text{dim}\
\text{ker}(\Phi)/\text{ran}(\widetilde{M})+\text{dim\ ran}\ \Phi$
$\displaystyle=|\text{kimura tangent
endpoints}|+2-\text{dim}(\text{ker}(E_{+})+\Lambda\text{ker}(E_{-})).$

so that

$\displaystyle\text{ind}(L_{\gamma})=$
$\displaystyle\text{ind}(M)=\text{ind}(\widetilde{M})$ $\displaystyle=$
$\displaystyle\text{dim}[\text{ker}(E_{-})\cap\Lambda^{-1}\text{ker}(E_{+})]-[2-\text{dim}(\text{ker}(E_{+})$
$\displaystyle+\Lambda\text{ker}(E_{-}))]-|\text{kimura tangent endpoints}|$
$\displaystyle=$ $\displaystyle\text{dim}\ \text{ker}(E_{+})+\text{dim}\
\text{ker}(E_{-})-2-|\text{Kimura tangent endpoints}|.$

From the definitions of $E_{\pm}$ we conclude that

$\displaystyle\text{dim}\ \text{ker}(E_{+})+\text{dim}\
\text{ker}(E_{-})=1+1+|\text{tangent endpoints}|,$

and hence

$\displaystyle\text{ind}(L_{\gamma})=1+1+|\text{tangent
endpoints}|-2-|\text{tangent kimura endpoints}|$ $\displaystyle=|\text{tangent
quadratic endpoints}|\qquad=\ \kappa^{+}+\kappa^{-}.$

##  4 Exponential convergence to invariant measures

In this section we study the convergence rate to the invariant measure in two
different cases. First, when there is at least one tangent boundary, then the
invariant measure is Dirac measure(s) at the tangent boundary(ies) and
quadratic endpoints (if any). To estimate the convergence rate we use Lyapunov
functions constructed in Section 4.1.2. These Lyapunov functions display
different asymptotic behaviors at different boundary endpoints, and turn out
to lie in the function spaces $C(\alpha,\beta)$ (see Definition 4.1). We show
that the growth bound of the generator of $\mathcal{Q}_{t}$ on
$C(\alpha,\beta)$ is not larger than the Lyapunov function rate (58), and thus
negative. From this, we conclude that the transition probability converges
exponentially to the corresponding invariant measure in a Wasserstein
distance. The main results are summarized in Theorem 4.1.

In the second case where both boundary points are transverse, the invariant
measures include a unique invariant measure $\mu$ supported on the whole
domain $[0,1]$ and Dirac measures at quadratic endpoints (if any). Motivated
by underdamped Langevin dynamics, we wish to prove and then apply a Poincaré
inequality in $L^{2}(\mu)$ to show exponential convergence. Using tools
developed in [3], we obtain a (W)-Lyapunov-Poincaré inequality; see Definition
4.2. The main tools we use are the local Lyapunov function, which is Lyapunov
except on a small subset $\mathcal{C}$ and a local Poincaré inequality on
$\mathcal{C}^{c}$; see Definition 127. Finally we apply the (W)-Lyapunov-
Poincaré inequality to show that $\mu$ satisfies the Poincaré inequality and
for any probability measure $v=h\mu$ with $h\in L^{2}(\mu)$,
$\mathcal{Q}^{*}_{t}v$ converges exponentially to $\mu$ in total variation
distance. The main results are summarized in Theorem 4.2.

###  4.1 Case of one/two tangent boundary points

####  4.1.1 Function Space $C(\alpha,\beta)$

When there is at least one tangent boundary point, we consider a function
space isometrically isomorphic to $C^{0}([0,1])$:

######  Definition 4.1.

For $\alpha,\beta\in\mathbb{R}$, we define the function space

$\displaystyle C(\alpha,\beta):=x^{\alpha}(1-x)^{\beta}C^{0}([0,1])$

equipped with norm

$\displaystyle||f||_{\alpha,\beta}:=\left|\left|\frac{f}{x^{\alpha}(1-x)^{\beta}}\right|\right|_{C^{0}}.$

We will specify $\alpha,\beta$ in different cases later (see Sec.4.1.2 for
more detail) but list their ranges as:

$\displaystyle\alpha,\beta=\begin{cases}\begin{array}[]{cc}\in(0,1)&\text{Kimura/quadratic
tangent}\\\ 0&\text{Kimura transverse}\\\ <0&\text{quadratic
transverse.}\end{array}\end{cases}$

######  Remark 4.1.

The solution operator $\mathcal{Q}_{t}$ constructed in Theorem 2.1 is also the
solution operator of the Cauchy problem on $C(\alpha,\beta)$. To see this, the
two local operators $\widetilde{Q}^{0}_{t},\widetilde{Q}^{1}_{t}$ naturally
act on $C(\alpha,\beta)$ and so does the perturbation operator.

We let $A$ be the generator of $\mathcal{Q}_{t}$ on $C(\alpha,\beta)$. We use
the notation $f(x_{0})\sim 0$ in $C(\alpha,\beta)$ if
$\frac{f}{x^{\alpha}(1-x)^{\beta}}(x_{0})=0$ and $f\succ 0(\succeq 0)$ in
$C(\alpha,\beta)$ if $\frac{f}{x^{\alpha}(1-x)^{\beta}}$ is strictly positive
(nonnegative).

######  Lemma 4.1 (Positive Minimum Principle).

For $\lambda\in\mathbb{R}$, the operator $B:=A-\lambda$ satisfies the positive
minimum principle on $C(\alpha,\beta)$, i.e.

$\displaystyle\text{for every $0\preceq f\in D(A)$ and $x\in[0,1]$, $f(x)\sim
0$ implies $(Bf)(x)\succeq 0$}.$

###### Proof.

We say $f\in C^{0}([0,1])$ is in $D^{2}([0,1])$ if $f$ is twice differentiable
up to Kimura endpoint and if at quadratic endpoint if any, say $x=0$, then
$x\partial_{x}f,x^{2}\partial_{x}^{2}f$ has a continuous limit $0$ at $x=0$.
Consider the space

$\displaystyle D([0,1]):=x^{\alpha^{*}}(1-x)^{\beta^{*}}g,\ g\in
D^{2}([0,1]),$ $\displaystyle\text{where}\ \alpha^{*}=\alpha,\beta^{*}=\beta\
\text{if x=0,1 is quadratic and 0 if Kimura}.$

Since $C^{2}([0,1])$ is dense in $C^{0}([0,1]$, then $D^{2}([0,1])\supset
C^{2}([0,1])$ is also dense in $C^{0}([0,1]$, and hence $D([0,1])$ is dense in
$C^{\alpha,\beta}([0,1])$. And for $t>0$, $\mathcal{Q}_{t}(D([0,1]))\subset
D([0,1])$, so by [1, A-I, Proposition 1.9], $D([0,1])$ is a core for $B$. So
it is sufficient to prove the result for $D([0,1])$.

Assume $f\in D([0,1])$ and $f(x_{0})=0$. If $x_{0}$ is an interior point, then
$\partial_{x}f(x_{0})=0,\partial^{2}_{x}f(x_{0})\geq 0$, so $Bf(x_{0})\geq 0$.
If $x_{0}$ is Kimura tangent, since $f$ is twice differentiable at $x_{0}$,
$Bf(x_{0})=0$. If $x_{0}$ is Kimura transverse, in this case,
$\partial_{x}f(x_{0})\geq 0,x\partial_{x}^{2}f(x_{0})=0$, and so
$Bf(x_{0})\geq 0$.

If $x_{0}$ is a quadratic endpoint, locally near $0$, $f=x^{\alpha}g$ for
$g\in D^{2}([0,1])$. As a function $g(x)\geq 0,g(0)=0$ where
$x\partial_{x}g,x^{2}\partial_{x}^{2}g$ have a continuous limit $0$ at $x=0$,
so $\underset{x\to 0}{\lim}\frac{Af}{x^{\alpha}}=0$. Thus, $(Bf)(x_{0})\sim
0$.

∎

To analyze the semigroup $Q_{t}$ generated by $A$, we first recall the
spectral bound $s$ and the growth bound $w_{1}$ for $A$ defined as

$\displaystyle s(A):=\sup\\{\text{Re}\ \lambda:\lambda\in\sigma(A)\\}$ (56)
$\displaystyle w_{1}:=\underset{w\in\mathbb{R}}{\inf}\\{||Q_{t}x||\leq
Me^{wt}||x||_{D(A)},\forall x\in D(A),t\geq 0\text{\ for suitable}\ M\\}$ (57)

and let

$\displaystyle\lambda_{0}:=\underset{\lambda\in\mathbb{R}}{\inf}\
\\{Af\leq\lambda f,\ 0\prec f\\}.$ (58)

$C(\alpha,\beta)$ equipped the norm $||\cdot||_{\alpha,\beta}$ makes it a
Banach lattice as defined in [1]. A first result relating $s(A)$ and $w_{1}$
is that

$s(A)=w_{1}(T)$

for $T$ being a positive $C_{0}$-semigroup on a Banach lattice with generator
$A$ (see [36, Theorem 1.4.1]). The second result is that:

######  Proposition 4.1.

$s(A)\leq\lambda_{0}.$

###### Proof.

Our proof is a modification of the proof of [1, B-II Corollary 1.14]. But
unlike $C^{0}([0,1])$, our underlying space $C(\alpha,\beta)$ present
different boundary behaviors which should be treated carefully. So we provide
a whole proof here.

Given $\lambda>\lambda_{0}$, by definition of $\lambda_{0}$, there exists
$0<u\in C(\alpha,\beta)$ such that $Au\leq\lambda u$. We define a strict half
norm $P_{u}$ on $C(\alpha,\beta)$:

$\displaystyle P_{u}(f)=\underset{x\in[0,1]}{\sup}\frac{f^{+}(x)}{u(x)},\qquad
f^{+}:=\max(f,0).$

Since $u>0$, $P_{u}$ is well defined on $C(\alpha,\beta)$. $P_{u}$ gives rise
to a norm

$\displaystyle||f||_{p}:=P_{u}(f)+P_{u}(-f)$

which is equivalent to the norm on $C(\alpha,\beta)$.

Let $B=\overline{L}-\lambda$, then $Bu\leq 0$. We now show $B$ is
$P_{u}$-dissipative ([1, A-II Definition 2.1]), i.e. for all $f\in D(B)$,
there exists a $\phi_{f}$ in the dual space $C(\alpha,\beta)^{\prime}$ such
that:

  1. 1.

$(Bf,\phi_{f})\leq 0$

  2. 2.

$\forall g\in C(\alpha,\beta)$, $(g,\phi_{f})\leq P_{u}(g)$. Especially
$(f,\phi_{f})=P_{u}(f)$.

Fix $f$. If $f\leq 0$, define $\phi_{f}:=0$. If $f$ is positive at least at
one point, denote by $x_{0}$ a point such that
$P_{u}(f)=\frac{f^{+}(x_{0})}{u(x_{0})}$. Now consider $\phi_{f}\in
C(\alpha,\beta)^{\prime}$ such that

$\displaystyle(g,\phi_{f})=\frac{g(x_{0})}{u(x_{0})}.$

Clearly such $\phi_{f}$ satisfies the second condition. Next we check the
first condition. Let $f\in D(B)$, if $P_{u}(f)=0$, then $f\leq 0$, so
$(Bf,\phi_{f})=(Bf,0)=0$. If $P_{u}(f)>0$, since
$P_{u}(f)=\frac{f(x_{0})}{u(x_{0})}\geq\frac{f(x)}{u(x)}$, then

$\displaystyle P_{u}(f)\cdot u-f\geq 0,\ (P_{u}(f)\cdot u-f)(x_{0})=0.$ (59)

By the positive minimum principle Lemma 4.1,

$\displaystyle Bf(x_{0})\leq P_{u}(f)(Bu)(x_{0})\leq 0$

i.e. $(Bf,\phi_{f})\leq 0$.

Let $P_{t}$ be the semigroup generated by $B$. We show that $P_{t}$ is
$P_{u}$-contraction, i.e. $P_{u}(P_{t}f)\leq P_{u}(f),\forall f\in D(B)$. Let
$f\in D(B),\ t>0$, then

$\displaystyle
P_{u}(f)=(f,\phi_{f})=(f-tBf+tBf,\phi_{f})\leq(f-tBf,\phi_{f})\leq
P_{u}(f-tBf).$

Thus for $\lambda>w_{1}$, $f\in C(\alpha,\beta)$,
$(\lambda-B)R(\lambda,B)f=f$, so $P_{u}(\lambda R(\lambda,B)f)\leq P_{u}(f)$,
then by the formula $P_{t}f=\underset{n\to\infty}{\lim}\left(\frac{n}{t}\cdot
R(\frac{n}{t},B)\right)^{n}f$, we have

$\displaystyle P_{u}(P_{t}f)\leq P_{u}(f).$

Thus the closure $\overline{B}$ generates a $P_{u}$-contraction semigroup,
hence the closure $\overline{A}=\overline{B}+\lambda$ of $A$ generates a
positive semigroup of type $w_{1}(\bar{A})\leq\lambda$. Hence we showed that
$\lambda_{0}\geq w_{1}(A)=s(A).$ ∎

####  4.1.2 Estimation of $\lambda_{0}$ and rate of convergence

In this part, we show that $\lambda_{0}<0$, which implies the exponential
convergence to $\delta$ measures at tangent boundaries. In order to do this,
we construct $u$ that satisfies $Lu<\lambda u$ for some $\lambda<0$. The
strategy of construction is to first use the behavior of $L$ in the vicinity
of two the boundary points to construct $u$ near the boundaries and then find
a connecting interior function.

##### Boundary Construction

$\mathit{1.\ Quadratic\ endpoint}$. In the vicinity of quadratic point,
$L=x^{2}\partial_{xx}+(b(x)+1)x\partial_{x}$ and recall ${b_{-}}=b(0)$. Let

$\displaystyle u=Ax^{c},A>0.$

Taking $c=\frac{-{b_{-}}}{2}$,

$\displaystyle\frac{Lu}{u}=(c+b(x))c,$

is negative in a neighborhood of 0.

$\mathit{2.Kimura\ endpoint}.$ In the vicinity of a Kimura point,
$L=x\partial_{xx}+b(x)\partial_{x}$ with $b=b(0)$. Let

$\displaystyle\mathit{u=\begin{cases}\begin{array}[]{cc}Ax^{c},\ 0<c<1&b=0\\\
A(1-x)^{c},\ c>0&b>0\end{array}\end{cases}}.$

In both cases

$\displaystyle\frac{Lu}{u}=(c-1+b(x))cx^{-1}$
$\displaystyle\frac{Lu}{u}=c(1-x)^{-2}[(c-1)x-b(x)(1-x)]$

are negative in a neighborhood of 0.

##### Interior Construction

Suppose we have constructed $u$ in $[0,x_{1}]$ and $[x_{2},1]$. We now need to
construct an interior function $u$ that satisfies the boundary conditions.
Inside $[x_{1},x_{2}]$, suppose $L$ has the form

$\displaystyle L=\partial_{xx}+b(x)\partial_{x},$

where $b(x)$ is smooth on $[x_{1},x_{2}]$. Let

$\displaystyle B(x)=\int_{x_{1}}^{x}b(t)dt,$ (60)

then $B$ is smooth on $[x_{1},x_{2}]$. We want $Lu$ strictly negative, that is

$\displaystyle u_{xx}+b(x)u_{x}<0.$

It is equivalent to $e^{B(x)}u_{x}$ being a decreasing function, i.e. there
exists some positive function $f$ on $[x_{1},x_{2}]$, such that

$\displaystyle(e^{B(x)}u_{x})^{\prime}=-f.$ (61)

##### Case I: one tangent, one transverse points

Without loss of generality, we assume a tangent boundary at $x=0$. Now by
construction of $u$ at both boundaries, $u_{x}(x_{1})>0$, $u_{x}(x_{2})>0$. We
first modify the constant $A$ in the construction near $x=1$ such that

$\displaystyle e^{B(x_{2})}u_{x}(x_{2})<e^{B(x_{1})}u_{x}(x_{1}).$

Next, by (61), $f(x_{1})=-e^{B(x_{1})}Lu(x_{1})>0,\
f(x_{2})=-e^{B(x_{2})}Lu(x_{2})>0$ are two positive fixed constants. Then we
can set $f$ to be a positive function with $f(x_{1}),\ f(x_{2})$ fixed and

$\displaystyle\int_{x_{1}}^{x_{2}}f(t)dt=e^{B(x_{1})}u_{x}(x_{1})-e^{B(x_{2})}u_{x}(x_{2}).$

Such an $f$ guarantees that $e^{B(x)}u_{x}(x)>e^{B(x_{2})}u_{x}(x_{2})>0$ for
$x\in[x_{1},x_{2}]$ which guarantees positiveness of $u_{x}$ and hence
positiveness of $u$.

##### Case II: two tangent boundaries

By construction of $u$ at both boundaries, $u_{x}(x_{1})>0,\ u_{x}(x_{2})<0$.
We directly get $e^{B(x_{2})}u_{x}(x_{2})<e^{B(x_{1})}u_{x}(x_{1})$.

Denoting $F(x)=\int_{x_{1}}^{x}f(t)dt$, then $F(x)$ is a differentiable
increasing function on $[x_{1},x_{2}]$ with $F(x_{1})=0$. Then by considering
derivatives of $u$

$\displaystyle u_{x}(x)=e^{-B(x)}\left[e^{B(x_{1})}u_{x}(x_{1})-F(x)\right]$

we know $u$ is first increasing then decreasing on $[x_{1},x_{2}]$, which
implies $u$ is lower bounded by $\min(u(x_{1}),u(x_{2}))$. Hence we only need
to assign a positive $f$ such that

$\displaystyle\int_{x_{1}}^{x_{2}}f(t)dt=e^{B(x_{1})}u_{x}(x_{1})-e^{B(x_{2})}u_{x}(x_{2}).$

######  Theorem 4.1.

If there is only one tangent endpoint $p$, then for any non-quadratic point
$x$, the transition probability $p_{t}(x,\cdot)$ converges exponentially to
$\delta(p)$ in Wasserstein distance.

If there are two tangent endpoints, then there exists $S_{0}$, satisfying
$S_{0}(0)=0$, $S_{0}(1)=1$ and $LS_{0}=0$, such that for any probability
measure $v$, $\mathcal{Q}^{*}_{t}v$ converges exponentially to
$\delta(0)\int_{0}^{1}v(1-S_{0})+\delta(1)\int_{0}^{1}vS_{0}$ in Wasserstein
distance.

###### Proof.

If there is only one tangent endpoint, say $x=0$, then for any $f\in
C^{0}([0,1])$ with $\text{Lip}(f)\leq 1$, $f$ can be decomposed as

$\displaystyle f=f_{0}+f(0),f_{0}\in C(\alpha,\beta).$

Then by Proposition 4.1,

$\displaystyle||\mathcal{Q}_{t}f-f(0)||_{C(\alpha,\beta)}=||\mathcal{Q}_{t}f_{0}||_{C(\alpha,\beta)}\leq
Me^{\lambda_{0}t}||f_{0}||_{C(\alpha,\beta)}$

with some constant $M>0$ independent of $f$. Since
$f_{0}(0)=0,\text{Lip}(f_{0})\leq 1$ and $0<\alpha<1,\beta\leq 0$, then
$||f_{0}||_{C(\alpha,\beta)}\leq C_{\alpha,\beta}$ for some constant
$C_{\alpha,\beta}>0$ only dependent on $\alpha,\beta$. Hence

$\displaystyle|\mathcal{Q}_{t}f(x)-f(0)|\leq
MC_{\alpha,\beta}e^{\lambda_{0}t}x^{\alpha}(1-x)^{\beta}.$ (62)

Notice that only when $x=1$ is a quadratic endpoint do we have $\beta<0$.
Hence for any non-quadratic point $x$, the transition probability
$q_{t}(x,\cdot)$ converges to $\delta(0)$ in the sense of Wasserstein distance
at an exponential rate. This convergence is uniform on any compact interval
away from quadratic transverse point. If $x=1$ is a quadratic point, then
$q_{t}(x,\cdot)=\delta(1)$ for $\forall t>0$.

If both endpoints are tangent, by Theorem 3.2, the kernel of $L$ is in the
linear space of $\\{1,S\\}$. Then we can take $S_{0}$ as a linear combination
of $1$ and $S$, such that $S_{0}(0)=0$ and $S_{0}(1)=1$. For any $f\in
C^{0}([0,1])$ with $\text{Lip}(f)\leq 1$, we decompose this as

$\displaystyle f=f_{0}+(1-S_{0})f(0)+S_{0}f(1),\quad f_{0}\in
C(\alpha,\beta).$

Again, by Proposition 4.1,

$\displaystyle||\mathcal{Q}_{t}f-(1-S_{0})f(0)-S_{0}f(1)||_{C(\alpha,\beta)}=||\mathcal{Q}_{t}f_{0}||_{C(\alpha,\beta)}\leq
Me^{\lambda_{0}t}||f_{0}||_{C(\alpha,\beta)}$

with $M>0$ independent of $f$. Since $0<\alpha,\beta<1,f_{0}(0)=f_{0}(1)=0$
and $\text{Lip}(f_{0})\leq 1$, then

$\displaystyle||\mathcal{Q}_{t}f-(1-S_{0})f(0)-S_{0}f(1)||_{0}\leq
M_{\alpha,\beta}||\mathcal{Q}_{t}f-(1-S_{0})f(0)-S_{0}f(1)||_{C(\alpha,\beta)},$
$\displaystyle||f_{0}||_{C(\alpha,\beta)}\leq C_{\alpha,\beta}$

for some constant $M_{\alpha,\beta},C_{\alpha,\beta}>0$ only dependent on
$\alpha,\beta$. Hence,

$\displaystyle||\mathcal{Q}_{t}f-(1-S_{0})f(0)-S_{0}f(1)||_{0}\leq
MC_{\alpha,\beta}M_{\alpha,\beta}e^{\lambda_{0}t},$

so that the transition probability $q_{t}(x,\cdot)$ converges uniformly on
$[0,1]$ to $(1-S_{0}(x))\delta(0)+S_{0}(x)\delta(1)$ at an exponential rate in
Wasserstein distance. Hence starting from any probability measure $v$,
$\mathcal{Q}_{t}^{*}v$ converges exponentially to $\delta(0)\int
v(1-S_{0})+\delta(1)\int vS_{0}$ in Wasserstein distance. ∎

######  Remark 4.2.

Consider as a the first example the case with one tangent boundary point, say
$x=0$, quadratic tangent while $x=1$ is Kimura transverse. For instance,

$\displaystyle L=x^{2}(1-x)\partial_{xx}+bx(a-x)\partial_{x}\ \text{on
[0,1]},\ ab<1,\ b(a-1)<0.$

Consider $f=x^{c}$. If we choose $0<c<1-ab$, then

$\displaystyle\frac{Lf}{f}=\frac{Lx^{c}}{x^{c}}=\left[(1-x)(c-1)+b(a-x)\right]c\leq
max(c-1+ab,b(a-1))c<0.$

So

$\displaystyle\lambda_{0}\leq\underset{0<c<1-ab}{\min}\max(c-1+ab,b(a-1))c<0.$

As a second example, consider two tangent boundary points , for instance

$\displaystyle L=x^{2}(1-x)\partial_{xx}\ \text{on}\ [0,1].$

In this case, we should expect
$p(t,x,\cdot)\to(1-S_{0})\delta_{0}+S_{0}\delta_{1}$, where $S_{0}$ in this
case is $x$. By setting $f=x^{c}(1-x),\ 0<c<1$, then

$\displaystyle Lf=(c(c-1)-c(c+1)x)x^{c}(1-x)\leq\lambda f.$

Here, $\lambda=c(c-1)<0$. So

$\displaystyle\lambda_{0}\leq\underset{0<c<1}{\min}c(c-1)=-\frac{1}{4}.$

###  4.2 Case with two transverse boundaries

Recall that we have constructed the invariant measure $\mu$ when both
endpoints are transverse in Section 3.2. We first introduce a (W)-Lyapunov-
Poincaré inequality that will be used to analyze the long time behavior of the
transition probability.

####  4.2.1 (W)-Lyapunov-Poincaré inequality

Consider $U(z)=\int_{0}^{z}b(s)ds$ with $b(s)$ negative when $s\to-\infty$ and
positive when $s\to\infty$. Then $U$ grows linearly to $+\infty$ near
$z=\pm\infty$. $L$ can be taken as

$\displaystyle L_{z}=\frac{1}{2}\partial_{zz}-\nabla
U(z)\partial_{z}=-\frac{1}{2}\partial_{z}^{*}\partial_{z}$

with $\partial^{*}_{z}=\partial_{z}+2\nabla U(z)$ on a probability space
$(X,\mu)$, with $\mu$ the invariant measure

$\displaystyle\mu(dz)=\frac{e^{-2U(z)}}{Z}dz$ (63)

where the normalizing constant $Z<\infty$ due to the linear growth of $U$ .
The main advantage of switching to $L^{2}(\mu)$ is $L$ is self adjoint on
$L^{2}(\mu)$. Indeed

$\displaystyle\int_{\mathbb{R}}(\partial_{z}\varphi)\phi
d\mu=\int_{\mathbb{R}}\varphi(-\partial_{z}\phi+2\phi\nabla
U(z))d\mu=\int_{\mathbb{R}}\varphi\partial^{*}_{z}\phi d\mu$

so that

$\displaystyle(\varphi,L\phi)_{L^{2}(\mu)}=(L\varphi,\phi)_{L^{2}(\mu)}.$

For some function $W\in D(L)$ with $W\geq 1$, let $I_{W}(t)=\int
Q^{2}_{t}fWd\mu$. Then [3]

$\displaystyle I^{\prime}_{W}(t)=-\int W|\nabla(Q_{t}f)|^{2}d\mu+\int
WLQ^{2}_{t}fd\mu.$ (64)

This leads to the definition (W)-Lyapunov-Poincaré inequality:

######  Definition 4.2.

We say that $\mu$ satisfies a (W)-Lyapunov-Poincaré inequality, if there
exists $W\in D(L),W\geq 1$ and a constant $C_{LP}$ such that for all nice $f$
with $\int fd\mu=0$,

$\displaystyle\int f^{2}Wd\mu\leq C_{LP}\int\left(W|\nabla
f|^{2}-WLf^{2}\right)d\mu.$ (65)

##### Local Poincaré inequality

In the transverse case, we can not find a global Lyapunov function. We have to
weaken the condition to define a local Lyapunov function $V$ as

$\displaystyle LV\leq-\alpha V+\beta\mathbf{1}_{\mathcal{C}},\ V\geq 1$ (66)

for some set $\mathcal{C}$.

######  Definition 4.3 (Local Poincaré inequality).

Let ${\Omega}$ be a subset of the whole space $X$ (we will use
$\mathcal{C}\subset{\Omega}$). We say that $\mu$ satisfies a local Poincaré
inequality on ${\Omega}$ if there exists some constant $\kappa_{\Omega}$ such
that for all nice $f$ with $\int_{X}fd\mu=0$,

$\displaystyle\int_{\Omega}f^{2}d\mu\leq\kappa_{U}\int_{X}|\nabla
f|^{2}d\mu+\frac{1}{\mu({\Omega})}\left(\int_{\Omega}fd\mu\right)^{2}.$ (67)

We now show the:

######  Proposition 4.2.

Assume that there is some local Lyapunov function $V\geq 1$ under some set
$\mathcal{C}$ such that (66) holds, and $\mu$ satisfies a local Poincaré
inequality on ${\Omega}\supset\mathcal{C}$ with moreover

$\displaystyle\beta\mu({\Omega}^{c})<\alpha\mu({\Omega}).$ (68)

Then we can find $\lambda\geq 0$ such that if $W=V+\lambda$, then $\mu$
satisfies a (W)-Lyapunov-Poincaré inequality.

###### Proof.

Multiply (66) by $f^{2}$ and integrate to get

$\displaystyle\int_{X}f^{2}LVd\mu\leq-\alpha\int_{X}f^{2}Vd\mu+\beta\int_{C}f^{2}d\mu.$

Let $\int_{X}fd\mu=0$. The local Poincaré inequality implies

$\displaystyle\int_{\Omega}f^{2}d\mu$
$\displaystyle\leq\kappa_{\Omega}\int_{X}|\nabla
f|^{2}d\mu+\frac{1}{\mu({\Omega})}\left(\int_{\Omega}fd\mu\right)^{2}$
$\displaystyle\leq\kappa_{\Omega}\int_{X}|\nabla
f|^{2}d\mu+\frac{1}{\mu({\Omega})}\left(-\int_{{\Omega}^{c}}fd\mu\right)^{2}$
$\displaystyle\leq\kappa_{\Omega}\int_{X}|\nabla
f|^{2}d\mu+\frac{\mu({\Omega}^{c})}{\mu({\Omega})}\left(\int_{{\Omega}^{c}}f^{2}d\mu\right),$

so that

$\displaystyle\int_{X}f^{2}\left(LV+\alpha V\right)d\mu$
$\displaystyle\leq\beta\int_{C}f^{2}d\mu\leq\beta\int_{\Omega}f^{2}d\mu$
$\displaystyle\leq\beta\kappa_{U}\int_{X}|\nabla
f|^{2}d\mu+\frac{\beta\mu({\Omega}^{c})}{\mu({\Omega})}\left(\int_{{\Omega}^{c}}f^{2}Vd\mu\right).$

We re-organize the inequality as,

$\displaystyle\alpha\left(1-\frac{\beta\mu({\Omega}^{c})}{\alpha\mu({\Omega})}\right)\int_{X}f^{2}Vd\mu$
$\displaystyle\leq\beta\kappa_{\Omega}\int_{X}|\nabla
f|^{2}d\mu-\int_{X}f^{2}LVd\mu$ $\displaystyle\leq\int_{X}|\nabla
f|^{2}(V+\lambda)d\mu-\int_{X}f^{2}L(V+\lambda)d\mu$
$\displaystyle=\int_{X}|\nabla
f|^{2}(V+\lambda)d\mu-\int_{X}Lf^{2}(V+\lambda)d\mu.$

Here we take$\ \lambda=(\beta\kappa_{U}-1)_{+}$ so that
$\beta\kappa_{\Omega}\leq V+\lambda$. Since,

$\displaystyle\int_{X}f^{2}(V+\lambda)d\mu\leq(1+\lambda)\int_{X}f^{2}Vd\mu,$

then

$\displaystyle\alpha\left(1-\frac{\beta\mu({\Omega}^{c})}{\alpha\mu({\Omega})}\right)\frac{1}{1+\lambda}\int_{X}f^{2}(V+\lambda)d\mu\leq\int_{X}\left(|\nabla
f|^{2}(V+\lambda)-Lf^{2}(V+\lambda)\right)d\mu$

and
$\frac{1}{C_{LP}}=\alpha\left(1-\frac{\beta\mu({\Omega}^{c})}{\alpha\mu({\Omega})}\right)/(1+\lambda)$.
∎

####  4.2.2 Long time behaviors

Let us come back to the operator $L=\frac{1}{2}\Delta-\nabla U\cdot\nabla$ on
$(X,\mu)$. We first construct the local Lyapunov function $V$ with
$\mathcal{C}$ satisfying (66).

######  Lemma 4.2.

There exists a local Lyapunov function.

###### Proof.

Let $X=[x_{l},x_{r}],\ -\infty\leq x_{l}<x_{r}\leq\infty$ and

$\displaystyle V=\begin{cases}\begin{array}[]{cc}e^{U}&x_{l}=-\infty\\\
2-(x(z)-x_{l})&x_{l}>-\infty.\end{array}\end{cases}$

We define $V$ similarly in the vicinity of $x_{r}$. We check that
$LV\leq-\alpha V,\ V>1$ for some $\alpha>0$ in a neighborhood $x_{l}$:

  1. 1.

When $x_{l}=-\infty$, for $V=e^{U}$,

$\displaystyle LV=V\Big{(}\frac{1}{2}\Delta U-\frac{1}{2}|\nabla
U|^{2}\Big{)}.$

By the transverse condition, $\underset{z\to-\infty}{\lim}\nabla U=-b_{-}<0$,
which implies $\underset{z\to\infty}{\lim}\Delta U=0$. So

$\displaystyle\underset{z\to-\infty}{\lim}\Big{(}\frac{1}{2}\Delta
U-\frac{1}{2}|\nabla U|^{2}\Big{)}=-\frac{1}{2}b_{-}^{2}<0.$

Since $\underset{z\to-\infty}{lim}U(z)=\infty$, for
$0<\alpha<\frac{b_{-}^{2}}{2}$, we have $LV\leq-\alpha V,\ V\geq 1$ in a
neighborhood of $x_{l}$.

  2. 2.

When $x_{l}>-\infty$, for $V=2-(x(z)-x_{l})$,

$\displaystyle LV=L_{x}(2-x)=-b(x).$

By the transverse condition, $b_{-}>0$, so $LV(0)<0$ and since $V(0)=2>0$,
$LV\leq-\alpha V,\ V\geq 1$ for some $\alpha>0$ in a neighborhood of $x_{l}$.

In all cases, $LV\leq-\alpha V,\ V>1$ for some $\alpha>0$ in a neighborhood
$x_{l},x_{r}$.

After constructing $V$ in a neighborhood of $x_{l}$ and $x_{r}$, we extend it
to a complement $\mathcal{C}$ of these two neighborhoods as a twice
differentiable function with $V\geq 1$. Since $\mathcal{C}$ is compact, $LV,V$
are bounded on $\mathcal{C}$, and we can find some $\beta>0$ large enough that
$V$ is a Lyapunov function for $\mathcal{C}$:

$\displaystyle LV\leq-\alpha V+\beta\mathbf{1}_{\mathcal{C}}.$

∎

Next we verify that $\mu$ satisfies a local Poincaré inequality (4.3). Indeed
by [32, Theorem 1], if $\mu$ is bounded and restricted to a Euclidean ball of
radius $R$, we know that it satisfies the Poincaré inequality, hence the local
Poincaré inequality on it.

We now state our main result of the section:

######  Theorem 4.2.

The invariant measure $\mu$ in (63) satisfies a Poincaré inequality. For $f\in
L^{2}(\mu)$,

$\displaystyle||\mathcal{Q}_{t}f-\int f\mu||_{L^{2}(\mu)}\leq
e^{-\frac{t}{2C_{LP}}}||f-\int f\mu||_{L^{2}(\mu)}.$ (69)

For any probability measure $v=h\mu$ with $h\in L^{2}(\mu)$,

$\displaystyle||\mathcal{Q}_{t}^{*}v-\mu||_{\text{TV}}\leq
e^{-\frac{t}{2C_{LP}}}||h-1||_{L^{2}(\mu)}.$

###### Proof.

By construction, $\mathcal{C}$ is a compact set in the interior of $X$. We
choose a larger set $U$ with $\mathcal{C}\subset U\subset int(X)$ such that
$\beta\mu(U^{c})<\alpha\mu(U)$. $\mu$ is bounded on $U$, so $\mu$ satisfies a
local Poincaré inequality on $U$. Applying Proposition 4.2, $\mu$ satisfies a
(W)-Lyapunov-Poincaré inequality. Thus by (64), for all $f$ such that $\int
f^{2}W\mu<\infty$ and $\int f\mu=0$,

$\displaystyle\int(\mathcal{Q}_{t}f)^{2}W\mu\leq e^{-\frac{t}{C_{LP}}}\int
f^{2}W\mu.$

Since $W\in L^{1}(\mu)$, by [3, Corollary 3.4], $\mu$ satisfies a Poincaré
inequality with Poincaré constant equal to $C_{LP}$. Hence (69) is derived by
Poincaré inequality.

For the second assertion, we use the symmetry property
$\mathcal{Q}_{t}^{*}(h\mu)=(\mathcal{Q}_{t}h)\mu$ and (69) to derive that

$\displaystyle||\mathcal{Q}^{*}_{t}v-\mu||_{\text{TV}}=||\mathcal{Q}_{t}h-1||_{L^{1}(\mu)}\leq||\mathcal{Q}_{t}h-1||_{L^{2}(\mu)}\leq
e^{-\frac{t}{2C_{LP}}}||h-1||_{L^{2}(\mu)}.$

∎

######  Remark 4.3.

Regarding logarithmic Sobolev inequalities, which are stronger than Poincaré
inequalities, we may use a criterion on a ball of radius $R$ ([2, Proposition
2.6]) to obtain when both of the endpoints are of Kimura type, that $\mu$
satisfies a logarithmic Sobolev inequality.

##  5 Long time behaviour in two dimension

In this section we analyze the long time behavior of transition probability on
a two-dimensional manifold with corners. We refer to [10] for an introduction
to 2-dimensional manifold with corners in our context and [22] for a more
complete introduction.

Let $P$ be a paracompact Hausdorff topological space. A chart of $p\in P$ is a
pair $(\mathcal{U}_{p},\psi_{p})$ where $\mathcal{U}_{p}$ is a neighborhood of
$p$ and $\psi_{p}$ is a homeomorphism with $\psi_{p}(p)=0$ from
$\mathcal{U}_{p}$ to a neighborhood of $0$ in
$\mathbb{R}^{l}_{+}\times\mathbb{R}^{2-l}$ for some $l\in\\{0,1,2\\}$. Two
charts $(\mathcal{U}_{p},\psi_{p}),(\mathcal{U}_{q},\psi_{q})$ are said to be
compatible if

$\psi_{p}\circ\psi_{q}^{-1}:\psi_{q}(\mathcal{U}_{p}\cap\mathcal{U}_{q})\longrightarrow\psi_{p}(\mathcal{U}_{p}\cap\mathcal{U}_{q})$

is a diffeomorphism. The codimension $l$ is well defined for $p\in P$. We say
that the point $p$ is an interior point if $l=0$, an edge point if $l=1$, a
corner if $l=2$. A two dimensional manifold with corners is paracompact
Hausdorff topological space equipped with a maximal compatible atlas.

###  5.1 Lyapunov function

We make the following assumption:

######  Assumption 5.1.

For $L$ on a 2 dimensional compact manifold with corners $P$, there is exactly
one tangent edge $H$, and when restricted to $H$, $L|_{H}$ is transverse to
both boundary points.

We say that $V$ is a Lyapunov function if $V$ is strictly positive except on
$H$, $V|_{H}\equiv 0$ and for some $\lambda_{0}<0$,

$\displaystyle LV\leq\lambda_{0}V.$ (70)

We make the following assumption:

######  Assumption 5.2.

There is a twice differentiable function $\rho(p)$ such that

  1. a.

$\rho>0$ except on $H$, $\rho|_{H}\equiv 0$;

  2. b.

$\nabla\rho\neq 0$ nowhere vanishing;

  3. c.

For $p$ in an edge $E$, if $\nabla_{E}\rho(p)=0$, then $\rho\equiv c$ in a
relative neighborhood of $p$ along $E$;

  4. d.

There is no local minimum point of $\rho$ other than on $H$.

The third assumption ensures that $\rho$ is constant or strictly monotonic
along any edge $E$. In the presence of such $\rho$, there exists a
stratification of $P$ so that:

  1. 1.

$P$ is covered by the layers:

$\displaystyle
P=\underset{i=0}{\overset{n}{\bigcup}}P_{[k_{i},k_{i+1}]}:=\underset{i=0}{\overset{n}{\bigcup}}\\{p\in
P:\rho(p)\in[k_{i},k_{i+1}]\\},$ (71) $\displaystyle
0=k_{0}<k_{1}<\cdot\cdot\cdot k_{n}=\underset{P}{\max}\ \rho.$

  2. 2.

Each layer $P_{[k_{i},k_{i+1}]}$ is covered by a finite collection of closed
sets ${U}_{i,p_{j}}$ , i.e.

$\displaystyle P_{[k_{i},k_{i+1}]}=\bigcup_{j}U_{i,p_{j}},$ (72)

where $p_{j}$ is a point in $U_{i,p_{j}}$ and the range of $\rho$ in each
$U_{i,p_{j}}$ is $[k_{i},k_{i+1}]$.

We leave the construction of the stratification to Lemma A.2 in the Appendix.
We then have the following result:

######  Theorem 5.1.

There exists a Lyapunov function $V$ on $P$.

###### Proof.

We first analyze the local behavior of $\rho,L$. Note that for

$\displaystyle
L=A(x,y)\partial_{xx}+B(x,y)\partial_{xy}+C(x,y)\partial_{yy}+D(x,y)\partial_{x}+E(x,y)\partial_{y},$

under the new coordinates $(\eta,\rho)$, the ordinary differential $\rho$-part
$L_{\rho}$ is

$\displaystyle
L_{\rho}=(A\rho_{xx}+B\rho_{xy}+C\rho_{yy})\partial_{\rho\rho}+(L\rho)\partial_{\rho}.$
(73)

Given any point $p\in P$, we discuss the local property of $L$ at point $p$
subject to the following cases.

##### 1\. $p$ is an interior point

There exists a closed set including $p$ equipped with coordinates $(x,\rho)$
that is diffeomorhic to the rectangle
$R:=[0,1]\times[\rho(p),\rho(p)+\epsilon]$. In this case, $L_{\rho}$ is
elliptic of the form

$\displaystyle
L_{\rho}=a(x,\rho)[\partial_{\rho\rho}+b(x,\rho)\partial_{\rho}],(x,\rho)\in
R,$ (74)

where $a(x,\rho)>0$.

##### 2\. $p$ is an edge point

Under local adapted coordinates $(x,y)$,

  1. a.

if $\rho\equiv\rho(p)$ along the edge $y=0$, then $(x,\rho)$ are local adapted
coordinates. There is a neighborhood of $p$ that is diffeomorphic to the
rectangle

$R:=[0,1]\times[\rho(p)-\epsilon,\rho(p)].$

$L_{\rho}$ is degenerate at $\rho(p)$ of the form: for $(x,\rho)\in R$,

$\displaystyle\textbf{Kimura}:L_{\rho}=a(x,\rho)\left[(\rho(p)-\rho)\partial^{2}_{\rho}+c(x,\rho)\partial_{\rho}\right]$
(75)
$\displaystyle\textbf{quadratic}:L_{\rho}=a(x,\rho)\left[(\rho-\rho(p))^{2}\partial_{\rho}^{2}+d(x,\rho)(\rho(p)-\rho)\partial_{\rho}\right]$
(76)

where $a(x,\rho)>0$.

  2. b.

If the gradient of $\rho$ along the edge $\nabla_{x}\rho(p)\neq 0$, then
$(y,\rho)$ are local adapted coordinates. There is a neighborhood of $p$ that
is diffeomorphic to $R:=[0,1]\times[\rho(p),\rho(p)+\epsilon]$ for some
$\epsilon>0$. In this case $L_{\rho}$ is elliptic of the same form with (74)
in $R$.

##### 3\. $p$ is a corner

Under local adapted coordinates $(x,y)$,

  1. a.

if along one (e.g. $x$) edge $\nabla_{x}\rho(p)=0$, then $\rho\equiv\rho(p)$
in a relative open neighborhood of $p$ on the $x$-edge. Since
$\nabla_{y}\rho(p)\neq 0$, then by rescaling $x$, $(x,\rho)$ are also local
adapted coordinates at $p$; this case is then the same as in (75), (76). See
the first picture in Figure 1.

  2. b.

If $\nabla_{x}\rho(p),\nabla_{y}\rho(p)\neq 0$, then $(x,\rho)$ are not
adapted coordinates. In this case, there exists a neighborhood of $p$ that is
diffeomorphic to an irregular area inside a rectangle

$T:=\\{(x,\rho):\rho\geq\rho(x)|_{y=0}\\}\subset[0,1]\times[\rho(p)-\epsilon_{1},\rho(p)+\epsilon_{2}].$

See the second and third pictures in Figure 1. $L_{\rho}$ is then degenerate
when $x=0$ and elliptic when $x\neq 0$.

If $p$ is neither a local maximum nor minimum, $T$ is made up of a rectangle
and an irregular area $T_{1}$, we can extend the coefficients of $L_{\rho}$ to
the rectangle supplemented by the dashed lines. Then in these two rectangles,
$L_{\rho}$ are both degenerate when $x=0$ and elliptic when $x\neq 0$.

Figure 1: $L_{\rho}$ at a corner

We now start the construction of $V$ iteratively in each layer. Such $V$
generally depends only on the variable $\rho$, i.e., $V(p)=V(\rho(p))$, except
near the corners.

##### Case $i=0$

$P_{[0,k_{1}]}$ is covered by finite rectangles $R:=[0,1]\times[0,k_{1}]$
where $L_{\rho}$ takes the form

$\displaystyle\textbf{Kimura}:L_{\rho}=a(x,\rho)\left(\rho\partial_{\rho}^{2}+b(x,\rho)\partial_{\rho}\right)$
(77)
$\displaystyle\textbf{quadratic}:L_{\rho}=a(x,\rho)(\rho^{2}\partial_{\rho}^{2}+c(x,\rho)\rho\partial_{\rho})$
(78)

where $a(x,\rho)>0,b(x,\rho)<1,c(x,\rho)<1$ since $H$ is a tangent edge. We
take $V=\rho^{\nu}$, where $\nu$ depends on the considered case. In case (77),
we choose $0<\nu<\underset{(x,\rho)}{\min}(1-b(x,\rho))$ so that
$L_{\rho}\rho^{\nu}=a\nu(\nu+b-1)\rho^{\nu-1}\leq-\lambda\rho^{\nu}$ for some
$\lambda>0$. In case (78), we choose
$0<\nu<\underset{(x,\rho)}{\min}(1-c(x,\rho))$ so that
$L_{\rho}\rho^{\nu}=a\nu(\nu+c-1)\rho^{\nu}\leq-\lambda\rho^{\nu}$ for some
$\lambda>0$.

##### Case $i\geq 1$

$P_{[k_{i},k_{i+1}]}$ is then covered by four kinds of disjoint collections of
open set $R$ as we analyzed before, namely:

  1. I.

$R$ does not contain any corner or local maximum point, $L_{\rho}$ is elliptic
for $\forall x$ of the form (74);

  2. II.

$R$ contains an edge and $\rho$ is constant along this edge, so $L_{\rho}$
takes the form (75), (76);

  3. III.

$R$ contains a corner that is a strict local maximum point of $\rho$,

  4. IV.

$R$ contains a corner that is neither a local maximum point nor local minimum
point of $\rho$.

At this layer $P_{[k_{i},k_{i+1}]}$, $R_{II}$ and $R_{III}$ are both
disconnected from the other three types. $R_{I}$ may only be connected with
$R_{IV}$. There maybe more than one $R$ of type II, but since $L_{\rho}$ takes
a uniform form in these $R_{II}$, in our construction below we may regard
these $R_{II}$ as a uniform $R$ and for convenience of notation we still let
$x\in[0,1]$. $R_{III}$ will be treated in this way as well. For $R_{I}$, we
can merge these $R_{I}$ as a uniform one as well. If there exists $R_{IV}$, we
construct $V$ separately in $R_{IV}$ so that it may be patched with $V$ in
$R_{I}$.

Thus, it suffices to construct $V$ in these four cases separately. Since
$k_{i}>0$ when $i\geq 1$, we assume that $\rho(p)=1$.

##### Case I:

Fix an arbitrary $x_{0}$, we choose $f(\rho)$ so that

$\displaystyle f^{\prime}(\rho)<b(x_{0},\rho)-b(x,\rho).$

Starting from $\rho=1$, we let, $V_{\rho}=\exp(f(\rho)-\int^{\rho}b(x,z)dz)$,
then
$\frac{L_{\rho}V_{\rho}}{V_{\rho}}=[f^{\prime}-b(x_{0},\rho)+b(x,\rho)]<0$.

##### Case II:

In the Kimura case (75), we fix $x_{0}$ so that
$c(x_{0},1)=\underset{x\in[0,1]}{\max}\ c(x,1)$. We choose $f(\rho)$ so that

$\displaystyle
f^{\prime}(\rho)<\frac{c+c(x_{0},\rho)-c(x,\rho)}{1-\rho},c=-c(x_{0},1)>0$

on $[1-\epsilon,1]$. Denote
$H^{x}(\rho)=\text{exp}\left[\int^{\rho}\frac{c(x,z)}{1-z}dz\right]\sim(1-\rho)^{-c(x,1)}$,
and let

$V_{\rho}=\frac{(1-\rho)^{c}e^{f(\rho)}}{H^{x_{0}}(\rho)}.$

Then,
$\frac{L_{\rho}V_{\rho}}{V_{\rho}}=[(1-\rho)f^{\prime}-c-c(x_{0},\rho)+c(x,\rho)]<0$.

In the quadratic case (76), we fix $x_{0}$ so that
$d(x_{0},1)=\underset{x\in[0,1]}{\max}\ d(x,1)$. We choose $f(\rho)$ so that

$\displaystyle(1-\rho)f^{\prime}-(d-1)-d(x_{0},\rho)+d(x,\rho)<0,d-1=-d(x_{0},1)-1>0$

on $[1-\epsilon,1]$. Denote
$H^{x}(\rho)=\text{exp}\left[\int^{\rho}\frac{d(x,z)}{1-z}dz\right]\sim(1-\rho)^{-d(x,1)}$,
let

$V_{\rho}=\frac{(1-\rho)^{d-1}e^{f(\rho)}}{H^{x_{0}}(\rho)}.$

Then,
$\frac{L_{\rho}V_{\rho}}{V_{\rho}}=(1-\rho)[(1-\rho)f^{\prime}-(d-1)-d(x_{0},\rho)+d(x,\rho)]<0$.

##### Case III:

If the corner $p$ is a strict local maximum point, we can extend the
coefficients of $L_{\rho}$ to the rectangle which is supplemented by the
dashed lines so that $L_{\rho}$ is has the same degree of degeneracy at
$\rho=\rho(p)$.

##### Case IV:

In this case, if $L$ has the same degeneracy type towards two edges
intersecting at the corner $p$. Let $U$ be a neighborhood of $p$ inside $T$
and introduce new coordinates $(x,z)$ so that $z=\rho$ in $T\setminus U$,
$z_{\rho}>0$ and $p$ attains local maximum of $z$. So in the
$(x,z)$-coordinate, $L_{z}$ has the same kind of degeneracy as $L_{\rho}$ in
an irregular area $T^{\prime}$. We extend the coefficients of $L_{z}$ to a
rectangle so that $L_{z}$ takes the form (75), (76). Taking the solutions $V$
to these two forms which we already constructed, we see $V$ is infinity at the
corner of two quadratic edges.

If $L$ has different types of degeneracy towards the two edges, then it takes
the form

$\displaystyle
L=\left[xa(x,\rho)+(\rho-1)^{2}\right]\partial_{\rho\rho}+xb(x,\rho)\partial_{xx}+c(x,\rho)\partial_{\rho}+d(x,y)\partial_{x}+x(\rho-1)e(x,y)\partial_{x\rho}$
(79)

on $[0,1]\times[0,1]$, $\rho\in[0,1]$, where $a,b,c,d>0$.

When $x=0$,
$L^{0}_{\rho}=(1-\rho)^{2}\partial_{\rho\rho}+c(x,\rho)\partial_{\rho}$.
Clearly a decreasing linear function $V$ would satisfy $L^{0}_{\rho}V<0$. But
such $V$ has negative derivative, which is not desired. So our strategy is to
first construct $f_{0}(\rho),f_{1}(\rho)$ at $x=0,1$ so that

$L^{0}_{\rho}f_{0},L^{1}_{\rho}f_{1}<0,\ f^{\prime}_{1}(\rho)>0.$

We then patch them together with the desired boundary conditions. We choose
two non-negative functions $\chi_{0}(x),\chi_{1}(x)$ so that

$\displaystyle\chi_{0}(0)>0,\chi_{0}=0\ \text{in}\ [1-\epsilon,1],$
$\displaystyle\chi_{1}(0)=0,\chi_{1}=\chi(1)\ \text{in}\ [1-\epsilon,1].$

Suppose that
$\underset{(x,\rho)}{\max}-\frac{xb\chi^{\prime\prime}_{1}+d\chi^{\prime}_{1}}{\chi_{1}}<M$
for some $M>0$ and let $c_{\min}=\underset{[0,1]\times[0,1]}{\min}c(x,\rho)$.
We choose another two non-negative functions $h_{0}(\rho),h_{\epsilon}(\rho)$
so that

$\displaystyle
h_{0}(0)=0,h^{\prime}_{0}(0),h^{\prime\prime}_{0}(0)=0,h_{0}(\rho)=\frac{c_{min}}{2}-M(1-\rho)\
\text{on}\ [1-\frac{c_{min}}{2M},1]$ $\displaystyle h_{\epsilon}\in
C_{c}^{2}([0,1-\frac{c_{min}}{2M}+\epsilon]),h_{\epsilon}(0)=0,h^{\prime}_{\epsilon}(0)=V^{\prime}(0),h^{\prime\prime}_{\epsilon}(0)=V^{\prime\prime}(0),$
$\displaystyle L_{\rho}h_{\epsilon}<0\ \text{on}\ [0,1-\frac{c_{min}}{2M}].$

Then for $\rho\in[1-\frac{c_{min}}{2M},1]$,

$\displaystyle\frac{(L_{\rho}+xe(\rho-1)\chi^{\prime}_{1}\partial_{\rho})h_{0}}{h_{0}}>-\frac{xb\chi^{\prime\prime}_{1}+d\chi^{\prime}_{1}}{\chi_{1}}.$
(80)

Indeed we can rescale $\chi_{1}$ so that the coefficient of $\partial_{\rho}$
in $L_{\rho}+xe(\rho-1)\chi^{\prime}_{1}\partial_{\rho}$ is bigger than
$\frac{c_{min}}{2}$, so the left hand side term is no less than $M$, hence
(80) is true.

We let

$\displaystyle V(x,\rho)=c_{0}f(x,\rho)h_{0}(\rho)+h_{\epsilon}(\rho)$
$\displaystyle\text{where}\
f(x,\rho)=\chi_{0}(x)f_{0}(\rho)+\chi_{1}(x)f_{1}(\rho),c_{0}>0\ \text{is a
constant}.$

We want to ensure that $L(fh_{0})<0$ on $\rho\in[1-\frac{c_{min}}{2M},1]$. The
coefficient of $f_{1}$ in $L(fh_{0})$ is

$\displaystyle(xb\chi^{\prime\prime}_{1}+d\chi^{\prime}_{1})h_{0}+\chi_{1}(x)L_{\rho}h_{0}+xe(x,\rho)(\rho-1)\chi_{1}^{\prime}(x)h^{\prime}_{0}(\rho)$
(81)

through a direct computation in(82) below, which is positive by (80). So we
can subtract a positive constant from $f_{1}$ so that $L(fh_{0})<0$ when
$\rho\in[1-\frac{c_{min}}{2M},1]$. We compute:

$\displaystyle Lf=$ $\displaystyle\chi_{0}(x)L\rho
f_{0}+\chi_{1}(x)L_{\rho}f_{1}+xb(x,\rho)\left[\chi^{\prime\prime}_{0}f_{0}+\chi^{\prime\prime}_{1}f_{1}\right]$
(82)
$\displaystyle+d(x,y)\left[\chi^{\prime}_{0}f_{0}+\chi^{\prime}_{1}f_{1}\right]+x(\rho-1)e(x,y)\left[\chi^{\prime}_{0}f^{\prime}_{0}+\chi^{\prime}_{1}f^{\prime}_{1}\right]$
$\displaystyle L(f(x,\rho)h_{0}(\rho))=$ $\displaystyle(Lf)\cdot h_{0}+f\cdot
L_{\rho}h_{0}+h^{\prime}_{0}(\rho)\left[xe(x,\rho)f_{x}+2(xa(x,\rho)+\rho^{2})f_{\rho}\right]$

Finally for $c_{0}>0$ small enough, $V$ is positive since $h_{\epsilon}>0$ and
$L_{\rho}h_{\epsilon}<0$ for $\rho\in[0,1-\frac{c_{min}}{2M}+\epsilon]$. We
adjust $h_{\epsilon}$ so that
$\underset{\rho\in[1-\frac{c_{min}}{2M},1]}{\max}L_{\rho}h_{\epsilon}<\underset{\rho\in[1-\frac{c_{min}}{2M},1]}{\max}-L_{\rho}(c_{0}fh_{0})$
and $LV<0$. Such $V$ is finite by construction, so $\frac{LV}{V}<0$. $V$
satisfies the boundary conditions at $\rho=0$ and when $x\in[1-\epsilon,1]$,
$V$ only depends on $\rho$, so this $V$ could be extended to a global function
in this layer.

By induction, we construct $V$ layer by layer, and such $V$ is positive on
$P\setminus H$ while $V$ is $\infty$ at corners which are the intersection of
two quadratic edges and on the quadratic edges where $\rho$ is constant. We
have finitely many layers and each layer is covered by finite closed
rectangles or irregular areas, and we showed that $\frac{LV}{V}<0$ in each of
this area, so $V$ is a Lyapunov function indeed satisfying
$LV\leq\lambda_{0}V$ for some $\lambda_{0}<0$. ∎

######  Remark 5.1.

The Lyapunov function $V$ we constructed above behaves like $\rho^{c}$ where
$\rho$ is the distance to the tangent edge and $c$ is some positive constant.

$V$ gives rise to a norm $||\cdot||_{P_{V}}$:

$\displaystyle||f||_{P_{V}}:=\underset{P}{\text{sup}}\
\frac{f^{+}}{V}+\underset{P}{\text{sup}}\ \frac{f^{-}}{V}.$

and we define the space $C^{V}(P)$ :

$\displaystyle C^{V}(P):=V\cdot C^{0}(P)$

equipped with the norm $||\cdot||_{P_{V}}$. By saying $f\succ 0$ if
$||f||_{P_{V}}>0$, let

$\displaystyle\lambda_{0}:=\underset{\lambda\in\mathbb{R}}{\inf}\
\\{Af\leq\lambda f,\ 0\prec f\\}.$

Clearly, $\lambda_{0}<0$. A similar argument of Proposition 4.1 gives rise to
the following result:

######  Proposition 5.1.

There exists some $M>0$ so that for $f\in C^{V}(P)$, $t>0$,

$||P_{t}f||_{P_{V}}\leq M||f||_{P_{V}}e^{\lambda_{0}t}.$

Fix a point $p\in P$ neither on $H$ nor any quadratic edge, we denote
$p_{t}(p,\cdot)$ the transition probability of the diffusion starting from
$p$. A corollary is

######  Corollary 5.1.

For $f\in C^{V}(P),t>0$

$\displaystyle|P_{t}f(p)|\leq V(p)||P_{t}f||_{P_{V}}\leq
MV(p)||f||_{P_{V}}e^{\lambda_{0}t}.$ (83)

######  Remark 5.2.

Consider the example in [6]: on the triangle $T=\\{(x,y):0\leq x,y,x+y\leq
1\\}$,

$\displaystyle
L=\gamma_{12}\left[xy(\partial_{x}-\partial_{y})^{2}+(y-x)(\partial_{x}-\partial_{y})\right]$
$\displaystyle+\gamma_{23}\left[y(x\partial_{x}+(y-1)\partial_{y})^{2}+(y-1)(x\partial_{x}+(y-1)\partial_{y})\right]$
$\displaystyle+\gamma_{13}\left[x((x-1)\partial_{x}+y\partial_{y})^{2}+(x-1)((x-1)\partial_{x}+y\partial_{y})\right].$

In this case $1-x-y=0$ is the only tangent edge. $V=1-x-y$, which measures the
scaled distance between $(x,y)$ and the tangent edge, is a Lyapunov function
since

$\displaystyle
LV=-(1-x)\gamma_{13}+(1-y)\gamma_{23})V\leq-\min(\gamma_{13},\gamma_{23})V.$

###  5.2 Exponential convergence to invariant measure

First, there exists a tubular neighborhood of $H$ that is diffeomorphic to
$U=[0,1]\times[0,1]$ with $H$ mapped to $[0,1]\times\\{0\\}$. We fix a non-
negative function $h\in C_{c}(U)$ satisfying $h|_{H}\equiv 1,h|_{y=1}\equiv
0$. Then we integrate the transition probability $p_{t}(p,x,y)$ with respect
to $h$ in $U$:

$\displaystyle q_{h}(t,x):=\int_{0}^{1}p_{t}(p,x,y)h(y)dy.$ (84)

It can be interpreted as the marginal density with respect to coordinate along
$H$ ($x$) and limited to some neighbourhood of $H$ (specified by $h$).

We already know that there exists a unique invariant measure $\mu_{0}$
supported on $H$. The total variation distance between $q_{h}(t,x)$ and
$\mu_{0}(x)$ can be bounded by $\chi^{2}$-divergence or Kullback-Leibler (KL)
divergence. To show the exponential decay of the $\chi^{2}$-divergence or the
Kullback-Leibler(KL), convergence requires a Poincaré inequality or a
logarithmic Sobolev inequality, respectively, for $\mu_{0}$. It turns out that
if $L|_{H}$ has a Kimura endpoint, then the $\chi^{2}$-divergence is not
finite. And if $L|_{H}$ has a quadratic endpoint, then $\mu_{0}$ does not
satisfy the logarithmic Sobolev inequality. So in the proof of the following
theorem, we employ an interpolation divergence as constructed in [8, Remark
3.2(3)].

For $1<r\leq 2$, define

$\displaystyle\psi(u)=3+\left(\frac{5}{2}-\frac{3}{r-1}\right)(u-3)+\frac{9}{r(r-1)}\left[\left(\frac{u}{3}\right)^{r}-1\right],u\geq
0.$ (85)

Such $\psi$ is convex, $\psi(1)=0$ and $\psi(u)\sim\frac{9}{r(r-1)3^{r}}u^{r}$
at $+\infty$.

######  Lemma 5.1.

[8, Lemma 1.1] There exists a constant $C_{\psi}>$ such that for two
probability measures $P,Q$,

$\displaystyle||P-Q||_{TV}\leq
C_{\psi}\sqrt{\int\psi\left(\frac{dP}{dQ}\right)dQ}.$

If $\mu$ satisfies a Poincaré inequality, then for $h\geq 0$ satisfying $\int
h\mu=1$, we have that $\mu$ satisfies the $I_{\psi}$-inequality:

$\displaystyle\int\psi(h)\mu\leq
C_{\psi}\int\psi^{\prime\prime}(h)|\partial_{x}h|^{2}\mu.$ (86)

######  Proposition 5.2.

Assume for some $0<A<1$ that

$\displaystyle h\geq 0,\ h|_{H\times[0,A]}\equiv 1\ \text{for}\ y\in[0,1],\
h|_{y=1},h^{\prime}|_{y=1}\equiv 0.$ (87)

Fix a point $p$ that is not on any quadratic transverse edge, then the total
variation between the $h$-weighted transition probability $q_{h}(t,x)$ defined
in (84) and the invariant measure $\mu_{0}$ has exponential decay:

$\displaystyle||q_{h}(t,x)-\mu_{0}||_{TV}\leq Me^{-\lambda t},$

where
$\lambda=\min\frac{1}{2}(\frac{1}{C_{\psi}}-\epsilon,-(1-\epsilon)\lambda_{0})$
$\forall\epsilon>0$.

###### Proof.

Three are three types of boundary of $L|_{H}$. We first take a coordinate
change at quadratic endpoints. By abuse of notation, we still denote $x$ the
coordinate along $H$ and $U$ is mapped to $I\times[0,1]$.

On $I\times[0,1]$, $L$ takes the form

$\displaystyle
L=a(x,y)\partial_{xx}+b(x,y)y^{m}\partial_{yy}+c(x,y)y\partial_{xy}+d(x,y)\partial_{x}+e(x,y)^{m-1}\partial_{y},\
\ m\in\\{1,2\\}.$ (88)

The proofs for the cases $H$ being a quadratic edge or Kimura edge are
essentially the same. So in the following proof, we assume that $m=2$. In
addition, we will prove below that in an appropriate set of variables, we may
choose

$\displaystyle c(x,1)=0.$ (89)

Regarding $q_{h}(t,x)$ as the marginal density limited to neighbourhood of
$H$, its total measure approaches 1 at an exponential rate of $\lambda_{0}$.
Indeed

$\displaystyle\int_{I}q_{h}(t,x)dx$
$\displaystyle=\int_{I}\int_{0}^{1}p_{t}(p,x,y)h(y)dxdy=1-\int_{I}\int_{0}^{1}p_{t}(p,x,y)[1-h(y)]dxdy$
(90) $\displaystyle\geq 1-Me^{\lambda_{0}t}$

for some $M>0$. Moreover, it satisfies the backwards equation:

$\displaystyle\partial_{t}q(t,x)=\int_{0}^{1}L^{*}p_{t}(p,x,y)h(y)dy=L_{t}^{*}q_{h}(t,x)+v(t,x)$
(91)

where by integrating by parts

$\displaystyle L_{t}^{*}q(t,x)=$
$\displaystyle\partial_{xx}\int_{0}^{1}a(x,y)p_{t}(p,x,y)h(y)dy-\partial_{x}\int_{0}^{1}c(x,y)yp_{t}(p,x,y)h^{\prime}(y)dy$
$\displaystyle-\partial_{x}\int_{0}^{1}d(x,y)p_{t}(p,x,y)h(y)dy$
$\displaystyle=:$
$\displaystyle\partial_{xx}(B(t,x)q_{h}(t,x))-\partial_{x}(A(t,x)q_{h}(t,x))$
$\displaystyle
v(t,x)=\int_{0}^{1}b(x,y)y^{2}p_{t}(p,x,y)h^{\prime\prime}(y)dy+\int_{0}^{1}e(x,y)yp_{t}(p,x,y))h^{\prime}(y)dy.$
(92)

We make a few remarks on these terms. First, $||v(t,x)||_{1}$ and
$||[A(t,x)-d(x,0)]q_{h}(t,x)||_{1}$ decay exponentially at the rate
$\lambda_{0}$. Second, $c(x,1)=0$ ensures that
$\frac{c(x,y)yh^{\prime}(y)}{h(y)}$ is bounded, so that by choosing $U$
sufficient small we may assume that

$\displaystyle\big{|}\partial_{x}[B(t,x)-a(x,0)]\big{|}+\big{|}\frac{\mu^{\prime}_{0}}{\mu_{0}}(B(t,x)-a(x,0))\big{|}+\big{|}(A(t,x)-d(x,0))\big{|}\leq\epsilon.$
(93)

We are now ready to estimate $||q_{h}(t,x)-\mu_{0}(x)||_{TV}$. Let
$f(t,x)=\frac{q_{h}(t,x)}{\mu_{0}(x)}$. Then by Lemma 5.1,

$\displaystyle||q_{h}(t,x)-\mu_{0}(x)||_{TV}\leq\frac{C_{\psi}}{P(t)}\sqrt{\int\psi(f(t,x))\mu}+Me^{-\lambda_{0}t},$
(94)

where $\mathbb{P}(t)=||q_{h}(t,x)||_{1}$. We differentiate the term

$\displaystyle\frac{d}{dt}(\psi(f(t,x)),\mu_{0})=$
$\displaystyle(\psi^{\prime}(f),\partial_{t}q_{h}(t,x))$ $\displaystyle=$
$\displaystyle(L_{0}\psi^{\prime}(f),f\mu_{0})+((L_{t}-L_{0})\psi^{\prime}(f),f\mu_{0})+(\psi^{\prime}(f),v(t,x))$
$\displaystyle=:$
$\displaystyle-\int_{I}\psi^{\prime\prime}(f)(\partial_{x}f)^{2}\mu_{0}+II+III.$

We will show below that

$\displaystyle II\leq
2\epsilon\int_{I}\psi^{\prime\prime}(f)(\partial_{x}f)^{2}\mu_{0}+\epsilon\int_{I}\psi(f)\mu_{0}dx+Me^{\lambda_{0}t}$
(95) $\displaystyle
III\leq\epsilon\int_{I}\psi(f)\mu_{0}dx+Me^{(1-\epsilon)\lambda_{0}t}.$ (96)

Therefore, we have

$\displaystyle\frac{d}{dt}(\psi(f(t,x)),\mu_{0})\leq-(1-3\epsilon)\int_{I}\psi^{\prime\prime}(f)(\partial_{x}f)^{2}\mu_{0}+2\epsilon\int_{I}\psi(f)\mu_{0}+Me^{\lambda_{0}t}$
$\displaystyle\leq-\left[\frac{1-3\epsilon}{C_{\psi}}-2\epsilon\right]\int_{I}\psi(f)\mu_{0}+Me^{\lambda_{0}t}.$

As a consequence,

$\displaystyle\int_{I}\psi(f(t,x))\mu_{0}\leq Me^{-at}$ (97)

where $a=\min(\frac{1-2\epsilon}{C_{\psi}}-2\epsilon,-\lambda_{0})$. Finally
combined with (94), we obtain that

$\displaystyle||q_{h}(t,x)-\mu_{0}(x)||_{TV}\leq Me^{-\frac{a}{2}t},$

for some constant $M>0$. This only leaves the proof of (95), (96), and (89).
We leave the proof of (96) to Lemma A.1 in the Appendix.

##### Proof of (95)

To see (95), we integrate by parts to see

$\displaystyle
II=-(\psi^{\prime\prime}(f)(\partial_{x}f)^{2},(B(t,x)-a(x,0))\mu_{0})-(C(t,x)\partial_{x}\psi^{\prime}(f),f\mu_{0})=:II_{1}+II_{2}$

where

$C(t,x)=\partial_{x}[B(t,x)-a(x,0)]+\frac{\mu^{\prime}_{0}}{\mu_{0}}(B(t,x)-a(x,0))-(A(t,x)-d(x,0)).$

It is easy to see that

$\displaystyle
II_{1}\leq\epsilon\int_{I}\psi^{\prime\prime}(f)(\partial_{x}f)^{2}\mu_{0}$
(98)

since by (93) $|B(t,x)-a(x,0)|\leq\epsilon$. We use Hölder and Cauchy-Schwarz
inequalities to bound

$\displaystyle|II_{2}|\leq(C(t,x)(\psi^{\prime\prime}(f))^{2}(\partial_{x}f)^{2}|f|^{2-p},\mu_{0})+(C(t,x)|f|^{p},\mu_{0})=:II_{21}+II_{22}.$

Since $\psi^{\prime\prime}(f)|f|^{2-p}$ is bounded by some constant
independent of $\epsilon$, by rescaling $\epsilon^{\prime}=\frac{\epsilon}{F}$
in (93) we obtain

$\displaystyle
II_{21}\leq\epsilon\int_{I}\psi^{\prime\prime}(f)(\partial_{x}f)^{2}\mu_{0}.$
(99)

We split $II_{22}$ into two parts

$\displaystyle\int_{f\leq 1}C(t,x)|f|^{p}\mu_{0}+\int_{f\geq
1}C(t,x)|f|^{p}\mu_{0}.$

The first term decreases exponentially as $e^{\lambda_{0}t}$ using
$|f|^{p}\leq f$ and that $(C(t,x),q_{h}(t,x))$ decreases exponentially. The
second term is bounded by $\int\psi(f)\mu_{0}$ multiplied by some constant. To
see this, when $0\leq f\leq 3$, because of the convexity of $\psi$,

$\psi(f)-\psi(1)-\psi^{\prime}(1)(f-1)\geq\frac{1}{2}\underset{0\leq f\leq
3}{\min}\psi^{\prime\prime}(f)(f-1)^{2},$

and $\psi(f)\sim f^{p}$ near $\infty$. Thus,

$\displaystyle(f-1)^{2}\leq
C(1+f^{2-p})(\psi(f)-\psi(1)-\psi^{\prime}(1)(f-1)).$

Hence

$\displaystyle\int_{I}\psi(f)\mu_{0}\geq\int\frac{(f-1)^{2}}{1+f^{2-p}}\mu_{0}\geq\int_{f\geq
1}\frac{(f-1)^{2}}{1+f^{2-p}}\mu_{0}\geq M\int_{f\geq 1}\psi(f)\mu_{0}.$ (100)

Gathering the above estimates,

$\displaystyle II\leq
II_{1}+II_{21}+II_{22}\leq\epsilon\int_{I}\psi^{\prime\prime}(f)(\partial_{x}f)^{2}\mu_{0}+\epsilon\int_{I}\psi(f)\mu_{0}+Me^{\lambda_{0}t}.$
(101)

##### Proof of (89)

If we consider the new coordinate $(t,y)$, then the coefficient of mixed
derivative $\partial_{tx}$ is

$A(x,y)=cyt_{x}+2by^{2}t_{y}$

To make $A(x,1)=0$, we choose

$t_{x}(x,1)=1,\qquad\mbox{ and then}\qquad
t_{y}(x,1)=-\frac{c(x,1)}{2b(x,1)}.$

Let $f(x,y)=\partial_{xy}t=\partial_{yx}t$, we choose $f(x,y)=f(x,1)h(y)$ and

$\displaystyle t_{x}(x,y)=\int_{0}^{y}f(x,z)dz+g(x),\
t_{y}(x,y)=\int_{0}^{x}f(z,y)dz+l(y).$ (102)

Plugging (102) into $t_{x},t_{y}$, we find

$\displaystyle\int_{0}^{1}f(x,y)dy+g(x)=1,\
\int_{0}^{x}f(z,1)dz+l(1)=-\frac{c(x,1)}{2b(x,1)},$

and hence

$f(x,1)=-\frac{\partial}{\partial x}\frac{c(x,1)}{2b(x,1)},\
g(x)=1-f(x,1)\int_{0}^{1}h(y)dy.$

Now we only need to make $t_{x}$ everywhere positive. Then, the coordinate
change $(x,y)\mapsto(t,y)$ would be bijective on $[0,1]^{2}$. Now by (102),

$\displaystyle
t_{x}(x,y)=f(x,1)\int_{0}^{y}h(z)dz+1-f(x,1)\int_{0}^{1}h(y)dy=1-f(x,1)\int_{y}^{1}h(z)dz.$

We can choose a smooth function $h(z)$ so that
$|f(x,1)|\cdot|\int_{y}^{1}h(z)dz|<1$ for all $(x,y)\in[0,1]^{2}$. Therefore
$t(x,y)$ exists since $t_{xy}=t_{yx}$ and $A(x,1)=0$. ∎ As a consequence we
have the following estimate in Wasserstein distance,

######  Theorem 5.2.

Fix a point $p$ that is not on any quadratic transverse edge. Then the
Wasserstein distance between the transition probability $p_{t}(p,\cdot)$ and
the invariant measure supported on the tangent edge converges exponentially:

$\displaystyle W(p_{t}(p,\cdot),\mu_{0}(x)\delta_{0}(y))\leq
Me^{-\frac{a}{2}t},t>0.$ (103)

###### Proof.

For any $f\in C^{0}(P)$ with $\text{Lip}(f)\leq 1$,

$\displaystyle\int_{P}fp_{t}-\int_{P}f\mu(x)\delta_{0}(y)=\int_{P\setminus
U}fp_{t}+\int_{U}fp_{t}-\int_{U}f\mu(x)\delta_{0}(y).$

$\int_{P\setminus U}fp_{t}$ is bounded by $Me^{-\lambda_{0}t}$with constant
$M$ independent of $f$. The second term is

$\displaystyle\int_{U}fp_{t}(p,x,y)dxdy-\int_{U}f\mu(x)\delta_{0}(y)$
$\displaystyle=$
$\displaystyle\int_{I}\int_{0}^{1}p_{t}(p,x,y)[f(x,y)-f(x,0)h(y)]dxdy$
$\displaystyle+\left[\int_{I}\int_{0}^{1}p_{t}(p,x,y)f(x,0)h(y)dxdy-\int_{I}f(x,0)\mu(x)dx\right].$

It is bounded by $||q_{h}(t,x)-\mu(x)||_{TV}$. The first term also has
exponential decay because $[f(x,y)-f(x,0)h(y)]\in C^{V}(P)$ with bounded norm
independent of $f$. Hence the Wasserstein distance between between
$p_{t}(p,\cdot)$ and $\mu(x)\delta_{0}(y)$ decays exponentially with the same
rate $\frac{a}{2}$. ∎

##  6 Numerical Experiments

In this section, we present numerical simulations that illustrates the long
time behaviors obtained in earlier sections. Since the distributions may
converge to lower dimensional manifolds at tangent boundaries, we adopt the
probabilistic representation [35] that considers the stochastic process
governed by the generator $L$. Comparing with solving PDE directly on
meshgrids, the probabilistic representation can be numerically implemented in
a way respecting the local coordinate chart at degenerated boundaries. See
[28, 18] for examples of stability of long time behaviour simulation by
numerical probabilistic approach.

Following the theoretical derivations, we start with examples in one space
dimension before considering a two dimensional case.

We present the Lagrangian scheme in detail in the one-dimensional setting.

###  6.1 One dimensional cases with mixed boundary conditions

We first consider a general model operator that is of Kimura type at $x=0$ and
of quadratic type at $x=1$. Specifically,

$\displaystyle
L=(c_{0}(1-x)^{2}-c_{1}x(1-x))\partial_{x}+x(1-x)^{2}\partial_{x}^{2},$ (104)

where $c_{0}$ and $c_{1}$ denote the strength of convection near boundaries.
We recall that a Kimura boundary at $x=0$ is transverse when $c_{0}>0$ and
tangent when $c_{0}=0$; the quadratic boundary at $x=0$ is transverse when
$c_{1}>1$ and tangent when $c_{1}<1$. When both boundaries are transverse, $L$
admits an invariant measure

$\displaystyle\mu_{0}=\frac{1}{Z}x^{c_{0}-1}(1-x)^{c_{1}-2}.$ (105)

To present the convergence result, we use particles to represent the
distribution $q(t,x)$ defined by

$\displaystyle q(t,x)=\int_{0}^{1}p_{t}(x,y)q_{0}(x)dy,$ (106)

where $q_{0}$ is the initial distribution and $p_{t}$ denotes the transition
probability of the process whose generator is $L$. We denote the process as
$X_{\cdot}$, where $X_{0}$ has initial distribution $q_{0}$, and $X_{\cdot}$
follows stochastic differential equation (SDE) in the Itô sense [35],

$\displaystyle dX=(c_{0}(1-X)^{2}-c_{1}X(1-X))dt+\sqrt{2X(1-X)^{2}}dW_{t},$
(107)

where $W_{t}$ is a standard Brownian motion. The probability density function
of $X_{t}$ is then $q(t,\cdot)$.

We now develop a modified Euler Maruyama (EM) scheme [24] to generate
approximated realizations of the process $X$ and then approximate $q$ by the
empirical distribution of such generated realizations. First, $X_{n}$ denotes
the approximated process at time $t_{n}=n\Delta t$. Here, the length of the
time interval $\Delta t$ is a fixed constant during the computation. Since the
degeneracy of $L$ only happens at boundaries, we apply a standard EM scheme
when $x$ is ‘far’ from the boundary.

In practice, when $X_{n}\in[0.01,0.99]$, starting from $X_{n}$, for one time
step $\Delta t$, we consider,

$\displaystyle X_{n+1}=X_{n}+(c_{0}(1-X_{n})^{2}-c_{1}X_{n}(1-X_{n}))\Delta
t+\sqrt{2X_{n}(1-X_{n})^{2}}\sqrt{\Delta t}\mathcal{N},$ (108)

where $\mathcal{N}$ denotes a generated standard Gaussian distribution.

When $X_{n}<0.01$, the particle is considered as approaching the Kimura
boundary at $x=0$. The leading order of SDE turns to be

$\displaystyle dX=c_{0}dt,$

which provides a strong forcing so that the position of particles remain
strictly positive. To this end, we consider an implicit scheme,

$\displaystyle X_{n+1}=X_{n}+(c_{0}(1-X_{n})^{2}-c_{1}X_{n}(1-X_{n}))\Delta
t+\sqrt{2X_{n+1}(1-X_{n})^{2}}\sqrt{\Delta t}\mathcal{N},$

which may be explicitly solved by

$\displaystyle\sqrt{X_{n+1}}=\frac{\sqrt{2\Delta t}\Delta
W(1-X_{n})+\sqrt{(\sqrt{2\Delta t}\Delta
W(1-X_{n}))+4(X_{n}+(c_{0}(1-X_{n})^{2}-c_{1}X_{n}(1-X_{n}))\Delta t)}}{2}.$
(109)

When $X_{n}>0.99$, the particle is considered as approaching the quadratic
boundary at $x=1$. The leading order of the SDE is

$\displaystyle dX=-c_{1}(1-X)dt+\sqrt{2}(1-X)dW_{t}.$

It is then natural to apply a logarithm transform over $1-X$, i.e., let
$y=-\log(1-x)$. Then by the Itô formula, $Y$ follows

$\displaystyle dY=(c_{0}(1-X)-c_{1}X)dt+\sqrt{2X}dW_{t}+Xdt,$ (110)

where $X=1-\exp(-Y)$. This is not a degenerate SDE since here $X$ is assumed
to be near $x=1$.

##### Case I: Transverse Kimura and Transverse Quadratic

In Fig.2, we show the empirical PDF (histogram) of the approximated
realizations obtained by the proposed algorithm. We start from a single-point
initial distribution, $q_{0}=\delta(x-0.5)$ with one million realizations. To
ensure transverse boundary condition on both side, we set
$(c_{0},c_{1})=(0.5,2)$. For each realization, we approximate the SDE with
$\Delta t=2^{-14}$ until $T=2^{1}$. As a reference, we plot the target
invariant measure $\mu_{0}$ with a solid red line.

Figure 2: Distribution of realizations when $x=0$ is of Kimura transverse type
and $x=1$ is of quadratic transverse type. $\mu_{0}$ is the solid red line in
each sub-graph

In addition, we compute the Wasserstein distance between to empirical
distribution of the simulation and the target invariant measure. By optimal
transport theory in one dimension, the optimal transport map in the
Wasserstein distance is given by the inverse cumulative distribution functions
[31]. More precisely, let $\\{x_{i}\\}_{i=1}^{N}\subset R^{1}$ be the position
of the particles and without loss of generality, assume $x_{1}\leq
x_{2}\leq\cdots\leq x_{N}$. Let $F_{\mu}$ be the cumulative distribution
functions of target distribution. Then the $W^{p}$ distance can be
approximated by

$\displaystyle
W^{p}=\Big{(}\frac{1}{N}\sum_{i=1}^{N}\big{(}x_{i}-F_{\mu}^{-1}(\frac{i-1/2}{N})\big{)}^{p}\Big{)}^{1/p}$
(111)

In Fig.3(a), we present the $W^{1}$ distance between the empirical
distribution of realization of SDE with $\Delta t=2^{-9}$ until $T=2^{1}$.

##### Case II: Transverse Kimura and Tangent Quadratic

In the second case, we investigate the case when one of the boundary
conditions becomes tangent. As an example, we set $(c_{0},c_{1})=(0.5,0.5)$.
Then the quadratic boundary on the right becomes tangent and the invariant
measure is $\delta(x-1)$. The $p$-Wasserstein distance in such case can be
computed by,

$\displaystyle
W^{p}(q(t,\cdot),\delta_{1})=\Big{(}\int(1-x)^{p}q(t,x)dx\Big{)}^{1/p}.$ (112)

In Fig.3(b), we present the $W^{1}$ distance between the empirical
distribution of realizations of the SDE with the invariant measure.

(a) $x=0$ is of Kimura transverse type and $x=1$ is of quadratic transverse
type.

(b) $x=0$ is of Kimura transverse type and $x=1$ is of quadratic tangent type

Figure 3: $W^{1}$ distance between the empirical distribution and invariant
measure

###  6.2 2D with tangent diagonal

We now turn to the long time behavior for a degenerate operator in two
dimension. Specifically we consider, $L$ defined in a triangular domain
$T=\\{x\geq 0,y\geq 0,x+y\leq 1\\}$,

$\displaystyle\mathcal{L}=$
$\displaystyle\gamma_{12}\left[xy\varphi_{3}\otimes\varphi_{3}:\nabla^{2}+\left(y-x\right)\varphi_{3}\cdot\nabla\right]+\gamma_{23}\left[y\varphi_{2}\otimes\varphi_{2}:\nabla^{2}+\left(y-1\right)\varphi_{2}\cdot\nabla\right]$
(113)
$\displaystyle+\gamma_{13}\left[x\varphi_{1}\otimes\varphi_{1}:\nabla^{2}+\left(x-1\right)\varphi_{1}\cdot\nabla\right]$

where,

$\displaystyle\varphi_{3}=\left(\begin{array}[]{c}1\\\
-1\end{array}\right),\quad\varphi_{2}=\left(\begin{array}[]{c}x\\\
y-1\end{array}\right),\quad\varphi_{1}=\left(\begin{array}[]{c}x-1\\\
y\end{array}\right).$ (120)

Such operator in Eq.(113) is the asymptotic generator of two reflection
coefficients in a stochastic homogenization regime [6]. Then we construct an
SDE $X=\left(\begin{array}[]{c}x\\\ y\end{array}\right)$ which follows,

$\displaystyle dX=$
$\displaystyle(\gamma_{12}(y-x)\varphi_{3}+\gamma_{23}(y-1)\varphi_{2}+\gamma_{13}(x-1)\varphi_{1})dt$
(121)
$\displaystyle+\sqrt{2\gamma_{13}x}\varphi_{1}dW_{1}+\sqrt{2\gamma_{23}y}\varphi_{2}dW_{2}+\sqrt{2\gamma_{12}xy}\varphi_{3}dW_{3},$

where $W_{i}$ for $i=1,2,3$ are independent standard Brownian motions and
$X(0)\in T$. We observe that $x=0$ and $y=0$ are transverse Kimura boundaries
while $x+y=1$ is tangent quadratic boundary. As in Theorem 5.2, the invariant
measure can be represented by $\mu(x)\delta_{x+y=1}$. $\mu$ may be computed by
restricting $L$ on $x+y=1$. More precisely,

$\displaystyle L|_{x+y=1}=$
$\displaystyle(\gamma_{12}(1-2x)+\gamma_{23}x^{2}+\gamma_{13}(1-x)^{2})\partial_{x}$
(122)
$\displaystyle+(1-x)x(\gamma_{12}+\gamma_{23}x+\gamma_{13}(1-x))\partial_{x}^{2},$
(123)

and $L$ admits an invariant measure density (along the diagonal edge),

$\displaystyle\mu=\frac{(\gamma_{12}+\gamma_{13})(\gamma_{12}+\gamma_{23})}{(\gamma_{12}+\gamma_{13}(1-x)+\gamma_{23}x)^{2}},\quad
x\in[0,1].$ (124)

Starting with $X(0)=(0.1,0.1)^{T}$, we numerically integrate the SDE (121)
with $\gamma_{12}=1$, $\gamma_{13}=2$ and $\gamma_{23}=1$ by a Euler method
with $\Delta t=2^{-9}$ until $T=2^{2}$ with $10^{6}$ realizations. To address
the degeneracies near the boundary, we apply a direct cutoff at $x=0$ and
$y=0$ at the transverse Kimura boundaries. The is due to the convection at
these boundaries points to interior of domain and does not degenerate,
comparing with quadratic cases. The process does not accumulate on these
boundaries. For boundary $x+y=1$, we first observe that the convection of
continuous dynamics are in the direction $(1,-1)^{T}$. So at the $n$th-step of
the Euler iteration starting from $(x_{n-1},y_{n-1})^{T}$, if $x_{n}+y_{n}>1$,
we apply a re-scale factor $\frac{x_{n-1}+y_{n-1}}{x_{n}+y_{n}}$ to $x_{n}$
and $y_{n}$ to ensure the dynamics only propagate in the $(1,-1)^{T}$
direction.

In Fig.4, we present the 2D histogram of the joint distribution of $(x,y)$ for
different times. This shows that the majority of particles first travel
rightwards along the $x$ axis before converging to the invariant distribution
along the diagonal $x+y=1$, as predicted theoretically.

Figure 4: 2D Histogram of the joint distribution of $(x,y)$ at different times

In Fig.5, we present the histogram of the marginal distribution of $x$. It is
found to converges to the invariant measure on the line $x+y=1$ given in
(124).

Figure 5: Histogram of the marginal distribution of $x$ at different times.
Reference pdf is the solid red line.

In Fig.6(a) we show the approximate $E(1-X-Y)=\int(1-x-y)q(t,x,y)dxdy$ against
time $t$. This confirms that the density of dynamics converges exponentially
fast to the diagonal $x+y=1$. To quantitatively describe the convergence to
the invariant measure, we also computed the Wasserstein distance between $x$
and $\mu(x)$ in Fig.6(b).

(a) $E(1-x-y)$ in 2D triangular example.

(b) $W^{1}$ distance between empirical $x$ and $\rho(x)$ in 2D triangular
example.

Figure 6: Convergence to degenerated invariant measure

##  Appendix A Appendix

######  Lemma A.1.

Fix $p\in P$ that is not on the quadratic edge or tangent Kimura edge. For
$q\in H\times[A,1]$, the transition probability $p_{t}(p,q)$ has a pointwise
upper bound:

$\displaystyle p_{t}(p,q)\leq
C\cdot\text{exp}\left[-\frac{x^{2}}{4C_{2}t}+C_{0}|x|+C_{1}t\right],$ (125)

for some constant $C,C_{0},C_{1},C_{2}>0$. Therefore

$\displaystyle\left|\int_{I}\psi^{\prime}(f)v(t,x)dx\right|\leq\epsilon\int_{I}\psi\mu+\sqrt{t}e^{(1-\epsilon)\lambda_{0}t}.$
(126)

###### Proof.

We can follow the proof of [34, Theorem 5.2.10] with several modifications.

Firstly, unlike the setting under the usual Rimmannian measure on $P$, our
proof introduces a weighted measure $d\mu$ on $P$. The principal symbol of $L$
induces a Riemannian metric $dV$ on $P$. For each Kimura boundary surface
$H_{i},1\leq i\leq\eta$, define $\rho_{i}(p)$ to be the Riemannian distance
from the point $p\in P$ to $H_{i}$. Then

$B_{i}|_{H_{i}}:=\frac{1}{4}L\rho_{i}^{2}|_{H_{i}}$

are coordinate-invariant quantities. We also let $B_{i}$ denote a smooth
extension from $H_{i}$ to $P$ of the coefficients. Then we define the weighted
measure $d\mu$ by

$\displaystyle
d\mu(p):=\underset{i=1}{\overset{\eta}{\Pi}}\rho_{i}(p)^{2B_{i}-1}dV.$

In an adapted system of local coordinates of a corner $p$, $L$ takes the form:
$m+n=2$,

$\displaystyle L=$
$\displaystyle\underset{i=1}{\overset{m}{\sum}}x_{i}\partial_{x_{i}x_{i}}+\underset{i,j=1}{\overset{m,n}{\sum}}b(x,y)x_{i}y_{j}\partial_{x_{i}y_{j}}+\underset{j=1}{\overset{n}{\sum}}c(x,y)y_{j}^{2}\partial_{y_{j}y_{j}}$
(127)
$\displaystyle+\underset{i=1}{\overset{m}{\sum}}d(x,y)\partial_{x_{i}}+\underset{j=1}{\overset{n}{\sum}}e(x,y)y_{j}\partial_{y_{j}}.$

The weighted measure $d\mu$ is a multiple by a smooth function of

$\displaystyle
d\mu(x,y)=\underset{i=1}{\overset{m}{\Pi}}x_{i}^{d_{i}(x,y)-1}dx_{i}\underset{j=1}{\overset{n}{\Pi}}\frac{1}{y_{j}}dy_{j}.$

For $\alpha\in\mathbb{R}$ and $\phi\in C^{\infty}(P)$ satisfying
$|\nabla\phi|\leq 1$ under the Riemannian metric on $P$, define
$H_{t}^{\alpha,\phi}(P)$ on $L^{2}(\mu)$ by

$\displaystyle
H_{t}^{\alpha,\phi}f(x)=e^{-\alpha\phi(x)}H_{t}[e^{\alpha\phi}f](x).$

Instead of the estimates of $||H_{t}^{\alpha,\phi}||_{2\to 2}$ in [34], we
have

$\displaystyle||H_{t}^{\alpha,\phi}||_{2\to 2}\leq
e^{(1+\epsilon)\alpha^{2}t+C_{0}|\alpha|t+C_{1}t}$ (128)

for some constants $C_{0},C_{1}\geq 0$.

##### Proof of (128)

To see this, compute the derivative:

$\displaystyle\frac{\partial}{\partial_{t}}||H_{t}^{\alpha,\phi}f||_{2}^{2}=2(H_{t}^{\alpha,\phi}f,e^{-\alpha\phi}Le^{\alpha,\phi}H_{t}^{\alpha,\phi}f)_{L^{2}(\mu)}.$

Taking $g=H_{t}^{\alpha,\phi}f$, to derive (128) it is sufficient to have

$\displaystyle(e^{-\alpha\phi}g,Le^{\alpha\phi}g)\leq[\alpha^{2}+C_{0}|\alpha|+C_{1}](g,g).$
(129)

We associate a bilinear form $Q(u,v)$ to $-(Lu,v)_{\mu}$ by letting

$\displaystyle Q(u,v)=(Lu,v)_{L^{2}(\mu)},$

which can be written as

$\displaystyle Q(u,v)=Q_{\text{sym}}+(Tu,v)_{L^{2}(d\mu)}$

where $Q_{\text{sym}}(u,v)$ is a symmetric bilinear form and $T$ is a first
order vector field on $P$. We need to estimate it in two cases.

##### Case I:

In a neighborhood that is away from any Kimura edge, then $L$ is uniformly
elliptic under $d\mu$ and of the form

$\displaystyle L=a\partial_{x}^{2}+2b\partial_{xy}+c\partial_{y}^{2}+V,$

where $V$ is a vector field. Then

$\displaystyle
Q_{\text{sym}}(u,v)=-\int[au_{x}v_{x}+cu_{y}v_{y}+b(u_{x}v_{y}+u_{y}v_{x})]d\mu.$

$T$ is a vector field with bounded smooth coefficients. Take
$u=e^{\alpha\phi}g,v=e^{-\alpha\phi}g$. Then

$\displaystyle