1 Introduction

Carr and Penrose (1998) introduced a linear version of the Lifschitz–Slyozov–Wagner (LSW) model (Lifschitz and Slyozov 1961; Wagner 1961). In this model, the density function \(c_0(x,t), \ x>0,t>0,\) evolves according to the system of equations,

$$\begin{aligned} \frac{\partial c_0(x,t)}{\partial t}= & {} \frac{\partial }{\partial x}\left\{ \left[ 1-\frac{x}{\Lambda _0(t)}\right] c_0 (x,t)\right\} , \quad x > 0, \end{aligned}$$
(1.1)
$$\begin{aligned} \int _0^\infty x c_0(x,t) \hbox {d}x= & {} 1. \end{aligned}$$
(1.2)

The parameter \(\Lambda _0(t) > 0\) in (1.1) is determined by the conservation law (1.2) and is therefore given by the formula,

$$\begin{aligned} \Lambda _0(t)=\int ^\infty _0 \ x c_0(x,t)\hbox {d}x \Big / \int ^\infty _0 c_0(x,t) \hbox {d}x. \end{aligned}$$
(1.3)

One can also see that the derivative of \(\Lambda _0(t)\) is given by

$$\begin{aligned} \frac{\hbox {d}\Lambda _0(t)}{\hbox {d}t} =c_0(0,t)\Big /\left[ \int ^\infty _0 c_0(x,t) \hbox {d}x\right] ^2 , \end{aligned}$$
(1.4)

whence \(\Lambda _0(\cdot )\) is an increasing function.

The system (1.1), (1.2) can be interpreted as an evolution equation for the probability density function (pdf) of random variables. Thus, let us assume that the initial data \(c_0(x)\ge 0, \ x>0,\) for (1.1), (1.2) satisfy \(\int _0^\infty c_0(x) \ \hbox {d}x<\infty \), and let \(X_0\) be the nonnegative random variable with pdf \(c_0(\cdot )/\int _0^\infty c_0(x) \ \hbox {d}x\). The conservation law (1.2) implies that the mean \(\langle X_0\rangle \) of \(X_0\) is finite, and this is the only absolute requirement on the variable \(X_0\). If for \(t>0\) the variable \(X_t\) has pdf \(c_0(\cdot ,t)/\int _0^\infty c(x,t) \ \hbox {d}x\), then (1.1) with \(\Lambda _0(t)=\langle X_t\rangle \) is an evolution equation for the pdf of \(X_t\). Equation (1.4) now tells us that \(\langle X_t\rangle \) is an increasing function of t.

There is an infinite one-parameter family of self-similar solutions to (1.1), (1.2). Using the normalization \(\langle X_0\rangle =1\), the initial data for these solutions are given by

$$\begin{aligned} P(X_0>x) ={\left\{ \begin{array}{ll} {[1-(1-\beta )x]^{\beta /(1-\beta )} , \ 0<x<1/(1-\beta )}, \quad &{}\text { if } \ 0<\beta <1, \\ e^{-x} &{}\text { if } \ \beta =1, \\ {[1+(\beta -1)x]^{\beta /(1-\beta )} , \ 0<x<\infty }, \quad &{} \ \text {if } \ \beta >1. \end{array}\right. } \end{aligned}$$
(1.5)

The random variable \(X_t\) corresponding to the evolution (1.1), (1.2) with initial data (1.5) is then given by

$$\begin{aligned} X_t =\langle X_t\rangle X_0 , \quad \frac{\hbox {d}}{\hbox {d}t} \langle X_t\rangle = \beta . \end{aligned}$$
(1.6)

The main result of Carr and Penrose (1998) (see also Carr 2006) is that a solution of (1.1), (1.2) converges at large time to the self-similar solution with parameter \(\beta \), provided the initial data and the self-similar solution of parameter \(\beta \) behave in the same way at the end of their supports. In Sect. 2, we give a simple proof of the Carr–Penrose convergence theorem using the beta function of a random variable introduced in Conlon (2011).

The large time behavior of the Carr–Penrose (CP) model is qualitatively similar to the conjectured large time behavior of the LSW model (Niethammer 1999), provided the initial data have compact support. In the LSW model, there is a one-parameter family of self-similar solutions with parameter \(\beta , 0<\beta \le 1\), all of which have compact support. The self-similar solution with parameter \(\beta <1\) behaves in the same way toward the end of its support as does the CP self-similar solution with parameter \(\beta \). It has been conjectured (Niethammer 1999) that a solution of the LSW model converges at large time to the LSW self-similar solution with parameter \(\beta \), provided the initial data and the self-similar solution of parameter \(\beta \) behave in the same way at the end of their supports. A weak version of this result has been proven in Conlon and Niethammer (2014).

It was already claimed in Lifschitz and Slyozov (1961) and Wagner (1961) that the only physically relevant self-similar LSW solution is the one with parameter \(\beta =1\). This has been explained in a heuristic way in several papers (Meerson 1999; Rubinstein and Zaltzman 2000; Velázquez 1998), by considering a model in which a second-order diffusion term is added to the first-order LSW equation. It is then argued that diffusion acts as a selection principle, which singles out the \(\beta =1\) self-similar solution as giving the large time behavior. In this paper, we study a diffusive version of the Carr–Penrose model, with the goal of understanding how a selection principle for the \(\beta =1\) self-similar solution (1.5) operates.

In our diffusive CP model, we simply add a second-order diffusion term with coefficient \(\varepsilon /2>0\) to the CP equation (1.1). Then, the density function \(c_\varepsilon (x,t)\) evolves according to a linear diffusion equation, subject to the linear mass conservation constraint as follows:

$$\begin{aligned}&\frac{\partial c_\varepsilon (x,t)}{\partial t}= \frac{\partial }{\partial x}\left\{ \left[ 1-\frac{x}{\Lambda _\varepsilon (t)}\right] c_\varepsilon (x,t)\right\} +\frac{\varepsilon }{2}\frac{\partial ^2 c_\varepsilon (x,t)}{\partial x^2}, \quad x>0, \end{aligned}$$
(1.7)
$$\begin{aligned}&\int _0^\infty x c_\varepsilon (x,t) \hbox {d}x = 1. \end{aligned}$$
(1.8)

We also need to impose a boundary condition at \(x=0\) to ensure that (1.7), (1.8) with given initial data \(c_\varepsilon (x,0) = c_0(x) \ge 0, \ x > 0\), satisfying the constraint (1.8) has a unique solution. We impose the Dirichlet boundary condition \(c_\varepsilon (0,t)=0, \ t>0,\) because in this case the parameter \(\Lambda _\varepsilon (t) > 0\) in (1.7) is given by the formula

$$\begin{aligned} \Lambda _\varepsilon (t) = \int ^\infty _0 \ x c_\varepsilon (x,t)\hbox {d}x \Big / \int ^\infty _0 c_\varepsilon (x,t) \hbox {d}x. \end{aligned}$$
(1.9)

Hence, the diffusive CP model is an evolution equation for the pdf \(c_\varepsilon (\cdot ,t)/\int _0^\infty c_\varepsilon (x,t) \hbox {d}x\) of a random variable \(X_{\varepsilon ,t}\) and \(\Lambda _\varepsilon (t)=\langle X_{\varepsilon ,t}\rangle \). Furthermore, it is easy to see from (1.7), (1.8) that

$$\begin{aligned} \frac{\hbox {d}\Lambda _\varepsilon (t)}{\hbox {d}t} =\frac{\varepsilon }{2}\frac{\partial c_\varepsilon (0,t)}{\partial x}\Big /\left[ \int ^\infty _0 c_\varepsilon (x,t) \hbox {d}x\right] ^2. \end{aligned}$$
(1.10)

It follows from (1.10) and the maximum principle (Protter and Weinberger 1984) applied to (1.7) that the function \(t\rightarrow \Lambda _\varepsilon (t)\) is increasing.

Smereka (2008) studied a discretized CP model and rigorously established a selection principle for arbitrary initial data with finite support. He also proved that the rate of convergence to the \(\beta =1\) self-similar solution (1.5) is logarithmic in time. Since discretization of a first-order PDE introduces an effective diffusion, one can just as well apply the discretization algorithm of Smereka (2008) to (1.7). In the discretized model of Smereka, time t is left continuous and the x discretization \(\Delta x\) for (1.7) is required to satisfy the condition \(\varepsilon =2\Delta x\). In Smereka (2008), the large time behavior of solutions to this discretized model is studied by using a Fourier method. The Fourier method cannot be implemented if the assumption \(\varepsilon =2\Delta x\) is dropped.

In Sect. 2, we discuss a variety of models, which in some sense interpolate between the CP model (1.1), (1.2) and the diffusive CP model (1.7), (1.8). We first show that the CP and the discretized diffusive model are algebraically equivalent if and only if \(\varepsilon =2\Delta x\). In particular, they are associated with the unique two-dimensional non-Abelian Lie algebra. Next we study the CP model (1.1), (1.2) with Gaussian initial data. It follows from Carr and Penrose (1998) that the solution converges at large time to the \(\beta =1\) self-similar solution (1.5). We prove that the rate of convergence is logarithmic in time as in the Smereka model. The solution \(c_\varepsilon (x,t)\) to (1.7) with initial data of compact support has the property that for \(t>0\) the function \(c_\varepsilon (x,t)\) is approximately Gaussian at large x. Based on this observation, we conjecture that for a large class of initial data with compact support, the solution \(c_\varepsilon (\cdot ,t)\) to the diffusive CP model (1.7), (1.8) converges as \(t\rightarrow \infty \) to the \(\beta =1\) self-similar solution (1.5). Furthermore, the rate of convergence is logarithmic in time.

In the remainder of the paper, we take some steps in the direction of proving this conjecture. First in Sect. 2, we study the semiclassical approximation to the solution of (1.7). This leads us, as in the classical works of Hopf (1950) and Lax (1973), to a Burgers’ equation describing the approximate evolution of the diffusive CP model. The solution \(v_\varepsilon (x,t), \ x,t>0,\) of the Burgers’ equation is related to the solution \(c_\varepsilon (x,t)\) of (1.7), (1.8) as follows: There exists a positive random variable \(X_t\) such that \(v_\varepsilon (x,t)=1/E[X_t-x \ | \ X_t>x]\) and \(X_t\) approximates the random variable with pdf \(c_\varepsilon (x,t)/\int _0^\infty c_\varepsilon (x',t) \ \hbox {d}x'\). We relate the Burgers’ model to the diffusive CP model by introducing an interpolating family of models with parameter \(\nu , \ 0\le \nu \le 1\). Each of these models is an evolution equation for a function \(x\rightarrow 1/E[X_t-x \ | \ X_t>x]\), where \(X_t, \ t>0,\) are positive random variables with the property that the function \(t\rightarrow \langle X_t\rangle \) is increasing. The evolution PDE is of viscous Burgers’ type (Hopf 1950) with viscosity coefficient proportional to \(\nu \). The \(\nu =1\) model is identical to the diffusive CP model (1.7), (1.8), while the \(\nu =0\) model is the Burgers’ model. We shall refer to the Burgers’ model as the inviscid CP model since its evolution PDE is an inviscid Burgers’ equation (Smoller 1994). Similarly, we refer to the model with \(0<\nu \le 1\) as the viscous CP model with viscosity \(\nu \). Our hierarchy of models can be summarized as follows:

CP model \((\varepsilon =\nu =0)\rightarrow \) inviscid CP model \((\varepsilon >0, \ \nu =0)\rightarrow \) viscous CP model \((\varepsilon >0,\ 0<\nu \le 1)\rightarrow \) diffusive CP model \((\varepsilon >0, \nu =1)\).

In Sect. 3, we study the large time behavior of the inviscid CP model and establish a version of our conjecture for this model. We obtain the following theorem:

Theorem 1.1

Suppose the initial data for the inviscid CP model correspond to the nonnegative random variable \(X_0\), and assume that \(X_0\) satisfies

$$\begin{aligned}&\varepsilon < \langle X_0 \rangle , \ \Vert X_0\Vert _\infty <\infty , \quad x\rightarrow E[X_0-x \ | \ X_0>x] \ \mathrm{is \ decreasing \ for \ } \nonumber \\&0\le x<\Vert X_0\Vert _\infty . \end{aligned}$$
(1.11)

Then, \(\lim _{t\rightarrow \infty } \langle X_t\rangle /t=1\), and for any \(\eta ,m,M>0\) with \(m<M\), there exists \(T>0\) such that

$$\begin{aligned} \exp \left[ -\frac{(1+\eta )x}{t}\right] \le P(X_t>x) \le \exp \left[ -\frac{x}{(1+\eta )t}\right] \quad \mathrm{for \ } t>T, \ mt\le x\le Mt . \end{aligned}$$
(1.12)

Assume in addition that the function \(x\rightarrow E[X_0-x \ | \ X_0>x] \) is \(C^1\) and convex for x close to \(\Vert X_0\Vert _\infty \) with

$$\begin{aligned} \liminf _{x\rightarrow \Vert X_0\Vert _\infty } \frac{\partial }{\partial x} \ \frac{1}{E[X_0-x \ | \ X_0>x] } > 0 . \end{aligned}$$
(1.13)

Then, there exists \(C,M,T>0\) such that the mean of \(X_t\) satisfies the inequality

$$\begin{aligned} 1-\frac{C}{\log t} \le \frac{\hbox {d}}{\hbox {d}t} \langle X_t\rangle \le 1 \quad \mathrm{for \ } \ t\ge T , \end{aligned}$$
(1.14)

and the distribution function of \(X_t\) the inequality

$$\begin{aligned}&\exp \left[ -\frac{x}{t}\left\{ 1+\frac{C}{\log t}\right\} \right] \le P(X_t>x) \nonumber \\&\quad \le \exp \left[ -\frac{x}{t}\left\{ 1-\frac{C}{\log t}\right\} \right] \quad \mathrm{for \ } t>T, \ 0\le x\le Mt\log t . \end{aligned}$$
(1.15)

Remark 1

Observe that a \(\beta <1\) self-similar solution (1.5) of the CP model has \(\Vert X_0\Vert _\infty =1/(1-\beta )\) and \(E[X_0-x \ | \ X_0>x]=1-(1-\beta )x, \ 0\le x<\Vert X_0\Vert _\infty \). Hence, the \(\beta <1\) self-similar solution satisfies all the conditions of Theorem 1.1 provided \(\varepsilon <1\). The condition \(\varepsilon <1\) is not crucial since for any \(\varepsilon >0\) one can rescale the initial data so that \(\varepsilon <\langle X_0\rangle \). Therefore, Theorem 1.1 proves a selection principle for the \(\beta =1\) self-similar solution (1.5) and establishes a rate of convergence which is logarithmic in time.

Remark 2

The condition (1.11) implies that the initial condition \(v_\varepsilon (x,0), \ x>0,\) for the inviscid Burgers’ equation is increasing. Hence, the Burgers’ equation may be solved by means of the method of characteristics. If \(v_\varepsilon (\cdot ,0)\) is not an increasing function, then the solution of the Burgers’ equation at some positive time contains shocks, corresponding to discontinuities of the function \(v_\varepsilon (\cdot ,t)\). A shock at time t gives rise to an atom in the distribution of the random variable \(X_t\).

The remainder of the present paper is devoted to the study of the diffusive CP model (1.7), (1.8). Since existence and uniqueness has already been proven for a diffusive version of the LSW model (Conlon 2010), we do not revisit this issue. In Sect. 6, we consider the problem of proving convergence of solutions of the diffusive CP model as \(\varepsilon \rightarrow 0\) to a solution of the CP model over some fixed time interval \(0\le t\le T\). Thus, we assume that the CP and diffusive CP models have the same initial data corresponding to a random variable \(X_0\). Let \(X_t, \ t>0,\) be the random variable corresponding to the solution of (1.1), (1.2) and \(X_{\varepsilon ,t}, t>0,\) the random variable corresponding to the solution of (1.7), (1.8). We show that \(X_{\varepsilon ,t}\) converges in distribution as \(\varepsilon \rightarrow 0\) to \(X_t\), uniformly in the interval \(0\le t\le T\). We also prove convergence of the diffusive coarsening rates (1.10) as \(\varepsilon \rightarrow 0\) to the CP coarsening rate (1.4). One easily sees from the formula (1.10) that a boundary layer analysis becomes necessary in this case. In the diffusive model, there exists a boundary layer with length of order \(\varepsilon \) so that \(c_\varepsilon (x,t)\simeq c_0(x,t)\) for \(x/\varepsilon >>1\). Since \(c_\varepsilon (0,t)=0\) one has that \(\partial c_\varepsilon (0,t)/\partial x\simeq 1/\varepsilon \), whence the RHS of (1.10) remains bounded above 0 as \(\varepsilon \rightarrow 0\) and in fact converges to the RHS of (1.4).

In Sect. 7, we concentrate on understanding large time behavior of the diffusive CP model for fixed \(\varepsilon >0\). We carry this out by relating it to the problem discussed in the previous paragraph of proving convergence as \(\varepsilon \rightarrow 0\) over a finite time interval. To see why the two problems are related, observe that one can always rescale \(\langle X_0\rangle \) to be equal to 1 in both the CP and diffusive CP models. Since the CP model is dilation invariant, the evolution PDE (1.1) remains the same. However, for the diffusive CP model the diffusion coefficient in (1.7) changes from \(\varepsilon \) to \(\varepsilon /\langle X_0\rangle \). Since \(\lim _{t\rightarrow \infty } \langle X_t\rangle =\infty \) in the diffusive CP model, an analysis of large time behavior can be made equivalent to an analysis of solutions to (1.7), (1.8) as \(\varepsilon \rightarrow 0\). In order to show that the conclusions of Theorem 1.1 also hold for the diffusive CP model, one needs to accomplish two objects: (a) prove that solutions to the viscous CP model are close to solutions of the inviscid model away from the boundary; (b) establish good control of the solution in the boundary layer. We do not address the problem of proving (a) in this paper. Instead, we prove in Sect. 7 the much simpler result that log-concave functions form an invariant set for the linear diffusion Eq. (1.7). To address (b) we apply the results of Sect. 5 which give bounds on the ratio of two Green’s functions for the PDE (1.7), the Dirichlet Green’s function for the half-line and the Green’s function for the full line. These bounds are uniform in \(\varepsilon \) as \(\varepsilon \rightarrow 0\). We also use the results of Sect. 5 in Sect. 6 to prove convergence of coarsening rates as \(\varepsilon \rightarrow 0\). However, this can be accomplished with less delicate estimates, as was carried out in Conlon (2010) for the case of the diffusive LSW model. Our main result on the large time behavior of the diffusive CP model is a bound on the coarsening rate, which is uniform is time:

Theorem 1.2

Suppose the initial data for the diffusive CP model (1.7), (1.8) correspond to the nonnegative random variable \(X_0\) with integrable pdf. Then, \(\lim _{t\rightarrow \infty } \langle X_t\rangle =\infty \). If in addition the function \( x\rightarrow E[X_0-x \ | \ X_0>x], \ 0\le x<\Vert X_0\Vert _\infty ,\) is decreasing, then there are constants \(C,T>0\) such that

$$\begin{aligned} 0 \le \frac{\mathrm{d}}{\mathrm{d}t} \langle X_t\rangle \le C \quad \mathrm{for \ } t\ge T . \end{aligned}$$
(1.16)

It is unnecessary to read Sects. 4, 5 in order to follow the proof of Theorem 1.2. However, these sections do begin to address a core difficulty in trying to extend the results of Theorem 1.1 to the viscous CP model. For the inviscid CP model, as for the CP model itself, there is no boundary condition at \(x=0\). In the case of the viscous CP model, a boundary condition at \(x=0\) becomes necessary in order to have uniqueness of the solution. In this paper, we have chosen the Dirichlet condition \(c_\varepsilon (0,t)=0, \ t>0\), for (1.7) because the function \(\Lambda _\varepsilon (\cdot )\) is given by the same formula as in the CP model and is increasing. Beyond this we do not have a justification for preferring the Dirichlet condition over other boundary conditions. That said, it seems likely that no matter what boundary condition at \(x=0\), a delicate boundary layer analysis will be required in order to understand large time behavior of the model. Our belief is that this boundary layer analysis is the simplest in the case of the Dirichlet condition.

To prove the ratio of Greens’ functions results in Sect. 5, we use Green’s function representations derived from semiclassical analysis. The derivations are carried out in Sect. 4 using two parallel approaches: the variational method of stochastic control theory and the probabilistic method using conditioned Markov processes. The variational method is a generalization of the approach of Hopf (1950) and Lax (1973) to the study of Burgers’ equation and hence shows the connection with the inviscid CP model studied in Sect. 3. In the probabilistic approach, we obtain a representation for a generalized Brownian bridge process in terms of Brownian motion. It is this representation which is used in Sect. 5 to prove the estimates on the ratio of Greens’ functions.

2 The Carr–Penrose Model and Extensions

The analysis of the CP model (Carr and Penrose 1998) is based on the fact that the characteristics for the first-order PDE (1.1) can be easily computed. Thus, let \(b:\mathbf {R}\times \mathbf {R}^+\rightarrow \mathbf {R}\) be given by \(b(y,s)=A(s)y-1, \ y\in \mathbf {R}, \ s\ge 0\), where \(A:\mathbf {R}^+\rightarrow \mathbf {R}^+\) is a continuous nonnegative function. We define the mapping \(F_A:\mathbf {R}\times \mathbf {R}^+\rightarrow \mathbf {R}\) by setting

$$\begin{aligned} F_A(x,t)=y(0), \quad \mathrm{where \ } \frac{\hbox {d}y(s)}{\hbox {d}s}=b(y(s),s), \ 0\le s\le t, \quad y(t)=x. \end{aligned}$$
(2.1)

From (2.1), we see that the function \(F_A\) is given by the formula

$$\begin{aligned} F_A(x,t)= & {} \frac{x+m_{2,A}(t)}{m_{1,A}(t)} , \quad \mathrm{where \ } \nonumber \\ m_{1,A}(t)= & {} \exp \left[ \int _0^tA(s) \ \hbox {d}s\right] , \quad m_{2,A}(t)=\int _0^t \exp \left[ \int _s^tA(s') \hbox {d}s'\right] \hbox {d}s . \end{aligned}$$
(2.2)

If we let \(w_0:\mathbf {R}^+\times \mathbf {R}^+\rightarrow \mathbf {R}^+\) be the function

$$\begin{aligned} w_0(x,t) =\int _x^\infty c_0(x',t) \hbox {d}x', \quad x,t\ge 0, \end{aligned}$$
(2.3)

where \(c_0(\cdot ,\cdot )\) is the solution to (1.1), then from the method of characteristics we have that

$$\begin{aligned} w_0(x,t) =w_0(F_{1/\Lambda _0}(x,t), 0) , \quad x,t\ge 0. \end{aligned}$$
(2.4)

The conservation law (1.2) can also be expressed in terms of \(w_0\) as

$$\begin{aligned} \int _0^\infty w_0(x,t) \ \hbox {d}x =\int _0^\infty w_0(F_{1/\Lambda _0}(x,t), 0) \ \hbox {d}x =1. \end{aligned}$$
(2.5)

Observe now that the functions \(m_{1,A}, \ m_{2,A}\) of (2.2) are related by the differential equation

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\left[ \frac{m_{2,A}(t)}{m_{1,A}(t)}\right] =\frac{1}{m_{1,A}(t)} . \end{aligned}$$
(2.6)

It follows from (2.5), (2.6) that if we define variables \([u(t),v(t)], t\ge 0,\) by

$$\begin{aligned} u(t) =\frac{1}{m_{1,1/\Lambda _0}(t)} , \quad v(t) =\frac{m_{2,1/\Lambda _0}(t)}{m_{1,1/\Lambda _0}(t)} , \end{aligned}$$
(2.7)

then the CP model (1.1), (1.2) with given initial data \(c_0(\cdot ,0)\) is equivalent to the two-dimensional dynamical system

$$\begin{aligned} \frac{\hbox {d}v(t)}{\hbox {d}t} = u(t) , \quad \frac{\hbox {d}}{\hbox {d}t} \int _0^\infty w_0\left( u(t)x+v(t),0\right) \hbox {d}x =0. \end{aligned}$$
(2.8)

Note, however, that the dynamical law for the two-dimensional evolution depends on the initial data for (1.1), (1.2), whereas the initial condition is always \(u(0)=1, \ v(0)=0\).

We can understand the two-dimensionality of the CP model and relate it to some other models of coarsening by using some elementary Lie algebra theory. Thus, observe that for operators \({\mathcal {A}}_0,\mathcal {B}_0\) acting on functions \(f:\mathbf {R}\rightarrow \mathbf {R}\), which are defined by

$$\begin{aligned} {\mathcal {A}}_0f(x) =\frac{\hbox {d}}{\hbox {d}x}f(x) , \quad \mathcal {B}_0f(x)=\frac{\hbox {d}}{\hbox {d}x}\left[ xf(x)\right] , \quad \mathrm{then \ } {\mathcal {A}}_0\mathcal {B}_0-\mathcal {B}_0{\mathcal {A}}_0 ={\mathcal {A}}_0 . \end{aligned}$$
(2.9)

The initial value problem (1.1) can be written in operator notation as

$$\begin{aligned} \frac{\partial c_0(\cdot ,t)}{\hbox {d}t} = \left[ {\mathcal {A}}_0-\frac{\mathcal {B}_0}{\Lambda _0(t)}\right] c_0(\cdot , t) \quad \mathrm{for }\; t>0, \quad c_0(\cdot ,0) =\mathrm{given}. \end{aligned}$$
(2.10)

It follows from (2.9) that the Lie algebra generated by \({\mathcal {A}}_0,\mathcal {B}_0\) is the unique two-dimensional non-Abelian Lie algebra. The corresponding two-dimensional Lie group is the affine group of the line (see Chapter 4 of Stillwell 2008). That is the Lie group consists of all transformations \(z\rightarrow az+b, \ z\in \mathbf {R},\) with \(a>0, \ b\in \mathbf {R}\). The solutions of Eq. (2.10) are a flow on this group. Hence, solutions of (2.10) for all possible functions \(\Lambda _0(\cdot )\) lie on a two-dimensional manifold.

Next we consider the discretized version of the CP model studied by Smereka (2008). Letting \(\Delta x\) denote space discretization, then a standard discretization of (1.7) with Dirichlet boundary condition is given by

$$\begin{aligned}&\frac{\partial c_\varepsilon (x,t)}{\partial t}+ \frac{J_\varepsilon (x,t)-J_\varepsilon (x-\Delta x,t)}{\Delta x} \nonumber \\&\quad = \frac{\varepsilon }{2}\frac{c_\varepsilon (x+\Delta x,t)+c_\varepsilon (x-\Delta x,t)-2c_\varepsilon (x,t)}{(\Delta x)^2} , \quad x=(n+1)\Delta x, \ \ n=0,1,2,\ldots ,\nonumber \\ \end{aligned}$$
(2.11)

where

$$\begin{aligned} J_\varepsilon (x,t) =\left[ \frac{x}{\Lambda _\varepsilon (t)}-1\right] c_\varepsilon (x,t) , \quad c_\varepsilon (0,t) =0. \end{aligned}$$
(2.12)

The backward difference approximation for the derivative of \(J_\varepsilon (x,t)\) is chosen in (2.11) to ensure stability of the numerical scheme for large x. Let \(D,D^*\) be the discrete derivative operators acting on functions \(u:(\Delta x)\mathbf {Z}\rightarrow \mathbf {R}\) defined by

$$\begin{aligned} Du(x) =\frac{u(x+\Delta x)-u(x)}{\Delta x} , \quad D^*u(x) =\frac{u(x-\Delta x)-u(x)}{\Delta x} . \end{aligned}$$
(2.13)

Then, using the notation of (2.13) we can rewrite (2.11) as

$$\begin{aligned} \frac{\partial c_\varepsilon (x,t)}{\partial t}-D^*J_\varepsilon (x,t) =\frac{\varepsilon }{2\Delta x} \left[ (D+D^*) c_\varepsilon (x,t)\right] . \end{aligned}$$
(2.14)

Observe that for operators \({\mathcal {A}}_{\Delta x},\mathcal {B}_{\Delta x}\) defined by

$$\begin{aligned} {\mathcal {A}}_{\Delta x} =D , \quad \mathcal {B}_{\Delta x}= -D^*x , \quad \mathrm{then \ } {\mathcal {A}}_{\Delta x}\mathcal {B}_{\Delta x}-\mathcal {B}_{\Delta x}{\mathcal {A}}_{\Delta x} ={\mathcal {A}}_{\Delta x} . \end{aligned}$$
(2.15)

Choosing \(\varepsilon = 2\Delta x\) in (2.14), we see that the equation can be expressed in terms of \({\mathcal {A}}_{\Delta x},\mathcal {B}_{\Delta x}\) as

$$\begin{aligned} \frac{\partial c_\varepsilon (\cdot ,t)}{\partial t} =\left[ {\mathcal {A}}_{\Delta x}-\frac{\mathcal {B}_{\Delta x}}{\Lambda _\varepsilon (t)}\right] c_\varepsilon (\cdot ,t) . \end{aligned}$$
(2.16)

Comparing (2.9), (2.10) to (2.15), (2.16), we see that we can obtain a representation for the solution to (2.15), (2.16) by using the fact that the solution to (2.9), (2.10) is given by (2.4). To see this, we use the fact that for \({\mathcal {A}}_0,\mathcal {B}_0\) as in (2.9) then

$$\begin{aligned} e^{{\mathcal {A}}_0s} f(x) =f(x+s) , \quad e^{\mathcal {B}_0s}f(x) =e^sf(e^s x) , \quad x,s\in \mathbf {R}. \end{aligned}$$
(2.17)

From (2.4), (2.7), (2.17), it follows that the solution to (2.9), (2.10) is given by

$$\begin{aligned} c_0(\cdot ,t) =u(t)^{\mathcal {B}_0} e^{{\mathcal {A}}_0v(t)} c_0(\cdot ,0) , \end{aligned}$$
(2.18)

where u(t), v(t) are given by (2.7). Hence, the solution to (2.15), (2.16) is given by

$$\begin{aligned} c_\varepsilon (\cdot ,t) =u(t)^{\mathcal {B}_{\Delta x}} e^{{\mathcal {A}}_{\Delta x}v(t)} c_\varepsilon (\cdot ,0) , \end{aligned}$$
(2.19)

where \({\mathcal {A}}_{\Delta x},\mathcal {B}_{\Delta x}\) are given by (2.15) and u(t), v(t) are given by (2.7) with \(\Lambda _\varepsilon \) in place of \(\Lambda _0\).

The operator \({\mathcal {A}}_{\Delta x}\) of (2.15) is the generator of a Poisson process. Thus,

$$\begin{aligned} \int _{(\Delta x)\mathbf {Z}} \hbox {d}x g(x)e^{{\mathcal {A}}_{\Delta x}s}f(x) =\int _{(\Delta x)\mathbf {Z}} \hbox {d}x E[g(X_s) \ | \ X_0=x] f(x) , \end{aligned}$$
(2.20)

where \(X_s\) is the discrete random variable taking values in \((\Delta x)\mathbf {Z}\) with pdf

$$\begin{aligned} P\left( X_s= y\ \big | \ X_0=x\right) =\frac{(s/\Delta x)^n}{n!}\exp \left[ -\frac{s}{\Delta x}\right] , \quad n =\frac{x-y}{\Delta x} , \quad n=0,1,2,\ldots \end{aligned}$$
(2.21)

If \(f,g:\mathbf {R}\rightarrow \mathbf {R}\) are continuous functions of compact support, then it is easy to see from (2.21) that

$$\begin{aligned} \lim _{\Delta x\rightarrow 0} \int _{(\Delta x)\mathbf {Z}} \hbox {d}x \ g(x)e^{{\mathcal {A}}_{\Delta x}s}f(x) =\int _{-\infty }^\infty \hbox {d}x \ g(x-s)f(x) , \end{aligned}$$
(2.22)

as we expect from (2.17). The operator \(\mathcal {B}_{\Delta x}\) of (2.15) is the generator of a Yule process (Karlin and Taylor 1975). Letting \(\mathbf {Z}^+\) denote the positive integers, the operator \(e^{-\mathcal {B}_{\Delta x}s}\) for \(s>0\) acts on functions \(f:(\Delta x)\mathbf {Z}^+\rightarrow \mathbf {R}\). Then,

$$\begin{aligned} \int _{(\Delta x)\mathbf {Z}^+} \hbox {d}x \ g(x)e^{-\mathcal {B}_{\Delta x}s}f(x) =\int _{(\Delta x)\mathbf {Z}^+} \hbox {d}x \ E[g(Y_s) \ | \ Y_0=x] f(x) , \end{aligned}$$
(2.23)

where \(Y_s\) is a discrete random variable taking values in \((\Delta x)\mathbf {Z}^+\). The pdf of \(Y_s\) conditioned on \(Y_0=\Delta x\) is given by

$$\begin{aligned} P\left( Y_s= y\ \big | \ Y_0=\Delta x\right) =e^{-s}\left\{ 1- e^{-s}\right\} ^{n-1} , \quad n =\frac{y}{\Delta x} , \quad n=1,2,\ldots \end{aligned}$$
(2.24)

Hence, \(Y_s\) conditioned on \(Y_0=\Delta x\) is a geometric variable. More generally, the variable \(Y_s\) conditioned on \(Y_0=m\Delta x\) with \(m\ge 2\) is a sum of m independent geometric variables with distribution (2.24) and is hence negative binomial. The identity (2.23) therefore implies that

$$\begin{aligned} \int _{(\Delta x)\mathbf {Z}^+} \hbox {d}x \ g(x)e^{-\mathcal {B}_{\Delta x}s}f(x) =\int _{(\Delta x)\mathbf {Z}^+} \hbox {d}x \ E\left[ g\left( Y_s^1+\cdots + Y_s^m\right) \ \big | \ m=x/\Delta x\right] f(x) , \end{aligned}$$
(2.25)

where the \(Y_s^j, \ j=1,2,\ldots ,\) are independent and have the distribution (2.24). Since the mean of \(Y_s\) is \(e^s\Delta x\), it follows from (2.25) that

$$\begin{aligned} \lim _{\Delta x\rightarrow 0} \int _{(\Delta x)\mathbf {Z}^+} \hbox {d}x \ g(x)e^{-\mathcal {B}_{\Delta x}s}f(x) =\int _0^\infty \hbox {d}x\ g(e^s x)f(x) , \end{aligned}$$
(2.26)

as we expect from (2.17).

The Smereka model consists of the evolution determined by (2.11) with \(\varepsilon =2\Delta x\) and the conservation law

$$\begin{aligned} \int _{(\Delta x)\mathbf {Z}^+} xc_\varepsilon (x,t) \hbox {d}x =1. \end{aligned}$$
(2.27)

We see from (2.19) that the model is equivalent to a two-dimensional dynamical system with dynamical law depending on the initial data. The first differential equation in this system is given by the first equation in (2.8). The second differential equation is determined by differentiating the expression on the LHS of (2.27) and setting it equal to zero. Using (2.21), (2.24), we can write the LHS of (2.27) in terms of u(t), v(t). In the case when the initial data are given by

$$\begin{aligned} c_\varepsilon (x,0) =0 \ \mathrm{if \ } x\ne \Delta x , \quad c_\varepsilon (\Delta x,0) =\frac{1}{(\Delta x)^2} , \end{aligned}$$
(2.28)

it has a simple form. Thus, from (2.19), (2.21), (2.24) we have that

$$\begin{aligned}&c_\varepsilon (\cdot ,t) =u(t)^{\mathcal {B}_{\Delta x}} e^{{\mathcal {A}}_{\Delta x}v(t)} c_\varepsilon (\cdot ,0) =u(t)^B \exp \left[ -\frac{v(t)}{\Delta x}\right] c_\varepsilon (\cdot ,0) , \nonumber \\&\quad \mathrm{so \ \ } c_\varepsilon (x,t) =u(t)\left[ 1-u(t)\right] ^{n-1}\exp \left[ -\frac{v(t)}{\Delta x}\right] \frac{1}{(\Delta x)^2} , \quad n= \frac{x}{\Delta x}. \end{aligned}$$
(2.29)

From (2.29), we see that the conservation law (2.27) becomes in this case

$$\begin{aligned} \frac{1}{u(t)} \exp \left[ -\frac{v(t)}{\Delta x}\right] =1. \end{aligned}$$
(2.30)

Hence, from the first equation of (2.8) and (2.30) we conclude that \(v(\cdot )\) is the solution to the initial value problem

$$\begin{aligned} \exp \left[ -\frac{v(t)}{\Delta x}\right] =\frac{\hbox {d}v(t)}{\hbox {d}t}, \quad v(0)=0. \end{aligned}$$
(2.31)

The initial value problem (2.31) was derived in §3 of Smereka (2008) by a different method. It can be solved explicitly, and so we obtain the formulas

$$\begin{aligned} u(t) = \frac{1}{1+t/\Delta x} , \quad v(t) =\Delta x \log \left[ 1+\frac{t}{\Delta x}\right] , \end{aligned}$$
(2.32)

when the initial data are given by (2.28). Hence, from (2.29), (2.32) we have an explicit expression for \(c_\varepsilon (\cdot ,t)\), and it is easy to see that this converges as \(t\rightarrow \infty \) to the self-similar solution corresponding to the \(\beta =1\) random variable defined by (1.5). It was also shown in Smereka (2008) that if the initial data have finite support then \(c_\varepsilon (\cdot ,t)\) converges as \(t\rightarrow \infty \) to the \(\beta =1\) self-similar solution.

The large time behavior of the CP model can be easily understood using the beta function of a random variable introduced in Conlon (2011). If X is a random variable with pdf \(c_X(\cdot )\), we define functions \(w_X(\cdot ), h_X(\cdot )\) by

$$\begin{aligned} w_X(x) = \int _x^\infty c_X(x') \hbox {d}x' , \quad h_X(x) =\int _x^\infty w_X(x') \ \hbox {d}x' , \quad x\in \mathbf {R}. \end{aligned}$$
(2.33)

Evidently, one has that

$$\begin{aligned} w_X(x) =P(X>x), \ \quad \frac{h_X(x)}{w_X(x)} =E\left[ X-x \ \big | \ X>x\right] \quad x\in \mathbf {R}. \end{aligned}$$
(2.34)

The beta function \(\beta _X(\cdot )\) of X is then defined by

$$\begin{aligned} \beta _X(x) =\frac{c_X(x)h_X(x)}{w_X(x)^2} =1+\frac{\hbox {d}}{\hbox {d}x} E\left[ X-x \ \big | \ X>x\right] \ \quad x\in \mathbf {R}. \end{aligned}$$
(2.35)

An important property of the beta function is that it is invariant under affine transformations. That is

$$\begin{aligned} \beta _X(\lambda x+\mu ) =\beta _{(X-\mu )/\lambda }(x) , \quad \lambda >0, \ \mu ,x\in \mathbf {R}. \end{aligned}$$
(2.36)

One can also see that the function \(h_X(\cdot )\) is log-concave if and only if \(\sup \beta _X(\cdot )\le 1\).

To understand the large time behavior of the CP model, we first observe that the rate of coarsening Eq. (1.4) can be rewritten as

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t} \langle X_t\rangle =\beta _{X_t}(0) , \quad t>0. \end{aligned}$$
(2.37)

Furthermore, the beta function of the self-similar variable \(X_0\) with pdf defined by (1.5) and parameter \(\beta >0\) is simply a constant \(\beta _{X_0}(\cdot )\equiv \beta \). We have already shown that the time evolution of the CP equation (1.1) is given by the affine transformation (2.4). It is also relatively simple to establish that for a random variable \(X_0\) corresponding to the initial data for (1.1), (1.2), then \(\lim _{t\rightarrow \infty } F_{1/\Lambda _0}(0,t)=\Vert X_0\Vert _\infty \). Hence, it follows from (2.35), (2.36) that if \(\lim _{x\rightarrow \Vert X_0\Vert _\infty }\beta _{X_0}(x)=\beta \) for the initial data random variable \(X_0\) of (1.1), (1.2), then the large time behavior of the CP model is determined by the self-similar solution (1.5) with parameter \(\beta \).

We have already observed from (1.4) that the function \(\Lambda _0(\cdot )\) in the CP model is increasing. If we assume that \(\inf \beta _{X_0}(\cdot )>0\), we can also see that \(\lim _{t\rightarrow \infty } \Lambda _0(t)=\infty \). Hence, in this case there exists a doubling time \(T_\mathrm{double}\) for which \(\Lambda _0(t)=2\Lambda _0(0)\) when \(t=T_\mathrm{double}\). Evidently, \(\inf \beta _{X_t}(\cdot )\ge \inf \beta _{X_0}(\cdot )\) and \(\sup \beta _{X_t}(\cdot )\le \sup \beta _{X_0}(\cdot )\). The notion of doubling time can be a useful tool in obtaining an estimate on the rate of convergence of the solution of the CP model to a self-similar solution at large time.

We illustrate this by considering the CP model with Gaussian initial data. In particular, we assume the initial data \(c_0(\cdot )\) are given by the formula

$$\begin{aligned} c_0(x) =K(L)\exp \left[ -a(L)x-\{a(L)x\}^2/2L\right] , \end{aligned}$$
(2.38)

where \(L\ge L_0>0\) and K(L), a(L) are uniquely determined by the requirement that (1.2) holds and the function \(\Lambda _0(\cdot )\) in (1.3) satisfies \(\Lambda _0(0)=1\). It is easy to see that the beta function for the initial data (2.38) is bounded above and below strictly larger than zero, uniformly in \(L\ge L_0\). Hence, from (2.37) there are constants \(T_1,T_2>0\) depending only on \(L_0\) such that \(T_1\le T_\mathrm{double}\le T_2\) for all \(L\ge L_0\). It follows also from (2.2) that there are constants \(\lambda _0,\lambda _1,\mu _0,\mu _1>0\) depending only on \(L_0\) such that \(F_{1/\Lambda _0}(x,T_\mathrm{double})=\lambda (L)x+\mu (L), \ x\in \mathbf {R},\) where \(0<\lambda _0\le \lambda (L)\le \lambda _1<1\) and \( 0<\mu _0\le \mu (L)\le \mu _1\) for \(L\ge L_0\). Since \(F_{1/\Lambda _0}\) is a linear function, \(c(\cdot ,T_\mathrm{double})\) is also Gaussian. Rescaling so that the mean of \(X_t\) is now 1 at time \(T_\mathrm{double}\), we see that \(c(x,T_\mathrm{double})\) is given by the formula (2.38) with L replaced by A(L), where

$$\begin{aligned} A(L) = L[1+\mu (L)a(L)/L]^2=L+2\mu (L)a(L)+\mu (L)^2a(L)^2/L . \end{aligned}$$
(2.39)

Since we are assuming that \(\Lambda _0(0)=1\), there are constants \(a_0,a_1>0\) depending only on \(L_0\) such that \(a_0\le a(L)\le a_1\) for \(L\ge L_0\). We conclude then from (2.39) that

$$\begin{aligned} L+\delta _0 \le A(L) \le L+\delta _1 , \quad \mathrm{where \ } \delta _0,\delta _1>0 \ \mathrm{depend \ only \ on \ }L_0 . \end{aligned}$$
(2.40)

It is easy to estimate from (2.40) the rate of convergence to the \(\beta =1\) self-similar solution \(c(x,t)=(1+t)^{-2}\exp [-x/(1+t)]\) for solutions to the CP model with Gaussian initial data. First, we estimate the beta function of a Gaussian random variable.

Lemma 2.1

Let \(L>0\) and \(Z_L\) be a positive random variable with pdf proportional to \(e^{-z-z^2/2L}, \ z>0\). Then, for any \(L_0>0\) there is a constant C depending only on \(L_0\) such that if \(L\ge L_0\) the beta function \(\beta _L\) for \(Z_L\) satisfies the inequality

$$\begin{aligned} \left| \ \beta _L(z)-1+\frac{1}{L(1+z/L)^2} \ \right| \le \frac{C}{L^2(1+z/L)^4} \quad \mathrm{for }\; z\ge 0. \end{aligned}$$
(2.41)

Proof

We use the formula for the beta function \(\beta (\cdot )\) of the pdf \(c(\cdot )\) given by (2.35). Thus,

$$\begin{aligned} \beta (z) =\frac{c(z)h(z)}{w(z)^2} , \quad w(z) = \int _0^\infty c(z+z') \ \hbox {d}z' , \ \ h(z) = \int _0^\infty z' c(z+z') \ \hbox {d}z'. \end{aligned}$$
(2.42)

Letting \(c(z)=e^{-z-z^2/2L}\), we have that

$$\begin{aligned} w(z) = c(z) \int _0^\infty e^{-z'[1+z/L]-z'^2/2L} \ \hbox {d}z' , \quad h(z) = c(z) \int _0^\infty z' e^{-z'[1+z/L]-z'^2/2L} \ \hbox {d}z' . \end{aligned}$$
(2.43)

It follows from (2.42), (2.43) on making a change of variable that

$$\begin{aligned} \beta _L(z) =\int _0^\infty xe^{-x-\delta x^2/2} \ \hbox {d}x \ \Big / \left[ \int _0^\infty e^{-x-\delta x^2/2} \ \hbox {d}x \ \right] ^2 , \end{aligned}$$
(2.44)

where \(\delta =1/L[1+z/L]^2\). It is easy to see that there is a universal constant K such that the RHS of (2.44) is bounded above by K for all \(\delta >0\). We also have by Taylor expansion in \(\delta \) that \(\beta _L(z)=1-\delta +O(\delta ^2)\) if \(\delta \) is small. The inequality (2.41) follows. \(\square \)

Proposition 2.1

Let \(c_0(\cdot ,\cdot )\) be the solution to the CP system (1.1), (1.2) with Gaussian initial data and \(\Lambda _0(\cdot )\) be given by (1.3). Then, there exists \(t_0>2\) and constants \(C_1,C_2>0\) such that

$$\begin{aligned} 1-\frac{C_1}{\log t} \le \frac{{\mathrm{d}}\Lambda _0(t)}{{\mathrm{d}}t} \le 1-\frac{C_2}{\log t} \quad \mathrm{for \ } \ t\ge t_0. \end{aligned}$$
(2.45)

Proof

The initial data can be written in the form \(c_0(x,0)=K_0\exp [-A_0(x+B_0)^2]\), where \(K_0,A_0,B_0\) are constants with \(K_0,A_0>0\). It follows from (2.2), (2.4) that for \(t>0\) one has \(c_0(x,t)=K_t\exp [-A_t(x+B_t)^2]\), where \(B_t=[B_0+F_{1/\Lambda _0}(0,t)](A_0/A_t)^{1/2}\). Since \(\lim _{t\rightarrow \infty } F_{1/\Lambda _0}(0,t)=\infty \), it follows that we may assume wlog that the initial data are of the form (2.38) and \(\Lambda _0(0)=1\). Evidently, then \(\beta _{X_0}(x)=\beta _L(a(L)x), \ x\ge 0,\) where \(\beta _L\) satisfies the inequality (2.41).

Assume now that the initial data for (1.1), (1.2) are given by (2.38) where \(L=L_0>0\), and let \(L_t\) be the corresponding value of L determined by \(c_0(\cdot ,t)\). We have then from (2.40) and the discussion preceding it that for \(N=1,2,\ldots ,\) there exist times \(t_N\) such that

$$\begin{aligned}{}[2^N-1]T_1 \le \ t_N \le [2^N-1]T_2, \quad L_0+N\delta _0\le L_{t_N}\le L_0+N\delta _1. \end{aligned}$$
(2.46)

Since \(L_t\) is an increasing function of t, it follows from (2.41), (2.46) that \(\beta _{X_t}(0)\) is bounded above and below as in (2.45). Now using the identity (2.37), we obtain the inequality (2.45). \(\square \)

We wish next to compare the foregoing to the situation of the diffusive CP model with Gaussian initial data. From (4.15), the solution to (1.7) with initial data \(c_0(\cdot )\) is given by

$$\begin{aligned} c_\varepsilon (x,t) =\int _0^\infty G_{\varepsilon ,D}(x,y,0,t) c_0(y) \ \hbox {d}y, \quad x>0, \ t>0, \end{aligned}$$
(2.47)

where \(G_{\varepsilon ,D}\) is the Dirichlet Green’s function for the half-space \(\mathbf {R}^+\) defined by (4.14) with \(A(s)=1/\Lambda _\varepsilon (s)\). If we replace the Dirichlet Green’s function \(G_{\varepsilon ,D}\) by the full-space Green’s function \(G_\varepsilon \) of (4.11), then the solution \(c_\varepsilon (\cdot ,t)\) is Gaussian for \(t>0\) provided \(c_\varepsilon (\cdot ,0)\) is Gaussian, just as in the CP model. We shall see in Sect. 5 that it is legitimate to approximate \(G_{\varepsilon ,D}(x,y,0,t)\) by \(G_\varepsilon (x,y,0,t)\) provided \(x,y\ge M\varepsilon \) for some large constant M. Making the approximation \(G_{\varepsilon ,D}\simeq G_\varepsilon \) in (2.47), we obtain a formula similar to (2.39) for the length scale \(A_\varepsilon (L)\) of the Gaussian at doubling time. It is given by

$$\begin{aligned} A_\varepsilon (L) =L[1+\mu _{\varepsilon }(L)a(L)/L]^2[1+\varepsilon a(L)^2\sigma ^2_\varepsilon (L)\lambda _{\varepsilon }(L)^2/L]^{-1} . \end{aligned}$$
(2.48)

As in (2.39), the functions \(\lambda _\varepsilon (L),\mu _\varepsilon (L)\) are obtained from the coefficients of the linear function \(F_{1/\Lambda _\varepsilon }\), when \(t=T_{\varepsilon ,\mathrm{double}},\) where \(T_{\varepsilon ,\mathrm{double}}\) denotes the doubling time for the diffusive model. The expression \(\sigma ^2_\varepsilon (L)\) is given by the formula for \(\sigma ^2_A(T)\) in (4.10) with \(A(s)=1/\Lambda _\varepsilon (s), \ s\le T,\) and \(T=T_{\varepsilon ,\mathrm{double}}\). In Sect. 6, we shall study the \(\varepsilon \rightarrow 0\) limit of the diffusive CP model. We prove that if the CP and diffusive CP models have the same initial data, then \(\lim _{\varepsilon \rightarrow 0}\Lambda _\varepsilon (t)=\Lambda _0(t)\) uniformly in any finite interval \(0\le t\le T\). It follows that \(\lim _{\varepsilon \rightarrow 0}A_\varepsilon (L)=A(L)\), where A(L) is defined by (2.39).

We wish next to try to understand the evolution of the diffusive CP model when initial data are non-Gaussian. Let \(w_\varepsilon (x,t), \ h_\varepsilon (x,t)\) be defined in terms of the solution \(c_\varepsilon (\cdot ,\cdot )\) to (1.7) by

$$\begin{aligned} w_\varepsilon (x,t) =\int _x^\infty c_\varepsilon (x',t) \ \hbox {d}x', \quad h_\varepsilon (x,t) =\int _x^\infty w_\varepsilon (x',t) \hbox {d}x' . \end{aligned}$$
(2.49)

Then, \(w_\varepsilon (\cdot ,t),h_\varepsilon (\cdot ,t)\) are proportional to the functions (2.33) corresponding to the random variable \(X_t\) with pdf \(c_\varepsilon (\cdot ,t)/ \int _0^\infty c_\varepsilon (x,t) \ \hbox {d}x\). Making the approximation \(G_{\varepsilon ,D}\simeq G_\varepsilon \), we see from (2.47), (2.49), (4.11) that

$$\begin{aligned} w_\varepsilon (x,t)= & {} \exp \left[ \int _0^t\frac{\hbox {d}s}{\Lambda _\varepsilon (s)}\right] \int _{-\infty }^\infty G_\varepsilon (x,y,0,t) w_\varepsilon (y,0) \hbox {d}y , \end{aligned}$$
(2.50)
$$\begin{aligned} h_\varepsilon (x,t)= & {} \exp \left[ 2\int _0^t\frac{\hbox {d}s}{\Lambda _\varepsilon (s)}\right] \int _{-\infty }^\infty G_\varepsilon (x,y,0,t) h_\varepsilon (y,0) \hbox {d}y . \end{aligned}$$
(2.51)

Writing \(h_\varepsilon (x,t)=\exp [-q_\varepsilon (x,t)], \ x,t>0,\) in (2.51), we see from (4.11) that the semiclassical approximation to \(q_\varepsilon (x,t)\) is given by the formula

$$\begin{aligned} q_\varepsilon (x,t)= & {} \frac{1}{2}\log 2\pi \varepsilon +\frac{1}{2}\log \sigma _{1/\Lambda _\varepsilon }^2(t)-2\int _0^t \frac{\hbox {d}s}{\Lambda _\varepsilon (s)}\nonumber \\&+\inf _{y} \left[ \frac{\{x+m_{2,1/\Lambda _\varepsilon }(t)-m_{1,1/\Lambda _\varepsilon }(t)y\}^2}{2\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)}+ q_\varepsilon (y,0) \right] . \end{aligned}$$
(2.52)

Let us assume that \(q_\varepsilon (\cdot ,0)\) is given similarly to (2.38) by

$$\begin{aligned} q_\varepsilon (y,0) =\mathrm{constant}+ a(L)y+ \{a(L)y\}^2/2L , \quad y>0. \end{aligned}$$
(2.53)

The minimizer in (2.52) is then \(y_\mathrm{min}(x,t)\) where

$$\begin{aligned} y_\mathrm{min}(x,t) = \frac{1}{1+\varepsilon a(L)^2\sigma _{1/\Lambda _\varepsilon }^2(t)/m_{1,1/\Lambda _\varepsilon }(t)^2L}\left[ \frac{x+m_{2,1/\Lambda _\varepsilon }(t)}{m_{1,1/\Lambda _\varepsilon }(t)}- \frac{\varepsilon a(L)\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \right] . \end{aligned}$$
(2.54)

If we substitute \(y=y_\mathrm{min}(x,t)\) into (2.52), we obtain a quadratic formula for \(q_\varepsilon (x,t)\) similar to (2.53). If \(t=T_{\varepsilon ,\mathrm{double}}\), then L in (2.53) is replaced by \(A_\varepsilon (L)\) as in (2.48).

More generally we can consider the case when \(q_\varepsilon (\cdot ,0)\) is convex so (2.52) is a convex optimization problem with a unique minimizer \(y=y_\mathrm{min}(x,t)\). In that case, it is easy to see that

$$\begin{aligned} \frac{\partial q_\varepsilon (x,t)}{\partial x}= & {} \frac{1}{m_{1,1/\Lambda _\varepsilon }(t)} \frac{\partial q_\varepsilon (y_\mathrm{min}(x,t),0)}{\partial y} , \nonumber \\ \frac{\partial ^2 q_\varepsilon (x,t)}{\partial x^2}= & {} \frac{1}{m_{1,1/\Lambda _\varepsilon }(t)^2} \frac{\partial ^2 q_\varepsilon (y_\mathrm{min}(x,t),0)}{\partial y^2} \ \Bigg / \ \left[ 1+\frac{\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \ \frac{\partial ^2 q_\varepsilon (y_\mathrm{min}(x,t),0)}{\partial y^2} \ \right] .\nonumber \\ \end{aligned}$$
(2.55)

It follows from (2.55) that if the inequality

$$\begin{aligned} \frac{\partial ^2 q_\varepsilon (x,t)}{\partial x^2} \le \left[ \ \frac{\partial q_\varepsilon (x,t)}{\partial x} \ \right] ^2 , \quad x\ge 0, \end{aligned}$$
(2.56)

holds at \(t=0\), then it holds for all \(t>0\). We define now the function \(\beta _\varepsilon :[0,\infty )\times \mathbf {R}^+\rightarrow \mathbf {R}\) in terms of \(q_\varepsilon \) by the formula

$$\begin{aligned} \beta _\varepsilon (x,t) =1- \frac{\partial ^2 q_\varepsilon (x,t)}{\partial x^2} \ \Bigg / \left[ \ \frac{\partial q_\varepsilon (x,t)}{\partial x} \ \right] ^2 . \end{aligned}$$
(2.57)

We can see from (2.35) that the function \(h_\varepsilon (\cdot ,t)=\exp [-q_\varepsilon (\cdot ,t)]\) is proportional to \(h_{X_t}(\cdot )\) for some random variable \(X_t\) if and only if \(\beta _\varepsilon (\cdot ,t)\) is nonnegative. Hence, by the remark after (2.56), if \(q_\varepsilon (\cdot ,0)\) corresponds to a random variable \(X_0\), then \(q_\varepsilon (\cdot ,t)\) corresponds to a random variable \(X_t\) for all \(t>0\). From (2.55), (2.57), we have that

$$\begin{aligned} 1-\beta _\varepsilon (x,t) =\left[ 1-\beta _\varepsilon (y_\mathrm{min}(x,t),0)\right] \ \Big / \left[ 1+\frac{\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \ \frac{\partial ^2 q_\varepsilon (y_\mathrm{min}(x,t),0)}{\partial y^2} \ \right] . \end{aligned}$$
(2.58)

It follows from (2.58) that if \(\sup \beta _\varepsilon (\cdot ,0)\le 1\), then \(\sup \beta _\varepsilon (\cdot ,t)\le 1\) for \(t>0\). Furthermore, (2.58) also indicates that \(\beta _\varepsilon (\cdot ,t)\) should increase toward 1 as \(t\rightarrow \infty \).

It is well known (Hopf 1950) that the solution \(q_\varepsilon \) to the optimization problem (2.52) satisfies a Hamilton–Jacobi PDE. We can derive this PDE by using (2.52), (2.55). From (2.52), we see that the minimizer \(y_\mathrm{min}(x,t)\) is the solution to the equation

$$\begin{aligned} m_{1,1/\Lambda _\varepsilon }(t)\frac{\{x+m_{2,1/\Lambda _\varepsilon }(t)-m_{1,1/\Lambda _\varepsilon }(t)y_\mathrm{min}(x,t)\}}{\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)} = \frac{\partial q_\varepsilon \left( y_\mathrm{min}(x,t),0\right) }{\partial y}. \end{aligned}$$
(2.59)

We also have on differentiating (2.52) the identity

$$\begin{aligned} \frac{\partial q_\varepsilon (x,t)}{\partial t}= & {} \frac{\hbox {d}}{\hbox {d}t}\left[ \frac{1}{2}\log \sigma _{1/\Lambda _\varepsilon }^2(t)\right] -\frac{2}{\Lambda _\varepsilon (t)} \nonumber \\&+\frac{\{x+m_{2,1/\Lambda _\varepsilon }(t)-m_{1,1/\Lambda _\varepsilon }(t)y\}}{\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)}\left[ \frac{d m_{2,1/\Lambda _\varepsilon }(t)}{\hbox {d}t}- \frac{d m_{1,1/\Lambda _\varepsilon }(t)}{\hbox {d}t}y \right] \nonumber \\&+\frac{\{x+m_{2,1/\Lambda _\varepsilon }(t)-m_{1,1/\Lambda _\varepsilon }(t)y\}^2}{2\varepsilon }\frac{\hbox {d}}{\hbox {d}t}\frac{1}{ \sigma _{1/\Lambda _\varepsilon }^2(t)} , \quad \mathrm{where \ } y=y_\mathrm{min}(x,t).\nonumber \\ \end{aligned}$$
(2.60)

Note that in (2.60) we have used (2.59) to conclude that in differentiating (2.52) with respect to t, the coefficient of \(\partial y_\mathrm{min}(x,t)/\partial t\) is 0. Using the fact that each of the functions \(m_{1,1/\Lambda _\varepsilon }(t), \ m_{2,1/\Lambda _\varepsilon }(t), \ \sigma ^2_{1/\Lambda _\varepsilon }(t)\) is solutions to linear first-order differential equations, we conclude from (2.55), (2.59), (2.60) that \(q_\varepsilon (x,t)\) is a solution to the Hamilton–Jacobi PDE

$$\begin{aligned} \frac{\partial q_\varepsilon (x,t)}{\partial t}+\left[ \frac{x}{\Lambda _\varepsilon (t)}-1\right] \frac{\partial q_\varepsilon (x,t)}{\partial x} +\frac{\varepsilon }{2}\left[ \frac{\partial q_\varepsilon (x,t)}{\partial x}\right] ^2 +\frac{1}{\Lambda _\varepsilon (t)}-\frac{1}{2\sigma ^2_{1/\Lambda _\varepsilon }(t)} = 0 . \end{aligned}$$
(2.61)

Differentiating (2.61) with respect to x and setting \(v_\varepsilon (x,t)=\partial q_\varepsilon (x,t)/\partial x\), we see that \(v_\varepsilon (x,t)\) is the solution to the inviscid Burgers’ equation with linear drift,

$$\begin{aligned} \frac{\partial v_\varepsilon (x,t)}{\partial t}+\left[ \frac{x}{\Lambda _\varepsilon (t)}-1+\varepsilon v_\varepsilon (x,t)\right] \frac{\partial v_\varepsilon (x,t)}{\partial x} +\frac{v_\varepsilon (x,t)}{\Lambda _\varepsilon (t)} = 0 . \end{aligned}$$
(2.62)

If \(q_\varepsilon (\cdot ,t)\) corresponds to the random variable \(X_t\), then \(v_\varepsilon (x,t)=E[X_t-x \ | \ X_t>x]^{-1}, \ x\ge 0\), and since \(\Lambda _\varepsilon (t)=E[X_t]\), we have that

$$\begin{aligned} v_\varepsilon (0,t) =\frac{1}{\Lambda _\varepsilon (t)} . \end{aligned}$$
(2.63)

The system (2.62), (2.63) is a model for the evolution of the pdf of a random variable \(X_t\) which is intermediate between the CP and diffusive CP models. To obtain the pdf of \(X_t\) from the function \(v_\varepsilon (\cdot ,t)\), we let \(c_\varepsilon (\cdot ,t)=c_{X_t}(\cdot ), \ w_\varepsilon (\cdot ,t)=w_{X_t}(\cdot ), \ h_\varepsilon (\cdot ,t)=h_{X_t}(\cdot )\) as in (2.33). Then, \(v_\varepsilon (x,t)=w_\varepsilon (x,t)/h_\varepsilon (x,t)\) and

$$\begin{aligned} \Gamma _\varepsilon (x,t) =v_\varepsilon (x,t)^2- \frac{\partial v_\varepsilon (x,t)}{\partial x} = \frac{c_\varepsilon (x,t)}{h_\varepsilon (x,t)}. \end{aligned}$$
(2.64)

We also have that

$$\begin{aligned} v_\varepsilon (x,t) =-\frac{\partial }{\partial x} \log h_\varepsilon (x,t) , \quad \mathrm{whence \ } h_\varepsilon (x,t) = A_\varepsilon (t)\exp \left[ -\int _0^x v_\varepsilon (x',t) \ \hbox {d}x' \right] , \end{aligned}$$
(2.65)

where \(A_\varepsilon (\cdot )\) can be an arbitrary positive function. Evidently, (2.64), (2.65) uniquely determine the pdf of \(X_t\) from the function \(v_\varepsilon (\cdot ,t)\).

We can do a more systematic derivation of the model (2.62), (2.63) by beginning with the solution \(c_\varepsilon \) to the diffusive CP model (1.7), (1.8). Setting \(w_\varepsilon ,h_\varepsilon \) to be given by (2.49), then we see on integration of (1.7) that \(w_\varepsilon \) is a solution to the PDE

$$\begin{aligned} \frac{\partial w_\varepsilon (x,t)}{\partial t}+\left[ \frac{x}{\Lambda _\varepsilon (t)}-1\right] \frac{\partial w_\varepsilon (x,t)}{\partial x} = \frac{\varepsilon }{2} \frac{\partial ^2 w_\varepsilon (x,t)}{\partial x^2} . \end{aligned}$$
(2.66)

If we integrate (2.66), then we obtain a PDE for \(h_\varepsilon \),

$$\begin{aligned} \frac{\partial h_\varepsilon (x,t)}{\partial t}+\left[ \frac{x}{\Lambda _\varepsilon (t)}-1\right] \frac{\partial h_\varepsilon (x,t)}{\partial x} -\frac{h_\varepsilon (x,t)}{\Lambda _\varepsilon (t)} = \frac{\varepsilon }{2} \frac{\partial ^2 h_\varepsilon (x,t)}{\partial x^2} . \end{aligned}$$
(2.67)

Setting \(h_\varepsilon (x,t)=\exp [-q_\varepsilon (x,t)]\), it follows from (2.67) that \(q_\varepsilon (x,t)\) is a solution to the PDE

$$\begin{aligned} \frac{\partial q_\varepsilon (x,t)}{\partial t}+\left[ \frac{x}{\Lambda _\varepsilon (t)}-1\right] \frac{\partial q_\varepsilon (x,t)}{\partial x} +\frac{\varepsilon }{2}\left[ \frac{\partial q_\varepsilon (x,t)}{\partial x} \right] ^2 +\frac{1}{\Lambda _\varepsilon (t)} = \frac{\varepsilon }{2} \frac{\partial ^2 q_\varepsilon (x,t)}{\partial x^2} . \end{aligned}$$
(2.68)

If we differentiate (2.68) with respect to x, we obtain a PDE for the function \(v_\varepsilon (x,t)=\partial q_\varepsilon (x,t)/\partial x\), whence we have

$$\begin{aligned} \frac{\partial v_\varepsilon (x,t)}{\partial t}+\left[ \frac{x}{\Lambda _\varepsilon (t)}-1+\varepsilon v_\varepsilon (x,t)\right] \frac{\partial v_\varepsilon (x,t)}{\partial x} +\frac{v_\varepsilon (x,t)}{\Lambda _\varepsilon (t)} = \frac{\varepsilon }{2} \frac{\partial ^2 v_\varepsilon (x,t)}{\partial x^2} . \end{aligned}$$
(2.69)

For \(0<\nu \le 1\), we define the viscous CP model with viscosity \(\nu \) as the solution to the PDE

$$\begin{aligned}&\frac{\partial v_{\varepsilon ,\nu }(x,t)}{\partial t}+\left[ \frac{x}{\Lambda _{\varepsilon ,\nu }(t)}-1+\varepsilon v_{\varepsilon ,\nu }(x,t)\right] \frac{\partial v_{\varepsilon ,\nu }(x,t)}{\partial x} +\frac{v_{\varepsilon ,\nu }(x,t)}{\Lambda _{\varepsilon ,\nu }(t)} \nonumber \\&\quad = \frac{\varepsilon \nu }{2} \frac{\partial ^2 v_{\varepsilon ,\nu }(x,t)}{\partial x^2} , \ \ x,t>0, \end{aligned}$$
(2.70)

with boundary condition

$$\begin{aligned} \frac{\partial v_{\varepsilon ,\nu }(0,t)}{\partial x} =v_{\varepsilon ,\nu }(0,t)^2 , \quad t>0, \end{aligned}$$
(2.71)

and with the constraint

$$\begin{aligned} v_{\varepsilon ,\nu }(0,t) =\frac{1}{\Lambda _{\varepsilon ,\nu }(t)} , \quad t\ge 0. \end{aligned}$$
(2.72)

Assuming that (2.70), (2.71) has a classical solution, we show that if the initial data for (2.70) correspond to a random variable \(X_0\), then \(v_{\varepsilon ,\nu }(\cdot ,t)\) corresponds to a random variable \(X_t\) for \(t>0\) in the sense that \(v_{\varepsilon ,\nu }(x,t)=E[X_t-x \ | \ X_t>x]^{-1}, \ x\ge 0\). To see this we define \(\Gamma _{\varepsilon ,\nu }\) similarly to \(\Gamma _\varepsilon \) in (2.64) but with \(v_\varepsilon \) replaced by \(v_{\varepsilon ,\nu }\) on the RHS. It follows from (2.70), (2.71) that \(\Gamma _{\varepsilon ,\nu }\) satisfies the PDE

$$\begin{aligned}&\frac{\partial \Gamma _{\varepsilon ,\nu }(x,t)}{\partial t}+\left[ \frac{x}{\Lambda _{\varepsilon ,\nu }(t)}-1+\varepsilon v_{\varepsilon ,\nu }(x,t)\right] \frac{\partial \Gamma _{\varepsilon ,\nu }(x,t)}{\partial x} +2\frac{\Gamma _{\varepsilon ,\nu }(x,t)}{\Lambda _\varepsilon (t)} \nonumber \\&\quad = \frac{\varepsilon \nu }{2}\frac{\partial ^2 \Gamma _{\varepsilon ,\nu }(x,t)}{\partial x^2}+ \varepsilon (1-\nu ) \left( \frac{\partial v_{\varepsilon ,\nu }(x,t)}{\partial x}\right) ^2, \end{aligned}$$
(2.73)

with Dirichlet boundary condition \(\Gamma _{\varepsilon ,\nu }(0,t)=0, \ t>0\). Hence by the maximum principle (Protter and Weinberger 1984), if \(\Gamma _{\varepsilon ,\nu }(\cdot ,0)\) is nonnegative, then \(\Gamma _{\varepsilon ,\nu }(\cdot ,t)\) is nonnegative for \(t>0\). We see from (2.64) that the nonnegativity of \(\Gamma _{\varepsilon ,\nu }(\cdot ,t)\) is equivalent to \(v_{\varepsilon ,\nu }(\cdot ,t)\) corresponding to a random variable \(X_t\). We have shown that for \(0<\nu \le 1\) the viscous CP model corresponds to the evolution of a random variable \(X_t, \ t\ge 0\). If \(\nu =1\) the model is identical to the diffusive CP model (1.7), (1.8) with Dirichlet condition \(c_\varepsilon (0,t)=0, \ t>0\).

We can think of the inviscid CP model (2.62), (2.63) as the limit of the viscous CP model (2.70), (2.71), (2.72) as the viscosity \(\nu \rightarrow 0\). It is not clear, however, what happens to the boundary condition (2.71) in this limit. Unless the initial data \(v_\varepsilon (\cdot ,0)\) for (2.62) are increasing, the solution \(v_\varepsilon (\cdot ,t)\) develops discontinuities at some finite time (Smoller 1994). For an entropy satisfying solution \(v_\varepsilon \), discontinuities have the property that the solution jumps down across the discontinuity. Hence, if \(v_\varepsilon (\cdot ,t)\) is discontinuous at the point z, then

$$\begin{aligned} \lim _{x\rightarrow z^-}v_\varepsilon (x,t) > \lim _{x\rightarrow z^+}v_\varepsilon (x,t) . \end{aligned}$$
(2.74)

Observe now that for a random variable X, the function \(x\rightarrow E[X-x \ | \ X>x]\) has discontinuities precisely at the atoms of X. In that case, the function jumps up across the discontinuity. Since \(v_\varepsilon (x,t)=E[X_t-x \ | \ X_t>x]^{-1}\) for some random variable \(X_t\), it follows that at discontinuities of \(v_\varepsilon (\cdot ,t)\) the function jumps down. Thus, discontinuities of \(v_\varepsilon (\cdot ,t)\) correspond to atoms of \(X_t\), and the entropy condition for (2.62) is automatically satisfied.

We have already observed that the function \(t\rightarrow \Lambda _0(t)\) in the CP model (1.1), (1.2) is increasing and that the function \(t\rightarrow \Lambda _\varepsilon (t)\) in the diffusive CP model (1.7), (1.8) is also increasing. To determine whether the function \(t\rightarrow \Lambda _{\varepsilon ,\nu }(t)\) in the viscous CP model (2.70)–(2.72) is increasing, we observe on setting \(x=0\) in (2.70) and using (2.72) that \(v_{\varepsilon ,\nu }(0,t)\) satisfies the equation

$$\begin{aligned} \frac{\partial v_{\varepsilon ,\nu }(0,t)}{\partial t}+\left\{ 1-\varepsilon (1-\nu ) v_{\varepsilon ,\nu }(0,t)\right\} \Gamma _{\varepsilon ,\nu }(0,t)+\varepsilon (1-\nu ) v_{\varepsilon ,\nu }(0,t)^3 +\frac{\varepsilon \nu }{2}\frac{\partial \Gamma _{\varepsilon ,\nu }(0,t)}{\partial x}=0. \end{aligned}$$
(2.75)

We have already seen that \(\Gamma _{\varepsilon ,\nu }(\cdot ,t)\) is a nonnegative function, and from (2.71), it follows that \(\Gamma _{\varepsilon ,\nu }(0,t)=0\) for \(t>0\). Hence, \(\partial \Gamma _{\varepsilon ,\nu }(0,t)/\partial x\ge 0\) for \(t>0\). We conclude then from (2.75) that the function \(t\rightarrow v_{\varepsilon ,\nu }(0,t)\) is decreasing provided

$$\begin{aligned} v_{\varepsilon ,\nu }(0,0) \le \frac{1}{\varepsilon (1-\nu )} . \end{aligned}$$
(2.76)

Thus, from (2.72) we see that if (2.76) holds, then the function \(t\rightarrow \Lambda _{\varepsilon ,\nu }(t)\) is increasing. Note that in the case of the diffusive CP model when \(\nu =1\) the condition (2.76) is redundant.

3 The Inviscid CP Model—Proof of Theorem 1.1

We shall restrict ourselves here to considering the solutions of (2.62), (2.63) when the initial data \(v_\varepsilon (\cdot ,0)\) are nonnegative, increasing and also the function \(\Gamma _\varepsilon (\cdot , 0)\) of (2.64) is nonnegative. The condition (2.76) becomes now \(v_\varepsilon (0,t)\le \varepsilon ^{-1}\), and assuming this holds also, we see that in this case (2.62) may be solved by the method of characteristics. To carry this out, we set \(\tilde{v}_\varepsilon (x,t)=m_{1,1/\Lambda _\varepsilon }(t)v_\varepsilon (x,t)\), where \(m_{1,A}(\cdot )\) is defined by (2.2). Then, (2.62) is equivalent to

$$\begin{aligned} \frac{\partial \tilde{v}_\varepsilon (x,t)}{\partial t}+\left[ \frac{x}{\Lambda _\varepsilon (t)}-1+\varepsilon \frac{\tilde{v}_\varepsilon (x,t)}{m_{1,1/\Lambda _\varepsilon }t)}\right] \frac{\partial \tilde{v}_\varepsilon (x,t)}{\partial x} = 0 . \end{aligned}$$
(3.1)

From (3.1), it follows that if \(x(s), \ s\ge 0,\) is a solution to the ODE

$$\begin{aligned} \frac{\hbox {d}x(s)}{\hbox {d}s} =\frac{x(s)}{\Lambda _\varepsilon (s)}-1+\varepsilon \frac{\tilde{v}_\varepsilon (x(s),s)}{m_{1,1/\Lambda _\varepsilon }(s)} , \end{aligned}$$
(3.2)

and characteristics do not intersect, then \(\tilde{v}_\varepsilon (x(t),t) =\tilde{v}_\varepsilon (x(0),0)\) for \(t\ge 0\). We can therefore calculate the characteristics of (3.1) by setting \(\tilde{v}_\varepsilon (x(s),s)=\tilde{v}_\varepsilon (x(0),0)=v_\varepsilon (x(0),0)\). We define the function \(F_{\varepsilon ,A}(x,t, v_0(\cdot ))\) depending on \(x,t\ge 0\) and increasing function \(v_0:[0,\infty )\rightarrow (0,\infty )\) by

$$\begin{aligned} z+\varepsilon \frac{\sigma _A^2(t)}{m_{1,A}(t)^2}v_0(z) =\frac{x+m_{2,A}(t)}{m_{1,A}(t)} =F_A(x,t) , \quad F_{\varepsilon ,A}(x,t, v_0(\cdot )) =z, \end{aligned}$$
(3.3)

where \(F_A\) is given by (2.2) and \(\sigma _A^2(\cdot )\) by (4.10). Since \(v_0(\cdot )\) is an increasing function, there is a unique solution z to (3.3) for all \(x\ge 0\) provided \(v_0(0)\le \varepsilon ^{-1} m_{1,A}(t)m_{2,A}(t)/\sigma _A^2(t)\). If this condition holds, then the method of characteristics now yields the solution to (2.62) as

$$\begin{aligned} v_\varepsilon (x,t) =\frac{1}{m_{1,A}(t)} v_\varepsilon \left( F_{\varepsilon ,A}(x,t, v_\varepsilon (\cdot ,0)), 0\right) \end{aligned}$$
(3.4)

with \(A=1/\Lambda _\varepsilon \). It follows from (3.3), (3.4) that \(v_\varepsilon (0,t) \le m_{2,A}(t)/\varepsilon \sigma _A^2(t)\le \varepsilon ^{-1}\).

We wish to prove a global existence and uniqueness theorem for solutions of (2.62), (2.63). To describe our assumptions on the initial data \(v_\varepsilon (\cdot ,0)\), we shall consider functions \(v_0:[0,x_\infty )\rightarrow \mathbf {R}^+\) with the properties:

$$\begin{aligned}&\mathrm{The \ function \ } x\rightarrow v_0(x) \ \mathrm{is \ increasing \ on \ the \ interval \ } [0,x_\infty ) \nonumber \\&\quad \mathrm{and \ } v_0(0)>0. \ \mathrm{If \ } x_\infty <\infty \ \mathrm{then \ } \lim _{x\rightarrow x_\infty } v_0(x)=\infty . \end{aligned}$$
(3.5)
$$\begin{aligned}&v_0(x_2)-v_0(x_1) \le \int _{x_1}^{x_2} v_0(x)^2 \ \hbox {d}x \quad \mathrm{for \ } 0\le x_1<x_2<x_\infty . \end{aligned}$$
(3.6)

Note that (3.6) implies that \(v_0(\cdot )\) is locally Lipschitz continuous in the interval \([0,x_\infty )\).

Lemma 3.1

Assume the function \( v_0(\cdot )=v_\varepsilon (\cdot ,0)\) satisfies (3.5), (3.6) and in addition that \((1+\delta _0)v_\varepsilon (0,0)<\varepsilon ^{-1}\) for some \(\delta _0>0\). Then, there exists \(\delta _1>0\) depending only on \(\delta _0\) such that there is a unique solution to (2.62), (2.63) for \(0\le t\le T=\delta _1/v_\varepsilon (0,0)\).

Proof

Let \(T,\delta _2>0\) and \(\mathcal {E}\) be the space of continuous functions \(V:[0,T]\rightarrow \mathbf {R}^+\) satisfying

$$\begin{aligned} V(0) =v_\varepsilon (0,0), \quad (1+\delta _2)^{-1}v_\varepsilon (0,0) \le V(t) \le (1+\delta _2) v_\varepsilon (0,0) \quad \mathrm{for \ } 0\le t\le T. \end{aligned}$$
(3.7)

For \(V\in \mathcal {E}\), we define a function \(\mathcal {B}V(t), \ 0\le t\le T,\) by \(\mathcal {B}V(t)=v_\varepsilon (0,t)\) where \(v_\varepsilon \) is the function (3.4) with \(A(s)=V(s), \ 0\le s\le T\). We shall show that if \(T>0\) is sufficiently small then \(V\in \mathcal {E}\) implies \(\mathcal {B}V\in \mathcal {E}\). To see this, we first observe from (3.3) that \(\mathcal {B}V(0)=v_\varepsilon (0,0)\). Next we note that for a function \(v_0(\cdot )\) satisfying (3.5), (3.6), then

$$\begin{aligned} v_0(0) \le v_0(z) \le \frac{v_0(0)}{1-zv_0(0)} , \quad \mathrm{for \ } 0 \le z \ < 1/v_0(0) . \end{aligned}$$
(3.8)

Since \(V\in \mathcal {E}\), it follows from (3.7) that with \(A(\cdot )=V(\cdot )\), then

$$\begin{aligned} t\exp \left[ -(1+\delta _2)v_\varepsilon (0,0)t\right] \le \frac{m_{2,A}(t)}{m_{1,A}(t)} \le t , \quad 0\le t\le T. \end{aligned}$$
(3.9)

Similarly, we have that

$$\begin{aligned} t\exp \left[ -2(1+\delta _2)v_\varepsilon (0,0)t\right] \le \frac{\sigma _A^2(t)}{m_{1,A}(t)^2} \le t , \quad 0\le t\le T. \end{aligned}$$
(3.10)

From (3.9), (3.10), we have that for \(\delta _1,\delta _2>0\) sufficiently small, depending only on \(\delta _0\), that

$$\begin{aligned} \varepsilon \frac{\sigma _A^2(t)}{m_{1,A}(t)^2}v_\varepsilon (0,0) \le \frac{m_{2,A}(t)}{m_{1,A}(t)} \quad \mathrm{for \ } 0\le t\le T. \end{aligned}$$
(3.11)

Hence, there is a unique solution \(z(t)\le m_{2,A}(t)/m_{1,A}(t)\) to (3.3) with \(x=0\) provided \(0\le t\le T\). Since \(\mathcal {B}V(t) =v_\varepsilon (z(t),0)/m_{1,A}(t)\), it follows from (3.8), (3.9) that on choosing \(\delta _1>0\) sufficiently small, depending only on \(\delta _2\), that the function \(\mathcal {B}V(t), \ 0\le t\le T\), also satisfies (3.7).

Next we show that \(\mathcal {B}\) is a contraction on the space \(\mathcal {E}\) with metric \(d(V_1,V_2)=\sup _{0\le t\le T}|V_2(t)-V_1(t)|\) for \(V_1,V_2\in \mathcal {E}\). To see this, let \(z_1(t), \ z_2(t), \ 0\le t\le T,\) be the solutions to (3.3) with \(x=0\) corresponding to \(V_1,V_1\in \mathcal {E}\), respectively. Then, from (3.4) we have

$$\begin{aligned} \mathcal {B}V_2(t)-\mathcal {B}V_1(t) = \frac{v_\varepsilon (z_2(t),0)-v_\varepsilon (z_1(t),0)}{m_{1,V_2}(t)}+ \left[ \frac{1}{m_{1,V_2}(t)}-\frac{1}{m_{1,V_1}(t)}\right] v_\varepsilon (z_1(t),0). \end{aligned}$$
(3.12)

The second term on the RHS of (3.12) is bounded as

$$\begin{aligned}&\left| \ \left[ \frac{1}{m_{1,V_2}(t)}-\frac{1}{m_{1,V_1}(t)}\right] v_\varepsilon (z_1(t),0) \ \right| \nonumber \\&\quad \le (1+\delta _2)v_\varepsilon (0,0)\int _0^t |V_2(s)-V_1(s)| \hbox {d}s , \quad 0\le t\le T. \end{aligned}$$
(3.13)

We use (3.6) to bound the first term in (3.12). Thus, we have that

$$\begin{aligned} \left| \ \frac{v_\varepsilon (z_2(t),0)-v_\varepsilon (z_1(t),0)}{m_{1,V_2}(t)} \ \right| \le (1+\delta _2)^2v_\varepsilon (0,0)^2 |z_2(t)-z_1(t)| , \quad 0\le t \le T. \end{aligned}$$
(3.14)

From (3.3), (3.5), it follows that

$$\begin{aligned} |z_2(t)-z_1(t)|\le & {} \left| \ \frac{m_{2,V_2}(t)}{m_{1,V_2}(t)}-\frac{m_{2,V_1}(t)}{m_{1,V_1}(t)} \ \right| \nonumber \\&+\left| \frac{\sigma _{V_2}^2(t)}{m_{1,V_2}(t)^2}-\frac{\sigma _{V_1}^2(t)}{m_{1,V_1}(t)^2}\right| \varepsilon (1+\delta _2)v_\varepsilon (0,0) , \quad 0\le t\le T.\nonumber \\ \end{aligned}$$
(3.15)

The RHS of (3.15) can be bounded similarly to (3.13), and so we obtain the inequality

$$\begin{aligned} |z_2(t)-z_1(t)| \le \left\{ 1+2\frac{1+\delta _2}{1+\delta _0}\right\} t\int _0^t |V_2(s)-V_1(s)| \ \hbox {d}s , \quad 0\le t\le T. \end{aligned}$$
(3.16)

It follows from (3.12)–(3.16) that

$$\begin{aligned} |\mathcal {B}V_2(t)-\mathcal {B}V_1(t) | \le 10 \delta _1\sup _{0\le t\le T}|V_2(t)-V_1(t)| , \quad 0\le t\le T=\delta _1/v_\varepsilon (0,0), \end{aligned}$$
(3.17)

provided \(\delta _1>0\) is chosen sufficiently small depending only on \(\delta _2\). Evidently, \(\mathcal {B}\) is a contraction mapping on \(\mathcal {E}\) and therefore has a unique fixed point if one also has \(10\delta _1<1\). \(\square \)

Lemma 3.2

Let \(v_\varepsilon (x,t), \ x\ge 0, \ 0\le t\le T\) be the solution to (2.62), (2.63) constructed in Lemma 3.1. Then, for any t satisfying \(0< t\le T\) the function \(v_0(\cdot )=v_\varepsilon (\cdot ,t)\) satisfies (3.5), (3.6) with \(x_\infty =\infty \). In addition, the function \(t\rightarrow v_\varepsilon (0,t), \ 0\le t\le T,\) is continuous and decreasing.

Proof

Since \(\varepsilon >0\), it follows from the fact that (3.5) holds for \(v_0(\cdot )=v_\varepsilon (\cdot ,0)\) that (3.3) has a unique solution \(z<x_\infty \) for any \(x>0\). Hence, \(x_\infty =\infty \) if \(t>0\), and it is also clear that the function \(x\rightarrow v_\varepsilon (x,t), \ x\ge 0,\) is increasing. We have therefore shown that (3.5) holds for \(v_0(\cdot )=v_\varepsilon (\cdot ,t)\) and \(x_\infty =\infty \) if \(0<t\le T\).

Next we wish to show that (3.6) holds for \(v_0(\cdot )=v_\varepsilon (\cdot ,t)\) with \(0<t\le T\). To see this, we observe from (3.4), (3.6) that for \(0\le x_1\le x_2<\infty \),

$$\begin{aligned} v_\varepsilon (x_2,t)-v_\varepsilon (x_1,t) =\frac{v_\varepsilon (z(x_2,t),0)-v_\varepsilon (z(x_1,t),0)}{m_{1,1/\Lambda _\varepsilon }(t)} \le \frac{1}{m_{1,1/\Lambda _\varepsilon }(t)} \int _{z(x_1,t)}^{z(x_2,t)} v_\varepsilon (z,0)^2 \ \hbox {d}z , \end{aligned}$$
(3.18)

where \(z(x,t)=F_{\varepsilon ,1/\Lambda _\varepsilon }(x,t,v_\varepsilon (\cdot ,0))\). We see from (3.3) that \(0\le \partial z(x,t)/\partial x \le 1/m_{1,1/\Lambda _\varepsilon }(t)\), whence

$$\begin{aligned} \frac{1}{m_{1,1/\Lambda _\varepsilon }(t)} \int _{z(x_1,t)}^{z(x_2,t)} v_\varepsilon (z,0)^2 \ \hbox {d}z =m_{1,1/\Lambda _\varepsilon }(t)\int _{x_1}^{x_2} v_\varepsilon (x,t)^2 \frac{\partial z(x,t)}{\partial x} \ \hbox {d}x \le \int _{x_1}^{x_2} v_\varepsilon (x,t)^2 \hbox {d}x . \end{aligned}$$
(3.19)

To show that the function \(t\rightarrow v_\varepsilon (0,t)\) is continuous and decreasing, we write \(v_\varepsilon (0,t)=v_0(z(t))/m_{1,1/\Lambda _\varepsilon }(t)\) where \(v_0(\cdot )=v_\varepsilon (\cdot ,0)\) and z(t) is the solution z to (3.3) with \(x=0\). We see from (3.3) that the function \(t\rightarrow z(t)\) is Lipschitz continuous, whence the function \(t\rightarrow v_\varepsilon (0,t)\) is continuous. If \(z\rightarrow v_0(z)\) is differentiable at \(z=z(t)\), then it follows from (2.2) that

$$\begin{aligned} \frac{ \partial v_\varepsilon (0,t)}{\partial t} = \frac{1}{m_{1,1/\Lambda _\varepsilon }(t)}\left[ v_0'(z(t))\frac{\hbox {d}z(t)}{\hbox {d}t}-\frac{v_0(z(t))}{\Lambda _\varepsilon (t)}\right] . \end{aligned}$$
(3.20)

Differentiating (3.3) with respect to t at \(x=0\), we obtain the equation

$$\begin{aligned} \left[ 1+\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2}v'_0(z(t))\right] \frac{\hbox {d}z(t)}{\hbox {d}t} = \frac{1}{m_{1,1/\Lambda _\varepsilon }(t)}\left[ 1-\frac{\varepsilon v_0(z(t))}{m_{1,1/\Lambda _\varepsilon }(t)}\right] . \end{aligned}$$
(3.21)

Hence, (3.20), (3.21) imply that

$$\begin{aligned}&\left[ 1+\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2}v'_0(z(t))\right] m_{1,1/\Lambda _\varepsilon }(t)\frac{ \partial v_\varepsilon (0,t)}{\partial t} \nonumber \\&\quad = \frac{v'_0(z(t))}{m_{1,1/\Lambda _\varepsilon }(t)}-\frac{v_0(z(t))}{\Lambda _\varepsilon (t)}-\frac{\varepsilon v_0'(z(t))v_0(z(t))}{m_{1,1/\Lambda _\varepsilon }(t)^2}\left[ 1+\frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{\Lambda _\varepsilon (t)}\right] . \end{aligned}$$
(3.22)

From (2.63), (3.4), (3.6), we have that

$$\begin{aligned} \frac{v'_0(z(t))}{m_{1,1/\Lambda _\varepsilon }(t)} \le \frac{v_0(z(t))^2}{m_{1,1/\Lambda _\varepsilon }(t)} =v_0(z(t))v_\varepsilon (0,t) =\frac{v_0(z(t))}{\Lambda _\varepsilon (t)} . \end{aligned}$$
(3.23)

We conclude from (3.22), (3.23) that \(\partial v_\varepsilon (0,t)/\partial t\le 0\). In the case when the function \(z\rightarrow v_0(z)\) is not differentiable at \(z=z(t)\), we can do an approximation argument to see that the function \(s\rightarrow v_\varepsilon (0,s)\) is decreasing close to \(s=t\). We have therefore shown that the function \(t\rightarrow v_\varepsilon (0,t)\) is decreasing, whence \(v_\varepsilon (0,t)\le v_\varepsilon (0,0)< \varepsilon ^{-1}\) for \(0\le t\le T\). Since the RHS of (3.21) is the same as \([1-\varepsilon v_\varepsilon (0,t)]/m_{1,1/\Lambda _\varepsilon }(t)\), this implies that the function \(t\rightarrow z(t)\) is increasing. \(\square \)

Proposition 3.1

Assume the initial data for (2.62), (2.63) satisfy the conditions of Lemma 3.1. Then, there exists a unique continuous solution \(v_\varepsilon (x,t), \ x,t\ge 0,\) globally in time to (2.62), (2.63). The solution \(v_\varepsilon (\cdot ,t)\) satisfies (3.5), (3.6) for \(t>0\) with \(x_\infty =\infty \), and the function \(t\rightarrow v_\varepsilon (0,t)\) is decreasing. Furthermore, there is a constant \(C(\delta _0)\) depending only on \(\delta _0\) such that \(\Lambda _\varepsilon (t)\le \Lambda _\varepsilon (0)+C(\delta _0)[\Lambda _\varepsilon (0)+t], \ t\ge 0\).

Proof

The global existence and uniqueness follow immediately from Lemmas 3.1, 3.2 upon using the fact that the function \(t\rightarrow v_\varepsilon (0,t)\) is decreasing. To get the upper bound on the function \(\Lambda _\varepsilon (\cdot )\), we observe that Lemmas 3.1 implies that with \(T_0=0,\)

$$\begin{aligned} \Lambda _\varepsilon (t) \le (1+\delta _2)\Lambda _\varepsilon (T_{k-1}) \ \mathrm{for \ } T_{k-1} \le t \le T_k, \quad T_k=T_{k-1}+\delta _1\Lambda _\varepsilon (T_{k-1}) , \ k=1,2,\ldots \end{aligned}$$
(3.24)

It follows from (3.24) that

$$\begin{aligned} \Lambda _\varepsilon (t) \le (1+\delta _2)(T_k-T_{k-1})/\delta _1 \quad \mathrm{for \ } T_{k-1}\le t\le T_k , \quad k=1,2,\ldots \end{aligned}$$
(3.25)

We also have that

$$\begin{aligned} T_k-T_{k-1} \le \delta _1(1+\delta _2)\Lambda _\varepsilon (T_{k-2}) \le (1+\delta _2)T_{k-1} , \quad k=2,3,\ldots \end{aligned}$$
(3.26)

From (3.25), (3.26), we conclude that \(\Lambda _\varepsilon (t) \le (1+\delta _2)^2t/\delta _1\) provided \(t\ge T_1\), whence the result follows. \(\square \)

The upper bound on the coarsening rate implied by Proposition 3.1 is independent of \(\varepsilon >0\) as \(\varepsilon \rightarrow 0\). We can see from (3.22) that a lower bound on the rate of coarsening depends on \(\varepsilon \). In fact, if we choose \(v_0(z)=1/[\Lambda _\varepsilon (0)-z], \ 0<z<\Lambda _\varepsilon (0),\) then \(v'_0(z)=v_0(z)^2\), and so at \(t=0\) the RHS of (3.22) is zero if \(\varepsilon =0\). The random variable \(X_0\) corresponding to this initial data is simply \(X_0\equiv \mathrm{constant}\), and it is easy to see that if \(\varepsilon =0\), then \(X_t\equiv X_0\) for all \(t>0\). For \(\varepsilon >0\), however, we have the following:

Lemma 3.3

Assume the initial data for (2.62), (2.63) satisfy the conditions of Lemma 3.1, and let z(t) be as in Lemma 3.2. Then, \(\lim _{t\rightarrow \infty } z(t)=x_\infty \), and if \(\varepsilon >0\), one has \(\lim _{t\rightarrow \infty }\Lambda _\varepsilon (t)=\infty \).

Proof

Let \(\lim _{t\rightarrow \infty }\Lambda _\varepsilon (t)=\Lambda _{\varepsilon ,\infty }\) and assume first that \(\Lambda _{\varepsilon ,\infty }<\infty \). In that case \(\lim _{t\rightarrow \infty } m_{1,\Lambda _\varepsilon }(t)=\infty \), and hence, (2.63), (3.4) imply that \(\lim _{t\rightarrow \infty }v_0(z(t))=\infty \). We conclude from (3.5), (3.6) that if \(\Lambda _{\varepsilon ,\infty }<\infty \), then \(\lim _{t\rightarrow \infty } z(t)=x_\infty \). We also have from (2.2) that

$$\begin{aligned} \limsup _{t\rightarrow \infty } \frac{m_{2,1/\Lambda _\varepsilon }(t)}{m_{1,1/\Lambda _\varepsilon }(t)} \le \Lambda _{\varepsilon ,\infty } , \quad \liminf _{t\rightarrow \infty } \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \ge \frac{\Lambda _\varepsilon (0)}{2} . \end{aligned}$$
(3.27)

It follows now from (3.3), (3.27) that if \(\varepsilon >0\), then \(\limsup _{t\rightarrow \infty } v_0(z(t))\le 2\Lambda _{\varepsilon ,\infty }/\varepsilon \Lambda _\varepsilon (0)<\infty \), which yields a contradiction.

Next we assume that \(\Lambda _{\varepsilon ,\infty }=\infty \), which we have just shown always holds if \(\varepsilon >0\). The function \(t\rightarrow z(t)\) is increasing, and let us suppose that \(\lim _{t\rightarrow \infty } z(t)=z_\infty <x_\infty \). Then, from (2.63), (3.4) we have that \(\lim _{t\rightarrow \infty } m_{1,1/\Lambda _\varepsilon }(t)=\infty \). We use the fact that for any \(T\ge 0,\) there exists a constant \(K_T\) such that

$$\begin{aligned} \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \le \frac{m_{2,1/\Lambda _\varepsilon }(t)}{m_{1,1/\Lambda _\varepsilon }(T)m_{1,1/\Lambda _\varepsilon }(t)}+K_T, \quad \mathrm{for \ } t\ge T. \end{aligned}$$
(3.28)

From (3.3), (3.28), we obtain the inequality

$$\begin{aligned} \frac{m_{2,1/\Lambda _\varepsilon }(t)}{m_{1,1/\Lambda _\varepsilon }(t)} \le z_\infty +\varepsilon \left[ \frac{m_{2,1/\Lambda _\varepsilon }(t)}{m_{1,1/\Lambda _\varepsilon }(T)m_{1,1/\Lambda _\varepsilon }(t)}+K_T \right] v_0(z_\infty ) \quad \mathrm{if \ }t\ge T. \end{aligned}$$
(3.29)

Choosing T sufficiently large so that \( m_{1,1/\Lambda _\varepsilon }(T)\ge 2\varepsilon v_0(z_\infty )\), we conclude from (3.28), (3.29) that there is a constant \(C_1\) such that \(\sigma _{1/\Lambda _\varepsilon }^2(t)/m_{1,1/\Lambda _\varepsilon }(t)^2\le C_1\) for \(t\ge T\). If we also choose T such that \(m_{1,1/\Lambda _\varepsilon }(t)\ge \varepsilon v_0(z(t))/2\) for \(t\ge T\), we have from (3.6), (3.21) that

$$\begin{aligned} \left[ 1+C_1\varepsilon v_0(z_\infty )^2\right] \frac{\hbox {d}z(t)}{\hbox {d}t} \ge \frac{1}{2m_{1,1/\Lambda _\varepsilon }(t)} \ \quad \mathrm{for \ } t\ge T. \end{aligned}$$
(3.30)

Since the function \(t\rightarrow z(t)\) is increasing and \(\lim _{t\rightarrow \infty } z(t)=z_\infty <\infty \), it follows from (2.63), (3.4), (3.30) that there is a constant \(C_2\) such that

$$\begin{aligned} \int _0^t \frac{\hbox {d}s}{\Lambda _\varepsilon (s)} \le v_0(z_\infty )\int _0^t \ \frac{\hbox {d}s}{m_{1,1/\Lambda _\varepsilon }(s)} \le C_2 \quad \mathrm{for \ } t \ge 0. \end{aligned}$$
(3.31)

However, (3.31) implies that \(\lim _{t\rightarrow \infty } m_{1,1/\Lambda _\varepsilon }(t)\le \exp [C_2]\) and so we have again a contradiction. We conclude that \(\lim _{t\rightarrow \infty } z(t)=x_\infty \). \(\square \)

Lemma 3.4

Assume the initial data for (2.62), (2.63) satisfy the conditions of Lemma 3.1 with \(x_\infty =\infty \), and that for \(0<\delta \le 1\), one has

$$\begin{aligned} \limsup _{x\rightarrow \infty } \frac{v_\varepsilon \left( x+\delta /v_\varepsilon (x,0),0\right) }{v_\varepsilon (x,0)} \le 1+\gamma (\delta ), \quad \mathrm{where \ } \lim _{\delta \rightarrow 0}\frac{\gamma (\delta )}{\delta }=0. \end{aligned}$$
(3.32)

Then, \(\lim _{t\rightarrow \infty } \Lambda _\varepsilon (t)/t=1\) for any \(\varepsilon \ge 0\).

Proof

The main point about the condition (3.32) is that it is invariant under the dynamics determined by (2.62), (2.63). It is easy to see this in the case \(\varepsilon =0\) since we have, on using the notation of (3.3), that

$$\begin{aligned} \frac{v_0\left( x+\delta /v_0(x,t),t\right) }{v_0(x,t)} = \frac{v_0\left( F_{1/\Lambda _0}(x,t)+\delta /v_0(F_{1/\Lambda _0}(x,t),0),0\right) }{v_0(F_{1/\Lambda _0}(x,t),0)}. \end{aligned}$$
(3.33)

For \(\varepsilon >0\), we have

$$\begin{aligned} \frac{v_\varepsilon \left( x+\delta /v_\varepsilon (x,t),t\right) }{v_\varepsilon (x,t)} = \ \frac{v_\varepsilon \left( z(x+\delta /v_\varepsilon (x,t),t),0\right) }{v_\varepsilon (z(x,t),0)}. \end{aligned}$$
(3.34)

Since the function \(x\rightarrow v_\varepsilon (x,0)\) is increasing, it follows from (3.3) that

$$\begin{aligned} z(x+\delta /v_\varepsilon (x,t),t) \le z(x,t)+\frac{\delta }{m_{1,1/\Lambda _\varepsilon }(t)v_\varepsilon (x,t)} . \end{aligned}$$
(3.35)

We conclude now from (3.4), (3.34), (3.35) that

$$\begin{aligned} \frac{v_\varepsilon \left( x+\delta /v_\varepsilon (x,t),t\right) }{v_\varepsilon (x,t)} \le \frac{v_\varepsilon \left( z(x,t)+\delta /v_\varepsilon (z(x,t),0),0\right) }{v_\varepsilon (z(x,t),0)} . \end{aligned}$$
(3.36)

From Lemma 3.3 and (3.36), there exists \(T_0\ge 0\) such that

$$\begin{aligned} \frac{v_\varepsilon \left( x+\delta /v_\varepsilon (x,t),t\right) }{v_\varepsilon (x,t)} \le 1+2\gamma (\delta ) \quad \mathrm{for \ } x\ge 0, \ t\ge T_0 . \end{aligned}$$
(3.37)

We use (3.37) to estimate \(\Lambda _\varepsilon (T_0+t)/\Lambda _\varepsilon (T_0)\) in the interval \(0\le t\le \delta /v_\varepsilon (0,T_0)\). Thus, we have

$$\begin{aligned} \frac{\Lambda _\varepsilon (T_0)}{\Lambda _\varepsilon (T_0+t)} =\frac{v_\varepsilon (z(t),T_0)m_{1,1/\Lambda _\varepsilon }(T_0)}{v_\varepsilon (0,T_0)m_{1,1/\Lambda _\varepsilon }(T_0+t)} , \quad \mathrm{where \ } 0\le z(t)\le t. \end{aligned}$$
(3.38)

We conclude from (3.37), (3.38) that

$$\begin{aligned} \frac{m_{1,1/\Lambda _\varepsilon }(T_0)}{\Lambda _\varepsilon (T_0)} \le \frac{dm_{1,1/\Lambda _\varepsilon }(T_0+t)}{\hbox {d}t} \le [1+2\gamma (\delta )]\frac{m_{1,1/\Lambda _\varepsilon }(T_0)}{\Lambda _\varepsilon (T_0)} \quad \mathrm{for \ } 0\le t\le \delta \Lambda _\varepsilon (T_0). \end{aligned}$$
(3.39)

On integrating (3.39), we have

$$\begin{aligned} 1+\frac{t}{\Lambda _\varepsilon (T_0)} \le \frac{m_{1,1/\Lambda _\varepsilon }(T_0+t)}{m_{1,1/\Lambda _\varepsilon }(T_0)} \le 1+\frac{[1+2\gamma (\delta )]t}{\Lambda _\varepsilon (T_0)} \quad \mathrm{for \ } 0\le t\le \delta \Lambda _\varepsilon (T_0). \end{aligned}$$
(3.40)

Hence, (3.38), (3.40) imply that

$$\begin{aligned} \frac{1}{1+2\gamma (\delta )}\left[ 1+\frac{t}{\Lambda _\varepsilon (T_0)}\right] \le \frac{\Lambda _\varepsilon (T_0+t)}{\Lambda _\varepsilon (T_0)} \le 1+\frac{[1+2\gamma (\delta )]t}{\Lambda _\varepsilon (T_0)} \quad \mathrm{for \ } 0\le t\le \delta \Lambda _\varepsilon (T_0). \end{aligned}$$
(3.41)

We define now \(T_1>T_0\) as the minimum time \(T_1=T_0+t\) such that \(\Lambda _\varepsilon (T_0+t)\ge (1+\delta /2)\Lambda _\varepsilon (T_0)\). The inequality (3.41) now yields bounds on \(T_1-T_0\) as

$$\begin{aligned} \frac{\delta /2}{1+2\gamma (\delta )} \le \frac{T_1-T_0}{\Lambda _\varepsilon (T_0)} \le [1+2\gamma (\delta )][1+\delta /2]-1 , \end{aligned}$$
(3.42)

provided the RHS of (3.42) is less than \(\delta \). In view of (3.32), this will be the case if \(\delta >0\) is sufficiently small. We can iterate the inequality (3.42) by defining \(T_k, \ k=1,2,\ldots ,\) as the minimum time such that \(\Lambda _\varepsilon (T_k)\ge (1+\delta /2)\Lambda _\varepsilon (T_{k-1})\). Thus, we have that

$$\begin{aligned} \frac{\delta /2}{1+2\gamma (\delta )} \le \frac{T_k-T_{k-1}}{(1+\delta /2)^{k-1}\Lambda _\varepsilon (T_0)}\le [1+2\gamma (\delta )][1+\delta /2]-1 , \quad k=1,2,\ldots \end{aligned}$$
(3.43)

On summing (3.43) over \(k=1,\ldots ,N\), we conclude that

$$\begin{aligned}&\left[ 1-\frac{1}{(1+\delta /2)^N}\right] \frac{1}{1+2\gamma (\delta )} \le \frac{T_N-T_{0}}{\Lambda _\varepsilon (T_N)} \nonumber \\&\quad \le \left[ 1-\frac{1}{(1+\delta /2)^N}\right] \frac{[1+2\gamma (\delta )][1+\delta /2]-1}{\delta /2}. \end{aligned}$$
(3.44)

It follows from (3.44) that

$$\begin{aligned}&\frac{1}{(1+\delta /2)[1+2\gamma (\delta )]} \le \liminf _{t\rightarrow \infty } \frac{t}{\Lambda _\varepsilon (t)} \nonumber \\&\quad \le \limsup _{t\rightarrow \infty } \frac{t}{\Lambda _\varepsilon (t)} \le (1+\delta /2)\frac{[1+2\gamma (\delta )][1+\delta /2]-1}{\delta /2} . \end{aligned}$$
(3.45)

Now using the fact that \(\lim _{\delta \rightarrow 0}\gamma (\delta )/\delta =0\), we conclude from (3.45) that \(\lim _{t\rightarrow \infty }\Lambda _\varepsilon (t)/t=1\). \(\square \)

Remark 3

Theorem 5.4 of Carr and Penrose (1998) implies in the case \(\varepsilon =0\) convergence to the exponential self-similar solution for initial data \(v_\varepsilon (x,0), \ x\ge 0,\) which has the property \(\lim _{x\rightarrow \infty } v_\varepsilon (x,0)/x^\alpha =v_{\varepsilon ,\infty }\) with \(0<v_{\varepsilon ,\infty }<\infty \) provided \(\alpha >-1\). It is easy to see that if \(\alpha \ge 0\), then such initial data satisfy the condition (3.32) of Lemma 3.4.

In Carr (2006), necessary and sufficient conditions—(5.18) and (5.19) of Carr (2006)—for convergence to the exponential self-similar solution are obtained in the case \(\varepsilon =0\). Note that (5.19) of Carr (2006) implies the condition (3.32) of Lemma 3.4.

Next we obtain a rate of convergence theorem for \(\lim _{t\rightarrow \infty }\Lambda _\varepsilon (t)/t\) which generalizes Proposition 2.1 to the system (2.62), (2.63). We assume that the function \(x\rightarrow v_\varepsilon (x,0)\) is \(C^1\) for large x, in which case the condition (3.32) becomes \(\lim _{x\rightarrow \infty } v_\varepsilon (x,0)^{-2}\partial v_\varepsilon (x,0)/\partial x=0\), or equivalently \(\lim _{x\rightarrow \infty }\beta _{X_0}(x)=1\) for the initial condition random variable \(X_0\). More precisely, we have that if \(v_\varepsilon (x,0)^{-2}\partial v_\varepsilon (x,0)/\partial x\le \eta \) for \(x\ge x_\eta \), then

$$\begin{aligned} \frac{1}{v_\varepsilon (x,0)} - \frac{1}{v_\varepsilon (x+\delta /v_\varepsilon (x,0),0)} \le \frac{\eta \delta }{v_\varepsilon (x,0)} \quad \mathrm{for \ } x\ge x_\eta . \end{aligned}$$
(3.46)

The inequality (3.46) implies (3.32) holds with \(\gamma (\delta )\le \eta \delta /(1-\eta \delta )\). We conclude that (3.32) holds if \(\lim _{x\rightarrow \infty } v_\varepsilon (x,0)^{-2}\partial v_\varepsilon (x,0)/\partial x=0\).

The condition on the initial data to guarantee a logarithmic rate of convergence for \(\Lambda _\varepsilon (t)/t\) is similar to (3.32). We require that there exists \(\delta ,\gamma (\delta ),x_\delta >0\) such that

$$\begin{aligned} \frac{v_\varepsilon \left( y,0\right) ^2}{\partial v_\varepsilon (y,0)/\partial y} \ge \frac{v_\varepsilon \left( x,0\right) ^2}{\partial v_\varepsilon (x,0)/\partial x} +\delta \gamma (\delta )\quad \mathrm{for \ } y=x+\delta /v_\varepsilon (x,0) , \ x\ge x_\delta . \end{aligned}$$
(3.47)

Observe that if (3.47) holds for arbitrarily small \(\delta >0\) and \(\liminf _{\delta \rightarrow 0}x_\delta =x_0\), then the function \(x\rightarrow 1/v_\varepsilon (x,0)\) is convex for \(x\ge x_0\). Furthermore, if the function \(x\rightarrow v_\varepsilon (x,0)\) is \(C^2\), then on taking \(\delta \rightarrow 0\) in (3.47) we obtain the second-order differential inequality

$$\begin{aligned} \frac{v_\varepsilon (x,0)\partial ^2 v_\varepsilon (x,0)/\partial x^2}{[\partial v_\varepsilon (x,0)/\partial x]^2} \le 2-\gamma (0) . \end{aligned}$$
(3.48)

Suppose now that (3.48) holds with \(\gamma (0) = \eta >0\) for \(x\ge x_0\). Then, we have that

$$\begin{aligned} \frac{\partial }{\partial x} \frac{v_\varepsilon (x,0)^2}{\partial v_\varepsilon (x,0)/\partial x} \ge \eta v_\varepsilon (x,0) \quad \mathrm{for \ } x\ge x_0 . \end{aligned}$$
(3.49)

On integrating (3.49) and using the fact that the function \(x\rightarrow v_\varepsilon (x,0)\) is increasing, we conclude that (3.47) holds for all \(\delta >0\) with \(\gamma (\delta )=\eta \) and \(x_\delta =x_0\).

It is easy to see that (3.48) is invariant under affine transformations. That is if the function \(x\rightarrow v_\varepsilon (x,0)\) satisfies (3.48) for all \(x>0\), then given any \(\lambda ,k>0\) so also does the function \(x\rightarrow \lambda v_\varepsilon (\lambda x+k,0)\). We can solve the differential equation determined by equality in (3.48). The solution is given by the formula

$$\begin{aligned} v_\varepsilon (x,0) = a[1+\lambda x]^\alpha , \quad \mathrm{where \ } \alpha =1/[\gamma (0)-1] . \end{aligned}$$
(3.50)

Since we require \(\gamma (0)>0\), it follows from (3.48) that \(\alpha \) must satisfy either \(\alpha > 0\) or \(\alpha <-1\). Note that the function \(x\rightarrow 1/v_\varepsilon (x,0)\) of (3.50) is convex precisely for this range of \(\alpha \) values.

Lemma 3.5

Assume the initial data \(x\rightarrow v_\varepsilon (x,0)\) for (2.62), (2.63) are \(C^1\) increasing and that the function \(x\rightarrow 1/v_\varepsilon (x,0)\) is convex for sufficiently large x. Assume further that there exists \(\delta ,\gamma (\delta ),x_\delta >0\) such that (3.47) holds. Then, there exists constant \(C_0,t_0>0\) such that

$$\begin{aligned} 1-\frac{C_0}{\log t} \le \frac{\mathrm{d}\Lambda _\varepsilon (t)}{{\mathrm{d}}t} \le 1 \ \quad \mathrm{for \ } t\ge t_0 . \end{aligned}$$
(3.51)

Proof

Since the inequality (3.47) is invariant under affine transformations, we see as in Lemma 3.4 that in the case \(\varepsilon =0\) there exists \(T_0>0\) such that if \(t\ge T_0\) the function \(x\rightarrow 1/v_\varepsilon (x,t)\) is convex for \(x\ge 0\), and

$$\begin{aligned} \frac{v_\varepsilon \left( y,t\right) ^2}{\partial v_\varepsilon (y,t)/\partial y} \ge \frac{v_\varepsilon \left( x,t\right) ^2}{\partial v_\varepsilon (x,t)/\partial x} +\delta \gamma (\delta )\quad \mathrm{for \ } y=x+\delta /v_\varepsilon (x,t) , \ x\ge 0, \ t\ge T_0 . \end{aligned}$$
(3.52)

Next observe that since \(\lim _{x\rightarrow \infty } v_\varepsilon (x,0)^{-2}\partial v_\varepsilon (x,0)/\partial x=0\), we may for any \(\nu >0\) choose \(T_0\) such that \(v_\varepsilon (x,T_0)^{-2}\partial v_\varepsilon (x,T_0)/\partial x\le \nu \) for \(x\ge 0\). It follows then from (3.46) that

$$\begin{aligned} \frac{v_\varepsilon (y,T_0)}{v_\varepsilon (0,T_0)} \le \frac{1}{1-\nu v_\varepsilon (0,T_0)y} \ \quad \mathrm{for \ } 0\le y< \frac{1}{\nu v_\varepsilon (0,T_0)} =\frac{\Lambda _\varepsilon (T_0)}{\nu } . \end{aligned}$$
(3.53)

Hence, as in (3.39) we see from (3.38), (3.53) that

$$\begin{aligned} \frac{dm_{1,1/\Lambda _\varepsilon }(T_0+t)}{\hbox {d}t} \le \frac{m_{1,1/\Lambda _\varepsilon }(T_0)}{[1-\nu v_\varepsilon (0,T_0)t]\Lambda _\varepsilon (T_0)} \quad \mathrm{for \ } 0\le t< \frac{\Lambda _\varepsilon (T_0)}{\nu }. \end{aligned}$$
(3.54)

Integrating (3.54), we conclude that for \(0\le t< \Lambda _\varepsilon (T_0)/\nu \),

$$\begin{aligned} \frac{m_{1,1/\Lambda _\varepsilon }(T_0)}{m_{1,1/\Lambda _\varepsilon }(T_0+t)} \ge \left[ 1-\frac{1}{\nu }\log \left\{ 1-\frac{\nu t}{\Lambda _\varepsilon (T_0)}\right\} \right] ^{-1} . \end{aligned}$$
(3.55)

Using the inequality \(-\log (1-z)\le 3z/2\) when \(0\le z\le 1/3\), we conclude from (3.55) that

$$\begin{aligned} \frac{m_{1,1/\Lambda _\varepsilon }(T_0)}{m_{1,1/\Lambda _\varepsilon }(T_0+t)} \ge \left[ 1+\frac{3 t}{2\Lambda _\varepsilon (T_0)}\right] ^{-1} \quad \mathrm{for \ } 0\le t\le \frac{\Lambda _\varepsilon (T_0)}{3\nu } . \end{aligned}$$
(3.56)

Similarly to (3.41), we have from (3.56) that

$$\begin{aligned} \frac{\Lambda _\varepsilon (T_0+t)}{\Lambda _\varepsilon (T_0)} \le 1+\frac{3 t}{2\Lambda _\varepsilon (T_0)} \quad \mathrm{for \ } 0\le t\le \frac{\Lambda _\varepsilon (T_0)}{3\nu } . \end{aligned}$$
(3.57)

In the case \(\varepsilon =0\), the LHS of (3.56) is \(\hbox {d}z(t)/\hbox {d}t\), so on integration we have that

$$\begin{aligned} z(t) \ge \frac{2\Lambda _\varepsilon (T_0)}{3}\log \left[ 1+\frac{3 t}{2\Lambda _\varepsilon (T_0)}\right] \quad \mathrm{for \ } 0\le t\le \frac{\Lambda _\varepsilon (T_0)}{3\nu } . \end{aligned}$$
(3.58)

We choose now \(\nu \) sufficiently small so that \(2\log [1+1/2\nu ]/3>\delta \), and let \(T_1\) be the minimum \(T_0+t\) such that \(z(t)\ge \delta /v_\varepsilon (0,T_0)\). Then, we have that

$$\begin{aligned} T_1-T_0 \le \frac{\Lambda _\varepsilon (T_0)}{3\nu } , \quad \frac{v_\varepsilon \left( 0,T_1\right) ^2}{\partial v_\varepsilon (0,T_1)/\partial x} \ge \frac{v_\varepsilon \left( 0,T_0\right) ^2}{\partial v_\varepsilon (0,T_0)/\partial x} +\delta \gamma (\delta ) . \end{aligned}$$
(3.59)

Furthermore, (3.57) implies that \(\Lambda _\varepsilon (T_1)/\Lambda _\varepsilon (T_0)\le 1+1/2\nu \). We now iterate the foregoing to yield a sequence of times \(T_k, \ k=1,2,\ldots ,\) with the properties that

$$\begin{aligned} T_k-T_{k-1} \le \frac{\Lambda _\varepsilon (T_{k-1})}{3\nu }, \quad \frac{\Lambda _\varepsilon (T_k)}{\Lambda _\varepsilon (T_{k-1})} \le 1+\frac{1}{2\nu }, \quad \frac{v_\varepsilon \left( 0,T_k\right) ^2}{\partial v_\varepsilon (0,T_k)/\partial x}\ge k\delta \gamma (\delta ). \end{aligned}$$
(3.60)

It follows from (3.60) that

$$\begin{aligned} T_N-T_0 \le \frac{2}{3} \left( 1+\frac{1}{2\nu }\right) ^N\Lambda _\varepsilon (T_0), \quad \frac{\hbox {d}\Lambda _\varepsilon (T_N)}{\hbox {d}t}\ge 1-\frac{1}{N\delta \gamma (\delta )}, \quad N=1,2,\ldots \end{aligned}$$
(3.61)

The inequality (3.61) implies the lower bound in (3.51) since the function \(t\rightarrow \hbox {d}\Lambda _\varepsilon (t)/\hbox {d}t\) is increasing for \(t\ge T_0\).

To deal with \(\varepsilon >0\), we first assume that the function \(x\rightarrow v_\varepsilon (x,0)\) is \(C^2\) for \(x>0\). Letting z(xt) be the solution to (3.3) we have that

$$\begin{aligned} \frac{\partial z(x,t)}{\partial x}= & {} \frac{1}{m_{1,1/\Lambda _\varepsilon }(t)} \left[ 1+\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \frac{\partial v_\varepsilon (z(x,t),0)}{\partial z}\right] ^{-1} , \end{aligned}$$
(3.62)
$$\begin{aligned} \frac{\partial ^2 z(x,t)}{\partial x^2}= & {} -\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \frac{\partial ^2 v_\varepsilon (z(x,t),0)}{\partial z^2} \nonumber \\&\quad \left[ 1+\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \frac{\partial v_\varepsilon (z(x,t),0)}{\partial z}\right] ^{-1}\left( \frac{\partial z(x,t)}{\partial x}\right) ^2 . \end{aligned}$$
(3.63)

We also have that

$$\begin{aligned} \frac{\partial v_\varepsilon (x,t)}{\partial x}= & {} \frac{1}{m_{1,1/\Lambda _\varepsilon }(t)} \frac{\partial v_\varepsilon (z(x,t),0)}{\partial z} \frac{\partial z(x,t)}{\partial x} , \end{aligned}$$
(3.64)
$$\begin{aligned} \frac{\partial ^2 v_\varepsilon (x,t)}{\partial x^2}= & {} \frac{1}{m_{1,1/\Lambda _\varepsilon }(t)} \left[ \frac{\partial ^2 v_\varepsilon (z(x,t),0)}{\partial z^2} \left( \frac{\partial z(x,t)}{\partial x}\right) ^2+ \frac{\partial v_\varepsilon (z(x,t),0)}{\partial z} \frac{\partial ^2 z(x,t)}{\partial x^2}\right] .\nonumber \\ \end{aligned}$$
(3.65)

It follows from (3.62)–(3.65) that the ratio (3.48) for the function \(x\rightarrow v_\varepsilon (x,t)\) is given by

$$\begin{aligned} \frac{v_\varepsilon (x,t)\partial ^2 v_\varepsilon (x,t)/\partial x^2}{[\partial v_\varepsilon (x,t)/\partial x]^2}= & {} \frac{v_\varepsilon (z(x,t),0)\partial ^2 v_\varepsilon (z(x,t),0)/\partial z^2}{[\partial v_\varepsilon (z(x,t),0)/\partial z]^2} \nonumber \\&\quad \left[ 1+\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \frac{\partial v_\varepsilon (z(x,t),0)}{\partial z}\right] ^{-1} . \end{aligned}$$
(3.66)

Hence, if (3.48) holds for all sufficiently large x, then Lemma 3.3 implies that there exists \(T_0>0\) such that the RHS of (3.66) is bounded above by 2 for \(x\ge 0,t\ge T_0\). It follows that the function \(x\rightarrow 1/v_\varepsilon (x,t)\) is convex for \(x\ge 0\) provided \(t\ge T_0\). Observe that if \(0<\gamma (0)\le 2\) in (3.48), then the RHS of (3.66) is bounded above by \(2-\gamma (0)\). However, if \(\gamma (0)>2\), then we can only bound the RHS above by 0. We have shown that if (3.48) holds for all sufficiently large x, then there exists \(T_0>0\) such that

$$\begin{aligned} \frac{v_\varepsilon (x,t)\partial ^2 v_\varepsilon (x,t)/\partial x^2}{[\partial v_\varepsilon (x,t)/\partial x]^2} \le \max [2-\gamma (0),0] \quad \mathrm{for \ } x\ge 0, \ t\ge T_0 . \end{aligned}$$
(3.67)

The identity (3.66) applied to the example (3.50) provides us with an illustration of how the selection principle operates. In the case \(1<\gamma (0)<2\) when \(\alpha >1\) in (3.50), then \(\lim _{z\rightarrow \infty }\partial v_\varepsilon (z,0)/\partial z=\infty \), in which case for \(t>0\) the RHS of (3.66) converges to 0 as \(x\rightarrow \infty \). Thus, if the initial data have \(\alpha >1\), then for \(t>0\) we expect the function \(v_\varepsilon (x,t)\) to behave linearly in x at large x.

In the case when we only assume that the function \(x\rightarrow v_\varepsilon (x,0)\) is \(C^1\) for \(x>0\), we can make a more careful version of the argument of the previous paragraph. We have now from (3.62), (3.64) that

$$\begin{aligned} \frac{v_\varepsilon \left( x,t\right) ^2}{\partial v_\varepsilon (x,t)/\partial x} =\frac{v_\varepsilon \left( z(x,t),0\right) ^2}{\partial v_\varepsilon (z(x,t),0)/\partial z} +\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2}v_\varepsilon (z(x,t),0)^2. \end{aligned}$$
(3.68)

Since the function \(x\rightarrow z(x,t)\) is increasing for \(x\ge 0\), it follows that the second function on the RHS of (3.68) is increasing for \(x\ge 0\). Since we are assuming that the function \(z\rightarrow 1/v_\varepsilon (z,0)\) is convex for all large z, it follows that the first function on the RHS of (3.68) is also increasing for \(x\ge 0\) provided \(t\ge T_0\) and \(T_0\) is sufficiently large. We conclude that the function \(x\rightarrow 1/v_\varepsilon (x,t)\) is convex for \(x\ge 0\) provided \(t\ge T_0\).

We can also obtain an inequality (3.52) for a \(\delta \) which is twice the \(\delta \) which occurs in (3.47). To show this, we consider two possibilities. In the first of these (3.52) follows from the monotonicity of the second function on the RHS of (3.68). We use the inequality

$$\begin{aligned}&\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2}v_\varepsilon \left( z\left( x+\frac{\eta }{v_\varepsilon (x,t)},t\right) ,0\right) ^2 \ \ge \varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2}v_\varepsilon (z(x,t),0)^2 \nonumber \\&\quad +2\varepsilon \eta \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2}\int _0^1d\rho \ \frac{\partial v_\varepsilon }{\partial z}\left( z\left( x+\frac{\rho \eta }{v_\varepsilon (x,t)},t\right) ,0\right) \nonumber \\&\qquad \left[ 1+ \varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2}\frac{\partial v_\varepsilon }{\partial z}\left( z\left( x+\frac{\rho \eta }{v_\varepsilon (x,t)},t\right) ,0\right) \ \right] ^{-1} , \end{aligned}$$
(3.69)

which follows from (3.4), (3.62), the monotonicity of the function \(z\rightarrow v_\varepsilon (z,0)\), and Taylor’s formula. Observe next from the convexity of the function \(z\rightarrow 1/v_\varepsilon (z,0)\) that for \(0<\rho '<\rho \),

$$\begin{aligned}&\partial v_\varepsilon \left( z\left( x+\frac{\rho \eta }{v_\varepsilon (x,t)},t\right) ,0\right) \Big /\partial z \nonumber \\&\qquad \le \left[ \frac{v_\varepsilon (x+\rho \eta /v_\varepsilon (x,t),t)}{v_\varepsilon (x+\rho '\eta /v_\varepsilon (x,t),t)}\right] ^2\frac{\partial v_\varepsilon }{\partial z}\left( z\left( x+\frac{\rho '\eta }{v_\varepsilon (x,t)},t\right) ,0\right) . \end{aligned}$$
(3.70)

Furthermore, we have similarly to (3.53) that for \(\nu >0\) there exists \(T_0>0\) such that for \(0<\rho '<\rho \) and \(t\ge T_0\),

$$\begin{aligned} 1 \le \frac{v_\varepsilon (x+\rho \eta /v_\varepsilon (x,t),t)}{v_\varepsilon (x+\rho '\eta /v_\varepsilon (x,t),t)} \le \frac{1-\nu \rho '\eta }{1-\nu \rho \eta } \quad \mathrm{provided \ } \rho <\frac{1}{\nu \eta } . \end{aligned}$$
(3.71)

Now let us assume that \(t\ge T_0\) and

$$\begin{aligned} \varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2}\frac{\partial v_\varepsilon }{\partial z}\left( z\left( x+\frac{\delta }{2v_\varepsilon (x,t)},t\right) ,0\right) \ge \frac{1}{4} . \end{aligned}$$
(3.72)

Then, (3.70), (3.71) imply upon setting \(\eta =\delta /2\) in (3.69) and choosing \(\nu \) less than some constant depending only on \(\delta \), that the second term on the RHS is bounded below by \(\delta /6\) for \(t\ge T_0\). This implies that (3.52) holds with \(\gamma (\delta )=1/6\).

Alternatively, we assume that (3.72) does not hold. Then on choosing \(\nu \) sufficiently small, depending only on \(\delta \), we see that (3.62), (3.70), (3.71) implies

$$\begin{aligned} z\left( x+\frac{2\delta }{v_\varepsilon (x,t)},t\right) \ge z\left( x+\frac{\delta }{2v_\varepsilon (x,t)},t\right) + \frac{11\delta }{10v_\varepsilon (z(x,t),0)} . \end{aligned}$$
(3.73)

Then, we use the first term on the RHS of (3.68) and (3.47), (3.73) to establish (3.52) with \(2\delta \) in place of \(\delta \). We have proved then that there exists \(T_0>0\) such that (3.52) holds (with \(\delta \) replaced by \(2\delta \)).

We wish next to establish an inequality like (3.59) in the case \(\varepsilon >0\), in which case we need to examine the terms of (3.21) that depend on \(\varepsilon \). Using the notation of (4.27), (4.28) the \(\varepsilon \) dependent coefficient on the LHS of (3.21) is given by

$$\begin{aligned}&\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(T_0,T_0+t)}{m_{1,1/\Lambda _\varepsilon }(T_0,T_0+t)^2}\frac{\partial v_\varepsilon (z(t),T_0)}{\partial z} \nonumber \\&\qquad = \varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(T_0,T_0+t)}{\Lambda _\varepsilon (T_0+t)^2}\left[ v_\varepsilon (z(t),T_0)^{-2}\frac{\partial v_\varepsilon (z(t),T_0)}{\partial z}\right] . \end{aligned}$$
(3.74)

The \(\varepsilon \) dependent coefficient on the RHS of (3.21) is given by

$$\begin{aligned} \varepsilon \frac{v_\varepsilon (z(t),T_0)}{m_{1,1/\Lambda _\varepsilon }(T_0,T_0+t)} =\frac{\varepsilon }{\Lambda _\varepsilon (T_0+t)} . \end{aligned}$$
(3.75)

We choose now \(T_0\) large enough so that \(\varepsilon /\Lambda _\varepsilon (T_0)<1/2\), whence (3.75) implies that the term in brackets on the RHS of (3.21) is at least 1/2. We also have from (3.74) that

$$\begin{aligned} \varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(T_0,T_0+t)}{m_{1,1/\Lambda _\varepsilon }(T_0,T_0+t)^2}\frac{\partial v_\varepsilon (z(t),T_0)}{\partial z} \le \frac{\nu }{2}\frac{\sigma _{1/\Lambda _\varepsilon }^2(T_0,T_0+t)}{\Lambda _\varepsilon (T_0+t)}. \end{aligned}$$
(3.76)

Now using (4.27), we conclude from (3.76) that for any \(K>0\),

$$\begin{aligned} \varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(T_0,T_0+t)}{m_{1,1/\Lambda _\varepsilon }(T_0,T_0+t)^2}\frac{\partial v_\varepsilon (z(t),T_0)}{\partial z} \le \frac{\nu K\exp (2K)}{2} \quad \mathrm{for \ } 0\le t\le K\Lambda _\varepsilon (T_0) . \end{aligned}$$
(3.77)

It follows from (3.52), (3.62), (3.77) that there exists \(T_1>T_0\) such that (3.59) holds. Therefore, we can define a sequence \( T_k, \ k=1,2,\ldots ,\) of times having the properties (3.60).

In order to estimate \(\hbox {d}\Lambda _\varepsilon (T_{k-1}+t)/\hbox {d}t\) for \(0\le t\le T_k-T_{k-1}\), we need to examine the terms of (3.22) that depend on \(\varepsilon \). Similarly to (3.77), we have from (2.63), (3.22) and the convexity of the function \(x\rightarrow 1/v_\varepsilon (x,T_{k-1})\) that

$$\begin{aligned} \left[ 1+\frac{\varepsilon \nu K\exp (2K)}{\Lambda _\varepsilon (T_{k-1})} \right] \frac{\hbox {d}\Lambda _\varepsilon (T_{k-1}+t)}{\hbox {d}t}\ge 1- \frac{\partial v_\varepsilon (0,T_{k-1})/\partial x}{v_\varepsilon (0,T_{k-1})^2} \quad \mathrm{for \ } 0\le t\le T_k-T_{k-1}. \end{aligned}$$
(3.78)

Noting from (3.41) that \(\Lambda _\varepsilon (T_k)\) grows exponentially in k, we conclude from (3.60), (3.61) and (3.78) that the lower bound in (3.51) holds. To obtain the upper bound in (3.51), we use the identity

$$\begin{aligned} \frac{\hbox {d}\Lambda _\varepsilon (t)}{\hbox {d}t} =1-\left[ 1-\varepsilon v_\varepsilon (0,t)\right] \frac{1}{v_\varepsilon (0,t)^2}\frac{\partial v_\varepsilon (0,t)}{\partial x} , \end{aligned}$$
(3.79)

obtained from (2.62), (2.63). Evidently, the RHS of (3.79) does not exceed 1. \(\square \)

Lemma 3.6

Assume \(\varepsilon >0\) and the initial data for (2.62), (2.63) satisfy the conditions of Lemma 3.1 with \(x_\infty <\infty \). Then, for any \(t>0\) the function \(x\rightarrow v_\varepsilon (x,t)\) satisfies \(\lim _{x\rightarrow \infty } v_\varepsilon (x,t)/x=v_\infty (t)\) for some \(v_\infty (t)>0\).

Assume in addition that the initial data are \(C^1\), the function \(x\rightarrow 1/v_\varepsilon (x,0)\) is convex for x sufficiently close to \(x_\infty \), and \(\liminf _{x\rightarrow x_\infty } \partial v_\varepsilon (x,0)/\partial x>0\). Then, for any \(t>0\) the function \(x\rightarrow 1/v_\varepsilon (x,t)\) is convex for x sufficiently large, and the inequality (3.47) holds for all \(\delta >0\).

Proof

From (3.3), (3.4), we have that

$$\begin{aligned} \frac{x+m_{2,1/\Lambda _\varepsilon }(t)-m_{1,1/\Lambda _\varepsilon }(t)x_\infty }{\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)} \le v_\varepsilon (x,t) \le \frac{x+m_{2,1/\Lambda _\varepsilon }(t)}{\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)} \quad \mathrm{for \ } x\ge 0. \end{aligned}$$
(3.80)

Hence, \(\lim _{x\rightarrow \infty } v_\varepsilon (x,t)/x=1/\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)\).

It is easy to see from our assumptions that (3.47) is satisfied if \(v_\varepsilon (x,0)\) is \(C^2\) for x sufficiently close to \(x_\infty \). In that case, it follows from the convexity of the function \(x\rightarrow 1/v_\varepsilon (x,0)\) close to \(x_\infty \) and (3.66) that there exists \(\eta (t)>0\) and \(x_\eta (t)\) with

$$\begin{aligned} \frac{v_\varepsilon (x,t)\partial ^2 v_\varepsilon (x,t)/\partial x^2}{[\partial v_\varepsilon (x,t)/\partial x]^2} \le 2\left[ 1+\varepsilon \frac{\sigma _{1/\Lambda _\varepsilon }^2(t)}{m_{1,1/\Lambda _\varepsilon }(t)^2} \frac{\partial v_\varepsilon (z(x,t),0)}{\partial z}\right] ^{-1} \le 2-\eta (t) \ \end{aligned}$$
(3.81)

for \(x>x_\eta (t)\). If we only assume the function \(x\rightarrow v_\varepsilon (x,t)\) is \(C^1\), then we use (3.68), whence (3.47) follows from (3.69). \(\square \)

Proof of Theorem 1.1:

Note the assumption that the function \(x\rightarrow E[X_0-x \ | \ X_0>x]\) is decreasing implies that the initial data \(v_\varepsilon (\cdot ,0)\) for (2.62), (2.63) are continuous and increasing. Now \(\lim _{t\rightarrow \infty }\langle X_t\rangle /t=1\) follows from Lemma 3.4, the remark following it and Lemma 3.6.

To prove (1.12), we first observe from (2.65) that

$$\begin{aligned} P(X_t>x) =\frac{v_\varepsilon (x,t)}{v_\varepsilon (0,t)}\exp \left[ -\int _0^x v_\varepsilon (x',t) \ \hbox {d}x' \ \right] . \end{aligned}$$
(3.82)

Since the function \(x\rightarrow v_\varepsilon (x,t)\) is increasing, it follows from (3.82) that

$$\begin{aligned} P(X_t>x)\ge \exp \left[ -xv_\varepsilon (x,t)\right] . \end{aligned}$$
(3.83)

The lower bound in (1.12) follows if we can show that there exists \(T>0\) such that we may take \(\delta =M\) in (3.37) with \(2\gamma (\delta )<\eta \) provided \(t>T\). This is a consequence of (3.36) since (3.80) implies that we may assume there are constants \(\lambda _0>0\) and \(k_1,k_2\) such that the initial data satisfy an inequality \(\lambda _0x+k_1\le v_\varepsilon (x,0)\le \lambda _0x+k_2\) at large x. The upper bound in (1.12) can be obtained similarly. Thus, we have from (3.82) that

$$\begin{aligned} P(X_t>x) \le \frac{v_\varepsilon (x,t)}{v_\varepsilon (0,t)}\exp \left[ -x v_\varepsilon (0,t) \ \right] . \end{aligned}$$
(3.84)

Now we argue as in the lower bound since \(x\ge mt\) implies that \(xv_\varepsilon (0,t)\ge m\).

To prove the logarithmic rate of convergence, we first observe that the inequality (1.14) follows from Lemmas 3.5 and 3.6. Now from the argument of Lemma 3.5, we see that there exists \(C,T>0\) such that for \(t\ge T\), the function \(x\rightarrow v_\varepsilon (x,t)^{-2}\partial v_\varepsilon (x,t)/\partial x\) is positive decreasing and satisfies the inequality

$$\begin{aligned} \frac{1}{v_\varepsilon (0,t)^2} \frac{\partial v_\varepsilon (0,t)}{\partial x} \le \frac{C}{\log t} , \quad t\ge T. \end{aligned}$$
(3.85)

From (3.85), we see that

$$\begin{aligned} 1 \le \frac{v_\varepsilon (x,t)}{v_\varepsilon (0,t)} \le \frac{1}{1-Cxv_\varepsilon (0,t)/\log t} , \quad 0\le x< \log t/Cv_\varepsilon (0,t), \ t\ge T . \end{aligned}$$
(3.86)

The inequality (1.15) follows from (3.86). \(\square \)

4 Representations of Green’s Functions

Let \(b:\mathbf {R}\times \mathbf {R}\rightarrow \mathbf {R}\) be a continuous function which satisfies the uniform Lipschitz condition

$$\begin{aligned} \sup \left\{ |\partial b(y,t)/\partial y| \ : \ y,t\in \mathbf {R}\right\} \le A_\infty \end{aligned}$$
(4.1)

for some constant \(A_\infty \). Then, the terminal value problem

$$\begin{aligned}&\frac{\partial u_\varepsilon (y,t)}{\partial t}+b(y,t)\frac{\partial u_\varepsilon (y,t)}{\partial y}+\frac{\varepsilon }{2}\frac{\partial ^2 u_\varepsilon (y,t)}{\partial y^2} =0, \quad y\in \mathbf {R}, \ t<T, \end{aligned}$$
(4.2)
$$\begin{aligned}&u_\varepsilon (y,T) =u_T(y), \quad y\in \mathbf {R}, \end{aligned}$$
(4.3)

has a unique solution \(u_\varepsilon \) which has the representation

$$\begin{aligned} u_\varepsilon (y,t) =\int _{-\infty }^\infty G_\varepsilon (x,y,t,T) u_T(x) \ \hbox {d}x, \quad y\in \mathbf {R}, \ t<T, \end{aligned}$$
(4.4)

where \(G_\varepsilon \) is the Green’s function for the problem. The adjoint problem to (4.2), (4.3) is the initial value problem

$$\begin{aligned}&\frac{\partial v_\varepsilon (x,t)}{\partial t}+\frac{\partial }{\partial x}\left[ b(x,t)v_\varepsilon (x,t)\right] =\frac{\varepsilon }{2}\frac{\partial ^2 v_\varepsilon (x,t)}{\partial x^2} , \quad x\in \mathbf {R}, \ t>0, \end{aligned}$$
(4.5)
$$\begin{aligned}&v_\varepsilon (x,0) =v_0(x), \quad y\in \mathbf {R}. \end{aligned}$$
(4.6)

The solution to (4.5), (4.6) is given by the formula

$$\begin{aligned} v_\varepsilon (x,T) =\int _{-\infty }^\infty G_\varepsilon (x,y,0,T) v_0(y) \ \hbox {d}y, \quad x\in \mathbf {R}, \ T>0. \end{aligned}$$
(4.7)

For any \(t<T\) let \(Y_\varepsilon (s), \ s>t,\) be the solution to the initial value problem for the SDE

$$\begin{aligned} \hbox {d}Y_\varepsilon (s) =b(Y_\varepsilon (s),s) \hbox {d}s+\sqrt{\varepsilon } \ \hbox {d}B(s), \quad Y_\varepsilon (t)=y, \end{aligned}$$
(4.8)

where \(B(\cdot )\) is Brownian motion. Then, \(G_\varepsilon (\cdot ,y,t,T)\) is the probability density for the random variable \(Y_\varepsilon (T)\). In the case when the function b(yt) is linear in y, it is easy to see that (4.8) can be explicitly solved. Thus, let \(A:\mathbf {R}\rightarrow \mathbf {R}\) be a continuous function and \(b:\mathbf {R}\times \mathbf {R}\rightarrow \mathbf {R}\) the function \(b(y,t)=A(t)y-1\). The solution to (4.8) is then given by

$$\begin{aligned} Y_\varepsilon (s)= & {} \exp \left[ \int _t^s A(s') \hbox {d}s'\right] y-\int _t^s \exp \left[ \int _{s'}^s A(s'') \hbox {d}s''\right] \hbox {d}s' \nonumber \\&+\sqrt{\varepsilon }\int _t^s \exp \left[ \int _{s'}^s A(s'') \hbox {d}s''\right] \hbox {d}B(s') . \end{aligned}$$
(4.9)

Hence, the random variable \(Y_\varepsilon (T)\) conditioned on \(Y_\varepsilon (0)=y\) is Gaussian with mean \(m_{1,A}(T)y-m_{2,A}(T)\) and variance \(\varepsilon \sigma _A^2(T)\), where \(m_{1,A},m_{2,A}\) are given by (2.2) and \(\sigma ^2_A\) by

$$\begin{aligned} \sigma _A^2(T)= \int _0^T \exp \left[ 2\int _{s}^T A(s') \hbox {d}s'\right] \hbox {d}s . \end{aligned}$$
(4.10)

The Green’s function \(G_\varepsilon (x,y,0,T)\) is therefore explicitly given by the formula

$$\begin{aligned} G_\varepsilon (x,y,0,T)=\frac{1}{\sqrt{2\pi \varepsilon \sigma _A^2(T)}}\exp \left[ -\frac{\{x+m_{2,A}(T)-m_{1,A}(T)y\}^2}{2\varepsilon \sigma _A^2(T)}\right] . \end{aligned}$$
(4.11)

To obtain the formula (4.11), we have used the fact that the solution to the terminal value problem (4.2), (4.3) has a representation as an expectation value \(u_\varepsilon (y,t)=E[u_0(Y_\varepsilon (T)) \ | \ Y(t)=y \ ]\), where \(Y_\varepsilon (\cdot )\) is the solution to the SDE (4.8). The initial value problem (4.5), (4.6) also has a representation as an expectation value in terms of the solution to the SDE

$$\begin{aligned} \hbox {d}X_\varepsilon (s) =b(X_\varepsilon (s),s) \hbox {d}s+\sqrt{\varepsilon } \ \hbox {d}B(s), \quad X_\varepsilon (T)=x, \ s<T. \end{aligned}$$
(4.12)

run backwards in time. Thus, in (4.12) \(B(s), \ s<T,\) is Brownian motion run backwards in time. The solution \(v_\varepsilon \) of (4.5), (4.6) has the representation

$$\begin{aligned} v_\varepsilon (x,T) = E\left[ \exp \left\{ -\int ^T_0\frac{\partial b(X_\varepsilon (s),s)}{\partial x} \ \hbox {d}s\right\} v_0(X_\varepsilon (0)) \ \ \Bigg | \ X_\varepsilon (T)=x \ \right] . \end{aligned}$$
(4.13)

Next we consider the terminal value problem (4.2), (4.3) in the half-space \(y>0\) with Dirichlet boundary condition \(u_\varepsilon (0,t)=0, \ t<T\). In that case, the solution \(u_\varepsilon (y,t)\) has the representation

$$\begin{aligned} u_\varepsilon (y,t) =\int _0^\infty G_{\varepsilon ,D}(x,y,t,T) u_T(x) \ \hbox {d}x, \quad y>0, \ t<T, \end{aligned}$$
(4.14)

in terms of the Dirichlet Green’s function \(G_{\varepsilon ,D}\) for the half-space. Similarly, the solution to (4.5), (4.6) in the half-space \(x>0\) with Dirichlet condition \(v_\varepsilon (0,t)=0, \ t>0,\) has the representation

$$\begin{aligned} v_\varepsilon (x,T) =\int _{0}^\infty G_{\varepsilon ,D}(x,y,0,T) v_0(y) \ \hbox {d}y, \quad x>0, \ T>0. \end{aligned}$$
(4.15)

The function \(G_{\varepsilon ,D}(\cdot ,y,t,T)\) is the probability density of the random variable \(Y_\varepsilon (T)\) for solutions \(Y_\varepsilon (s), \ s>t,\) to (4.8) which have the property that \(\inf _{t\le s\le T} Y_\varepsilon (s)>0\). No explicit formula for \(G_{\varepsilon ,D}(x,y,0,T)\) in the case of linear \(b(y,t)=A(t)y-1\) is known except when \(A(\cdot )\equiv 0\). In that case, the method of images yields the formula

$$\begin{aligned} G_{\varepsilon ,D}(x,y,0,T)=\frac{1}{\sqrt{2\pi \varepsilon T}} \left\{ \exp \left[ -\frac{(x-y+T)^2}{2\varepsilon T}\right] - \exp \left[ -\frac{2x}{\varepsilon }-\frac{(x+y-T)^2}{2\varepsilon T}\right] \right\} . \end{aligned}$$
(4.16)

It follows from (4.11), (4.16) that

$$\begin{aligned} G_{\varepsilon ,D}(x,y,0,T)/G_{\varepsilon }(x,y,0,T) = 1- \exp [-2xy/\varepsilon T] . \end{aligned}$$
(4.17)

We may interpret the formula (4.17) in terms of conditional probability for solutions \(Y_\varepsilon (s), \ s\ge 0,\) of (4.8) with \(b(\cdot ,\cdot )\equiv -1\). Thus, we have that

$$\begin{aligned} P\left( \inf _{0\le s\le T} Y_\varepsilon (s)>0 \ | \ Y_\varepsilon (0)=y, \ Y_\varepsilon (T)=x\right) =1-\exp [-2xy/\varepsilon T] . \end{aligned}$$
(4.18)

We wish to generalize (4.18) to the case of linear \(b(y,t)=A(t)y-1\) in a way that is uniform as \(\varepsilon \rightarrow 0\). To see what conditions on the function \(A(\cdot )\) are needed, we consider for \(x,y\in \mathbf {R},t<T,\) the function q(xyt) defined by the variational formula

$$\begin{aligned} q(x,y,t,T) = \min _{y(\cdot )} \left\{ \frac{1}{2} \int ^T_t \left[ \frac{\hbox {d}y(s)}{\hbox {d}s} - b(y(s),s)\right] ^2\hbox {d}s \ \Big | \ y(t) = y, \ y(T) = x \right\} . \end{aligned}$$
(4.19)

The Euler–Lagrange equation for the minimizing trajectory \(y(\cdot )\) of (4.19) is

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}s} \left[ \frac{\hbox {d}y(s)}{\hbox {d}s} - b(y(s),s) \right] + \frac{\partial b}{\partial y}(y(s),s) \left[ \frac{\hbox {d}y(s)}{\hbox {d}s} - b(y(s),s) \right] = 0, \quad t \le s \le T, \end{aligned}$$
(4.20)

and we need to solve (4.20) for the function \(y(\cdot )\) satisfying the boundary conditions \(y(t)=y, \ y(T)=x\). In the case \(b(y,t)=A(t)y-1\) Eq. (4.20) becomes

$$\begin{aligned} \left[ -\frac{\hbox {d}^2}{\hbox {d}s^2}+A'(s)+A(s)^2\right] y(s) =A(s), \quad t \le s \le T. \end{aligned}$$
(4.21)

It is easy to solve (4.21) with the given boundary conditions explicitly. In fact, taking \(t=0\) we see from (4.20) that

$$\begin{aligned} \frac{\hbox {d}y(s)}{\hbox {d}s} - b(y(s),s) = C(x,y,T)\exp \left[ \int _s^T A(s')\hbox {d}s'\right] , \quad 0\le s\le T, \end{aligned}$$
(4.22)

where the constant C(xyT) is given by the formula

$$\begin{aligned} C(x,y,T) = [x+m_{2,A}(T)-m_{1,A}(T)y]/\sigma _A^2(T) , \end{aligned}$$
(4.23)

with \(m_{1,A}(T),m_{2,A}(T)\) as in (2.2) and \(\sigma _A^2(T)\) as in (4.10). It follows from (4.11), (4.19), (4.22), (4.23) that the Green’s function \(G_\varepsilon (x,y,0,T)\) is given by the formula

$$\begin{aligned} G_\varepsilon (x,y,0,T)=\frac{1}{\sqrt{2\pi \varepsilon \sigma _A^2(T)}}\exp \left[ -q(x,y,0,T)/\varepsilon \right] . \end{aligned}$$
(4.24)

The minimizing trajectory \(y(\cdot )\) for (4.19) has probabilistic significance as well as the function q(xytT). One can easily see that for solutions \(Y_\varepsilon (s), \ 0\le s\le T,\) of (4.8), the random variable \(Y_\varepsilon (s)\) conditioned on \(Y_\varepsilon (0)=y, \ Y_\varepsilon (T)=x,\) is Gaussian with mean and variance given by

$$\begin{aligned}&E[ Y_\varepsilon (s) \ | \ Y_\varepsilon (0)=y, \ Y_\varepsilon (T)=x] =y(s), \quad 0\le s\le T, \end{aligned}$$
(4.25)
$$\begin{aligned}&\mathrm{Var}[ Y_\varepsilon (s) \ | \ Y_\varepsilon (0)=y, \ Y_\varepsilon (T)=x] = \varepsilon \sigma _A^2(0,s)\sigma _A^2(s,T)/\sigma _A^2(T) , \end{aligned}$$
(4.26)

where the function \(\sigma _A^2(s,t)\) is defined by

$$\begin{aligned} \sigma _A^2(s,t) =\int _s^t \exp \left[ 2\int _{s'}^t A(s'') \hbox {d}s''\right] \hbox {d}s' \quad \mathrm{for \ } s\le t. \end{aligned}$$
(4.27)

Let \(m_{1,A}(s,t), \ m_{2,A}(s,t)\) be defined by

$$\begin{aligned} m_{1,A}(s,t)=\exp \left[ \int _{s}^t A(s') \hbox {d}s'\right] , \quad m_{2,A}(s,t)= \int _s^t \exp \left[ \int _{s'}^t A(s'') \hbox {d}s''\right] \hbox {d}s' \quad \mathrm{for \ } s\le t. \end{aligned}$$
(4.28)

The minimizing trajectory \(y(\cdot )\) for the variational problem (4.19) is explicitly given by the formula

$$\begin{aligned} \sigma _A^2(T)y(s)= & {} xm_{1,A}(s,T)\sigma _A^2(0,s) +ym_{1,A}(0,s)\sigma _A^2(s,T) \nonumber \\&+ \ m_{1,A}(s,T)m_{2,A}(s,T)\sigma _A^2(0,s)-m_{2,A}(0,s)\sigma _A^2(s,T) . \end{aligned}$$
(4.29)

Now the process \(Y_\varepsilon (s), \ 0\le s\le T,\) conditioned on \(Y_\varepsilon (0)=y, \ Y_\varepsilon (T)=x,\) is in fact a Gaussian process with covariance independent of xy,

$$\begin{aligned} \mathrm{Covar}[ Y_\varepsilon (s_1),Y_\varepsilon (s_2) \ | \ Y_\varepsilon (0)=y, \ Y_\varepsilon (T)=x] = \varepsilon \Gamma _A(s_1,s_2) , \quad 0\le s_1, s_2 \le T, \end{aligned}$$
(4.30)

where the symmetric function \(\Gamma :[0,T]\times [0,T]\rightarrow \mathbf {R}\) is given by the formula

$$\begin{aligned} \Gamma _A(s_1,s_2) =\frac{m_{1,A}(s_1,s_2)\sigma _A^2(0,s_1)\sigma _A^2(s_2,T)}{\sigma _A^2(T)} , \quad 0\le s_1\le s_2\le T. \end{aligned}$$
(4.31)

The function \(\Gamma _A\) is the Dirichlet Green’s function for the operator on the LHS of (4.21). Thus, one has that

$$\begin{aligned} \left[ -\frac{\hbox {d}^2}{\hbox {d}s_1^2}+A'(s_1)+A(s_1)^2\right] \Gamma _A(s_1,s_2) =\delta (s_1-s_2), \quad 0<s_1,s_2 <T, \end{aligned}$$
(4.32)

and \(\Gamma _A(0,s_2)=\Gamma _A(T,s_2)=0\) for all \(0<s_2<T\).

We can obtain a representation of the conditioned process \(Y_\varepsilon (\cdot )\) in terms of the white noise process, which is the derivative \(dB(\cdot )\) of Brownian motion, by obtaining a factorization of \(\Gamma \) corresponding to the factorization

$$\begin{aligned} -\frac{\hbox {d}^2}{\hbox {d}s^2}+A'(s)+A(s)^2 = \left[ -\frac{\hbox {d}}{\hbox {d}s}-A(s)\right] \left[ \frac{\hbox {d}}{\hbox {d}s}-A(s)\right] . \end{aligned}$$
(4.33)

To do this, we note that the boundary value problem

$$\begin{aligned} \left[ \frac{\hbox {d}}{\hbox {d}s}-A(s)\right] u(s) = v(s), \quad 0<s<T, \quad u(0)=u(T)=0, \end{aligned}$$
(4.34)

has a solution if and only if the function \(v:[0,T]\rightarrow \mathbf {R}\) satisfies the orthogonality condition

$$\begin{aligned} \int _0^T \frac{v(s)}{m_{1,A}(s)} \hbox {d}s =0. \end{aligned}$$
(4.35)

Hence, it follows from (4.33) that we can solve the boundary value problem

$$\begin{aligned} \left[ -\frac{\hbox {d}^2}{\hbox {d}s^2}+A'(s)+A(s)^2 \right] u(s) = f(s), \quad 0<s<T, \quad u(0)=u(T)=0, \end{aligned}$$
(4.36)

by first finding the solution \(v:[0,T]\rightarrow \mathbf {R}\) to

$$\begin{aligned} \left[ -\frac{\hbox {d}}{\hbox {d}s}-A(s)\right] v(s) = f(s), \quad 0<s<T, \end{aligned}$$
(4.37)

which satisfies the orthogonality condition (4.35). Then, we solve the differential equation in (4.34) subject to the condition \(u(0)=0\).

The solution to (4.35), (4.37) is given by an expression

$$\begin{aligned} v(s) = K^*f(s) =\int _0^T k(s',s)f(s') \ \hbox {d}s' , \quad 0\le s\le T, \end{aligned}$$
(4.38)

where the kernel \(k:[0,T]\times [0,T]\rightarrow \mathbf {R}\) is defined by

$$\begin{aligned} k(s',s)= & {} \frac{m_{1,A}(s,s')\sigma ^2(s',T)}{\sigma _A^2(T)} \quad \mathrm{if} \ s'>s, \nonumber \\ k(s',s)= & {} \frac{\sigma _A^2(s',T)}{m_1(s',s)\sigma _A^2(T)}-\frac{1}{m_{1,A}(s',s)} \quad \mathrm{if} \ s'<s. \end{aligned}$$
(4.39)

If \(v:[0,T]\rightarrow \mathbf {R}\) satisfies the condition (4.35) then

$$\begin{aligned} u(s) = Kv(s) =\int _0^T k(s,s')v(s') \hbox {d}s' , \quad 0\le s\le T, \end{aligned}$$
(4.40)

is the solution to (4.34). It follows that the kernel \(\Gamma _A\) of (4.31) has the factorization \(\Gamma _A=KK^*\), and so the conditioned process \(Y_\varepsilon (\cdot )\) has the representation

$$\begin{aligned} Y_\varepsilon (s) =y(s)+ \sqrt{\varepsilon }\int _0^T k(s,s') \hbox {d}B(s') , \quad 0\le s\le T, \end{aligned}$$
(4.41)

where \(y(\cdot )\) is the function (4.29). In the case \(A(\cdot )\equiv 0\) Eq. (4.41) yields the familiar representation

$$\begin{aligned} Y_\varepsilon (s) = \frac{s}{T} x+\left( 1-\frac{s}{T}\right) y+\sqrt{\varepsilon }\left[ B(s)-\frac{s}{T}B(T)\right] , \quad 0\le s\le T, \end{aligned}$$
(4.42)

for the Brownian bridge process.

We can obtain an alternative representation of the conditioned process \(Y_\varepsilon (\cdot )\) in terms of Brownian motion by considering a stochastic control problem. Let \(Y_\varepsilon (\cdot )\) be the solution to the stochastic differential equation

$$\begin{aligned} \hbox {d}Y_\varepsilon (s) = \lambda _\varepsilon (\cdot ,s)\hbox {d}s + \sqrt{\varepsilon } \; \hbox {d}B(s), \end{aligned}$$
(4.43)

where \(\lambda _\varepsilon (\cdot ,s)\) is a non-anticipating function. We consider the problem of minimizing the cost function given by the formula

$$\begin{aligned} q_\varepsilon (x,y,t,T) = \min _{\lambda _\varepsilon } E \left[ \frac{1}{2} \int ^T_t \left[ \lambda _\varepsilon (\cdot ,s) - b( Y_\varepsilon (s),s) \right] ^2 \; \hbox {d}s \ \Big | \ Y_\varepsilon (t)= y,\; Y_\varepsilon (T) = x\right] . \end{aligned}$$
(4.44)

The minimum in (4.44) is to be taken over all non-anticipating \(\lambda _\varepsilon (\cdot ,s)\), \(t \le s < T\), which have the property that the solutions of (4.43) with initial condition \(Y_\varepsilon (t) = y\) satisfy the terminal condition \(Y_\varepsilon (T) = x\) with probability 1. Formally, the optimal controller \(\lambda ^*\) for the problem is given by the expression

$$\begin{aligned} \lambda _\varepsilon (\cdot ,s) = \lambda ^*_\varepsilon ( x, Y_\varepsilon (s), s) = b( Y_\varepsilon (s), s) - \frac{\partial q_\varepsilon }{\partial y} \; ( x, Y_\varepsilon (s), s,T). \end{aligned}$$
(4.45)

Evidently, in the classical control case \(\varepsilon =0\) the solution to (4.43), (4.44) is the solution to the variational problem (4.19). If \(b(y,t)=A(t)y-1\) is a linear function of y, then one expects as in the case of LQ problems that the difference between the cost functions for the classical and stochastic control problems is independent of y. Therefore, from (4.11), (4.24) we expect that

$$\begin{aligned} \lambda ^*_\varepsilon ( x, y, t) =b(y,t)-\frac{\partial q(x,y,t,T)}{\partial y} =A(t)y-1-\frac{\partial }{\partial y}\frac{\{x+m_{2,A}(t,T)-m_{1,A}(t,T)y\}^2}{2\sigma _A^2(t,T)} . \end{aligned}$$
(4.46)

It is easy to see that if we solve the SDE (4.43) with controller given by (4.46) and conditioned on \(Y_\varepsilon (t)=y\), then \(Y_\varepsilon (T)=x\) with probability 1 and in fact the process \(Y_\varepsilon (s), \ t\le s\le T,\) has the same distribution as the process \(Y_\varepsilon (s), \ t\le s\le T,\) satisfying the SDE (4.8) conditioned on \(Y_\varepsilon (t)=y, \ Y_\varepsilon (T)=x\). Thus, we have obtained the Markovian representation for the conditioned process of (4.8). Note, however, that the stochastic control problem with cost function (4.44) does not have a solution since the integral in (4.44) is logarithmically divergent at \(s=T\) for the process (4.43) with optimal controller (4.46).

Solving (4.43) with drift (4.46) and \(Y_\varepsilon (0)=y\), we see on taking \(t=0\) that (4.41) holds with kernel \(k:[0,T]\times [0,T]\rightarrow \mathbf {R}\) given by

$$\begin{aligned} k(s,s') =\frac{m_{1,A}(s',s)\sigma _A^2(s,T)}{\sigma _A^2(s',T)} \quad \mathrm{if \ } s'<s, \quad k(s,s') =0 \quad \mathrm{if } s'>s. \end{aligned}$$
(4.47)

Observe that the kernel (4.47) corresponds to the Cholesky factorization \(\Gamma _A=KK^*\) of the kernel \(\Gamma _A\) (Ciarlet 1989). In the case \(A(\cdot )\equiv 0\) Eq. (4.47) yields the Markovian representation

$$\begin{aligned} Y_\varepsilon (s) = \frac{s}{T} x+\left( 1-\frac{s}{T}\right) y+\sqrt{\varepsilon }(T-s)\int _0^s \frac{\hbox {d}B(s')}{T-s'} , \quad 0\le s\le T, \end{aligned}$$
(4.48)

for the Brownian bridge process.

We can also express the ratio (4.17) of Green’s functions for the linear case \(b(y,t)=A(t)y-1\) in terms of the solution to a PDE. Thus, we assume \(x>0\) and define

$$\begin{aligned} u(y,t) = P\left( \inf _{t\le s\le T} Y_\varepsilon (s) > 0 \ | \ Y_\varepsilon (t)=y\right) , \quad y>0,t<T, \end{aligned}$$
(4.49)

where \(Y_\varepsilon (\cdot )\) is the solution to the SDE (4.43) with drift (4.46). Then, u(yt) is the solution to the PDE

$$\begin{aligned} \frac{\partial u(y,t)}{\partial t} +\lambda _\varepsilon ^*(x,y,t)\frac{\partial u(y,t)}{\partial y}+\frac{\varepsilon }{2}\frac{\partial ^2 u(y,t)}{\partial y^2} =0, \quad y>0,t<T, \end{aligned}$$
(4.50)

with boundary and terminal conditions given by

$$\begin{aligned} u(0,t)=0 \ \mathrm{for} \ t<T, \quad \lim _{t\rightarrow T} u(y,t) =1 \ \mathrm{for} y>0. \end{aligned}$$
(4.51)

In the case \(A(\cdot )\equiv 0\) the PDE (4.50) becomes

$$\begin{aligned} \frac{\partial u(y,t)}{\partial t} +\left( \frac{x-y}{T-t}\right) \ \frac{\partial u(y,t)}{\partial y}+\frac{\varepsilon }{2}\frac{\partial ^2 u(y,t)}{\partial y^2} =0, \quad y>0,t<T. \end{aligned}$$
(4.52)

Evidently, the function u defined by

$$\begin{aligned} u(y,t) =1- \ \exp \left[ -\frac{2xy}{\varepsilon (T-t)}\right] , \quad t<T, y>0, \end{aligned}$$
(4.53)

is the solution to (4.51), (4.53). Observe that the RHS of (4.53) at \(t=0\) is the same as the RHS of (4.17).

5 Estimates on the Dirichlet Green’s Function

In this section, we shall obtain estimates on the ratio of the Dirichlet to the full-space Green’s function in the case of linear drift \(b(y,t)=A(t)y-1\). In particular, we shall prove a limit theorem which generalizes the formula (4.17):

Proposition 5.1

Assume \(b(y,t)=A(t)y-1\) where (4.1) holds and the function \(A(\cdot )\) is nonnegative. Then, for \(\lambda ,y, T>0\) the ratio of the Dirichlet to full-space Green’s function satisfies the limit

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \frac{G_{\varepsilon ,D}(\lambda \varepsilon ,y,0,T)}{G_{\varepsilon }(\lambda \varepsilon ,y,0,T)} = 1- \exp \left[ -2\lambda \left\{ 1-\frac{m_{2,A}(T)}{\sigma _A^2(T)}+\frac{m_{1,A}(T)y}{\sigma _A^2(T)}\right\} \ \right] , \end{aligned}$$
(5.1)

where \(m_{1,A}(T),m_{2,A}(T)\) are given by (2.2) and \(\sigma _A^2(T)\) by (4.10).

Note that since we are assuming \(A(\cdot )\) is nonnegative in the statement of the proposition, it follows from (4.10) that \(m_{2,A}(T)/\sigma _A^2(T)\le 1\). Hence, the RHS of (5.1) always lies between 0 and 1. We can see why (5.1) holds from the representation (4.39), (4.41) for the conditioned process \(Y_\varepsilon (s), \ 0\le s\le T\). Thus, we have that

$$\begin{aligned} Y_\varepsilon (s) =y(s)+\sqrt{\varepsilon }\left[ \ \frac{m_{1,A}(s)\sigma _A^2(s,T)}{\sigma _A^2(T)}\int _0^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')}-m_{1,A}(s)\int _s^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} \ \right] . \end{aligned}$$
(5.2)

Since \(\sigma _A^2(s,T)=O(T-s)\), the conditioned process \(Y_\varepsilon (s)\) close to \(s=T\) is approximately the same as

$$\begin{aligned} Y_\varepsilon (s) =\lambda \varepsilon -y'(T)(T-s)-\sqrt{\varepsilon }\int _s^T \ \hbox {d}B(s') . \end{aligned}$$
(5.3)

Observe now from (4.29) that

$$\begin{aligned} -y'(T) =O(\varepsilon )+1-\frac{m_{2,A}(T)}{\sigma _A^2(T)}+\frac{m_{1,A}(T)y}{\sigma _A^2(T)} . \end{aligned}$$
(5.4)

Hence, for s close to T the process \(Y_\varepsilon (s), \ s<T,\) is approximately Brownian motion with a constant drift. Thus, let \(Z_\varepsilon (t), \ t>0,\) be the solution to the initial value problem for the SDE

$$\begin{aligned} \hbox {d}Z_\varepsilon (t) = \mu \hbox {d}t+\sqrt{\varepsilon } \ \hbox {d}B(t), \quad Z_\varepsilon (0)=\lambda \varepsilon , \end{aligned}$$
(5.5)

where we assume the drift \(\mu \) is positive. Then, from (5.3), (5.4) we see that \(Y_\varepsilon (T-t)\simeq Z_\varepsilon (t)\) if \(\mu \) is given by the formula

$$\begin{aligned} \mu =1-\frac{m_{2,A}(T)}{\sigma _A^2(T)}+\frac{m_{1,A}(T)y}{\sigma _A^2(T)} . \end{aligned}$$
(5.6)

Observe now that \(P(\inf _{t>0} Z_\varepsilon (t)<0)=e^{-2\lambda \mu }\), whence the RHS of (5.1) is simply \(P(\inf _{t>0} Z_\varepsilon (t)>0)\) when \(\mu \) is given by (5.6). Since the time for which \(Z_\varepsilon (t)\) is likely to become negative is \(t\simeq O(\varepsilon )\), the approximations above are justified and so we obtain (5.1).

Proof of Proposition 5.1

Let \(Y_\varepsilon (s), \ 0\le s\le T,\) be given by (5.2) where \(y(T)=\lambda \varepsilon \). Then, we have that for \(0<a\varepsilon \le T\),

$$\begin{aligned} P\left( \inf _{0\le s\le T} Y_\varepsilon (s)>0\right) \le P\left( \inf _{0\le t\le a\varepsilon } Y_\varepsilon (T-t)>0\right) =P\left( \inf _{0<t<a\varepsilon } [Z_\varepsilon (t)+\tilde{Z}_\varepsilon (t)]>0\right) , \end{aligned}$$
(5.7)

where \(Z_\varepsilon (\cdot )\) is the solution to (5.5) with \(\mu \) given by (5.6) and \(\tilde{Z}_\varepsilon (\cdot )\) is given by the formula

$$\begin{aligned} \tilde{Z}_\varepsilon (t)= & {} y(T-t)-y(T)+y'(T)t\nonumber \\&+\,\sqrt{\varepsilon }\left[ \ \frac{m_{1,A}(T-t)\sigma _A^2(T-t,T)}{\sigma _A^2(T)}\int _0^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')}\right. \nonumber \\&\left. +\,\int _{T-t}^T\left[ 1-\frac{m_{1,A}(T-t)}{m_{1,A}(s')}\right] \ \hbox {d}B(s') \ \right] . \end{aligned}$$
(5.8)

We use the inequality

$$\begin{aligned}&P\left( \inf _{0<t<a\varepsilon } [Z_\varepsilon (t)+\tilde{Z}_\varepsilon (t)]>0\right) \le P\left( \inf _{0<t<a\varepsilon } Z_\varepsilon (t)>-b\lambda \varepsilon \right) \nonumber \\&\quad +\,P\left( \sup _{0<t<a\varepsilon } \tilde{Z}_\varepsilon (t)>b\lambda \varepsilon \right) , \end{aligned}$$
(5.9)

which holds for any \(a,b>0\) satisfying \(a\varepsilon \le T\).

To estimate the first term on the RHS of (5.9), we observe by the method of images that

$$\begin{aligned}&P\left( \inf _{0<t<a\varepsilon } Z_\varepsilon (t)<-b\lambda \varepsilon \right) \nonumber \\&\qquad =e^{-2\mu (1+b)\lambda }\frac{1}{\sqrt{2\pi }}\int _{[(1+b)\lambda -\mu a]/ \sqrt{a}}^\infty e^{-z^2/2} \hbox {d}z + \frac{1}{\sqrt{2\pi }}\int ^{-[(1+b)\lambda +\mu a]/\sqrt{a}}_{-\infty } e^{-z^2/2} \hbox {d}z.\nonumber \\ \end{aligned}$$
(5.10)

To estimate the second term, we write \(\tilde{Z}_\varepsilon (t)\) in (5.8) as a sum of three quantities. The first of these is bounded as

$$\begin{aligned} \sup _{0\le t\le a\varepsilon } |y(T-t)-y(T)+y'(T)t| \le C[\lambda \varepsilon +y+1]a^2\varepsilon ^2, \quad 0<a\varepsilon \le T, \end{aligned}$$
(5.11)

for a constant C depending only on \(A_\infty ,T\). The second is bounded as

$$\begin{aligned} \sup _{0\le t\le a\varepsilon }\left| \sqrt{\varepsilon } \ \frac{m_{1,A}(T-t)\sigma _A^2(T-t,T)}{\sigma _A^2(T)}\int _0^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} \right| \le Ca\varepsilon ^{3/2}\left| \int _0^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} \right| , \end{aligned}$$
(5.12)

where C depends only on \(A_\infty ,T\). Finally, the third quantity is bounded as

$$\begin{aligned}&\sup _{0\le t\le a\varepsilon }\left| \ \int _{T-t}^T\left[ 1-\frac{m_{1,A}(T-t)}{m_{1,A}(s')}\right] \ \hbox {d}B(s') \ \right| \le \sup _{0\le t\le a\varepsilon }\left| \ \int _{T-t}^T\left[ 1-\frac{m_{1,A}(T)}{m_{1,A}(s')}\right] \ \hbox {d}B(s') \ \right| \nonumber \\&\quad +\, Ca\varepsilon \sup _{0\le t\le a\varepsilon }\left| \int _{T-t}^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} \right| , \quad \mathrm{where \ } C \ \mathrm{depends \ only \ on \ } A_\infty ,T. \end{aligned}$$
(5.13)

We can estimate probabilities for the terms on the RHS of (5.12), (5.13) by using Martingale properties. Thus, if \(g:(-\infty ,T)\rightarrow \mathbf {R}\) is a continuous function we define \(X(t), \ t\ge 0,\) by

$$\begin{aligned} X(t) =\int _{T-t}^T g(s) \ \hbox {d}B(s) . \end{aligned}$$
(5.14)

Then, for \(\theta \in \mathbf {R}\)

$$\begin{aligned} X_\theta (t) =\exp \left[ \theta X(t)-\frac{\theta ^2}{2}\int _{T-t}^T \hbox {d}s \ g(s)^2 \ \right] \quad \mathrm{is \ a \ Martingale \ and \ } E[X_\theta (t)]=1. \end{aligned}$$
(5.15)

Using the inequality

$$\begin{aligned} P\left( \ |X(0)|>M \ \right) \le 2\exp \left[ -\theta M+\frac{\theta ^2}{2}\int _0^T \hbox {d}s \ g(s)^2 \ \right] \quad \mathrm{for \ } M,\theta >0, \end{aligned}$$
(5.16)

and optimizing the RHS of (5.16) with respect to \(\theta >0\), we conclude that

$$\begin{aligned} P\left( \ a\varepsilon ^{3/2}\left| \int _0^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} \right| > b\lambda \varepsilon /4 \ \right) \le 2\exp \left[ -Cb^2\lambda ^2/a^2\varepsilon \right] , \end{aligned}$$
(5.17)

where the constant \(C>0\) depends only on \(A_\infty ,T\). We use Doob’s inequality to estimate probabilities for the terms on the RHS of (5.13). Thus, we have for \(\theta >0\) that

$$\begin{aligned} P\left( \ \sup _{0\le t\le t_0} X(t)>M \ \right)\le & {} P\left( \ \sup _{0\le t\le t_0} X_\theta (t)>\exp \left[ \theta M- \frac{\theta ^2}{2}\int _{T-t_0}^T \hbox {d}s \ g(s)^2 \ \right] \ \right) \nonumber \\\le & {} \exp \left[ -\theta M+ \frac{\theta ^2}{2}\int _{T-t_0}^T \hbox {d}s \ g(s)^2 \ \right] . \end{aligned}$$
(5.18)

Optimizing the term on the RHS of (5.18) with respect to \(\theta >0\), we conclude that

$$\begin{aligned} P\left( \ \sup _{0\le t\le t_0} |X(t)|>M \ \right) \le 2\exp \left[ - M^2\bigg /2\int _{T-t_0}^T \hbox {d}s \ g(s)^2 \ \right] . \end{aligned}$$
(5.19)

Hence, we have from (5.19) for the first term on the RHS of (5.13) that

$$\begin{aligned} P\left( \ \sup _{0\le t\le a\varepsilon }\left| \ \int _{T-t}^T\left[ 1-\frac{m_{1,A}(T)}{m_{1,A}(s')}\right] \ \hbox {d}B(s') \ \right| > b\lambda \varepsilon /4 \ \right) \le 2\exp \left[ -Cb^2\lambda ^2/a^3\varepsilon \right] , \end{aligned}$$
(5.20)

where the constant \(C>0\) depends only on \(A_\infty ,T\). Similarly, we have that if \(C_1\) depends only on \(A_\infty ,T\) then

$$\begin{aligned} P\left( C_1a\varepsilon \sup _{0\le t\le a\varepsilon }\left| \int _{T-t}^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} \right| > b\lambda \varepsilon /4 \ \right) \le 2\exp \left[ -C_2b^2\lambda ^2/a^3\varepsilon \right] , \end{aligned}$$
(5.21)

where the constant \(C_2>0\) also depends only on \(A_\infty ,T\).

We choose now \(a=\varepsilon ^{-\alpha }, b=\varepsilon ^\beta \) for some \(\alpha ,\beta >0\). Since \(\mu >0\), it follows from (5.10) that the first term on the RHS of (5.9) converges to \(1-e^{-2\lambda \mu }\) as \(\varepsilon \rightarrow 0\). We also see from the estimates of the previous paragraph that the second term on the RHS of (5.9) converges to 0 as \(\varepsilon \rightarrow 0\) provided \(3\alpha +2\beta <1\). We have therefore shown that \(\limsup _{\varepsilon \rightarrow 0}P\left( \inf _{0\le s\le T} Y_\varepsilon (s)>0\right) \) is bounded above by the RHS of (5.1).

To obtain the corresponding lower bound, we use the inequality

$$\begin{aligned} P\left( \inf _{0\le s\le T} Y_\varepsilon (s)>0\right) \ge P\left( \inf _{T-a\varepsilon \le s\le T} Y_\varepsilon (s)>0\right) -P\left( \inf _{0\le s\le T-a\varepsilon } Y_\varepsilon (s)<0\right) . \end{aligned}$$
(5.22)

Next we use the inequality similar to (5.9) that

$$\begin{aligned} P\left( \inf _{T-a\varepsilon \le s \le T} Y_\varepsilon (s)>0\right) \ge P\left( \inf _{0<t<a\varepsilon } Z_\varepsilon (t)>b\lambda \varepsilon \right) -P\left( \inf _{0<t<a\varepsilon } \tilde{Z}_\varepsilon (t)<-b\lambda \varepsilon \right) . \end{aligned}$$
(5.23)

Arguing as previously we see from (5.23) on choosing \(a=\varepsilon ^{-\alpha },b=\varepsilon ^\beta \) with \(3\alpha +2\beta <1\) that \( \liminf _{\varepsilon \rightarrow 0}P\left( \inf _{T-\varepsilon ^{1-\alpha }\le s\le T} Y_\varepsilon (s)>0\right) \) is bounded below by the RHS of (5.1). Next we need to obtain a bound on the second term on the RHS of (5.22) when \(a=\varepsilon ^{-\alpha }\) which vanishes as \(\varepsilon \rightarrow 0\). Since \(A(\cdot )\) is nonnegative, there is a positive constant C depending only on \(A_\infty ,T\) such that the function \(y(\cdot )\) of (4.29) satisfies an inequality \(y(s)\ge C(T-s)y\) for \(0\le s\le T\). Hence, there is a positive constant c depending only on \(A_\infty ,T\) such that

$$\begin{aligned} P\left( \inf _{0\le s\le T-\varepsilon ^{1-\alpha }} Y_\varepsilon (s)<0\right)\le & {} P\left( \ \left| \int _0^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} \right| > \frac{cy}{\sqrt{\varepsilon }} \ \right) \nonumber \\&+\,P\left( \ \sup _{\varepsilon ^{1-\alpha }\le t\le T}\left| \frac{1}{t}\int _{T-t}^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} \right| > \frac{cy}{\sqrt{\varepsilon }} \ \right) .\nonumber \\ \end{aligned}$$
(5.24)

We can bound the first term on the RHS of (5.24) similarly to (5.17). We bound the second term by using the inequality

$$\begin{aligned} P\left( \sup _{\varepsilon ^{1-\alpha }\le t\le T} |X(t)|>cy/\sqrt{\varepsilon }\right) \le \sum _{k\ge 1} P\left( \sup _{k\varepsilon ^{1-\alpha }\le t\le (k+1)\varepsilon ^{1-\alpha }} |X(t)|>cy/\sqrt{\varepsilon }\right) . \end{aligned}$$
(5.25)

From (5.19), we see that for \(k\ge 1\),

$$\begin{aligned} P\left( \sup _{k\varepsilon ^{1-\alpha }\le t\le (k+1)\varepsilon ^{1-\alpha }} \left| \frac{1}{t}\int _{T-t}^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} \right| \ >cy/\sqrt{\varepsilon }\right) \le \exp \left[ -\frac{c_1ky^2}{\varepsilon ^\alpha } \ \right] , \end{aligned}$$
(5.26)

where \(c_1>0\) depends only on \(A_\infty ,T\). We conclude that the second term on the RHS of (5.22) converges when \(a=\varepsilon ^{-\alpha }\) with \(\alpha >0\) to zero as \(\varepsilon \rightarrow 0\). Hence, \(\liminf _{\varepsilon \rightarrow 0}P\left( \inf _{0\le s\le T} Y_\varepsilon (s)>0\right) \) is bounded below by the RHS of (5.1). \(\square \)

Next we wish to obtain estimates on the LHS of (5.1) which are uniform as \(\lambda \rightarrow 0\).

Lemma 5.1

Assume the function \(A(\cdot )\) is nonnegative and that \(0<\lambda \le 1, \ 0<\varepsilon \le T, \ y>0\). Let \(\Gamma :\mathbf {R}^+\times \mathbf {R}^+\rightarrow \mathbf {R}^+\) be the function \(\Gamma (a,b)=1\) if \(b>a^{-1/4}\) and otherwise \(\Gamma (a,b)=a^{1/8}\). Then, there is a constant C depending only on \(A_\infty T\) such that

$$\begin{aligned} \frac{G_{\varepsilon ,D}(\lambda \varepsilon ,y,0,T)}{G_{\varepsilon }(\lambda \varepsilon ,y,0,T)}\le & {} 1- \exp \left[ -2\lambda \left\{ 1-\frac{m_{2,A}(T)}{\sigma _A^2(T)}+ \frac{m_{1,A}(T)y}{\sigma _A^2(T)}\right\} \ \right] \nonumber \\&+\,C\lambda \Gamma \left( \frac{\varepsilon }{T},\frac{y}{T}\right) \left[ 1+\frac{y}{T}\right] . \end{aligned}$$
(5.27)

Proof

We make the change of variable \(s\leftrightarrow t\) in which

$$\begin{aligned} \frac{\hbox {d}s}{\hbox {d}t} = -\left[ \ \frac{m_{1,A}(s)}{m_{1,A}(T)} \ \right] ^2 , \quad s(0) = T. \end{aligned}$$
(5.28)

Hence, \(s\simeq T-t\) if t is small and

$$\begin{aligned} m_{1,A}(T) \int _s^T \ \frac{\hbox {d}B(s')}{m_{1,A}(s')} = \int _0^t \hbox {d}\tilde{B}(t') \quad \mathrm{where }\;\tilde{B}(\cdot ) \ \mathrm{is \ a \ Brownian \ motion.} \end{aligned}$$
(5.29)

Letting \(s(\tilde{T})=0\), we see from (5.2), (5.29) that \(Y_\varepsilon (s)=\tilde{Y}_\varepsilon (t)\) where

$$\begin{aligned} \tilde{Y}_\varepsilon (t) = \tilde{y}(t)\,+\,\sqrt{\varepsilon }\left[ \ \frac{m_{1,A}(s)\sigma _A^2(s,T)}{m_{1,A}(T)\sigma _A^2(T)}\int _0^{\tilde{T}} \hbox {d}\tilde{B}(t') -\frac{m_{1,A}(s)}{m_{1,A}(T)}\int _0^t \hbox {d}\tilde{B}(t') \ \right] , \end{aligned}$$
(5.30)

and \(\tilde{y}(t)=y(s)\), where \(y(\cdot )\) is the function (4.29). We consider any a for which \(0<a\varepsilon \le \tilde{T}\) and observe as in (5.7) that if \(M>0\), then

$$\begin{aligned} P\left( \inf _{0\le s\le T} Y_\varepsilon (s)>0\right)\le & {} P\left( \inf _{0\le t\le a\varepsilon } \tilde{Y}_\varepsilon (t)>0\right) \nonumber \\\le & {} P\left( \inf _{0\le t\le a\varepsilon } \tilde{Y}_\varepsilon (t)>0; \ \sup _{0\le t\le a\varepsilon }\left| \int _0^t \hbox {d}\tilde{B}(t') \right| \le M \right) \nonumber \\&+\, P\left( \inf _{0\le t\le a\varepsilon } \tilde{Y}_\varepsilon (t)>0; \ \sup _{0\le t\le a\varepsilon }\left| \int _0^t \hbox {d}\tilde{B}(t') \right| > M \right) .\nonumber \\ \end{aligned}$$
(5.31)

The first term on the RHS of (5.31) is bounded above by \( P\left( \inf _{0\le t\le a\varepsilon } \tilde{Y}_{0,\varepsilon }(t)>0\right) \) where \(\tilde{Y}_{0,\varepsilon }(t)\) is given from (5.30) by the formula

$$\begin{aligned} \tilde{Y}_{0,\varepsilon }(t) =\tilde{y}(t)+\frac{C\sqrt{\varepsilon }Mt}{T}+\sqrt{\varepsilon } \ \frac{m_{1,A}(s)\sigma _A^2(s,T)}{m_{1,A}(T)\sigma _A^2(T)} \int _{a\varepsilon }^{\tilde{T}} \hbox {d}\tilde{B}(t') -\sqrt{\varepsilon }\frac{m_{1,A}(s)}{m_{1,A}(T)} \int _0^t \hbox {d}\tilde{B}(t'), \end{aligned}$$
(5.32)

with C in (5.32) depending only on \(A_\infty T\). To estimate the second term on the RHS of (5.31), we introduce the stopping time \(\tau \) defined by

$$\begin{aligned} \tau = \inf \left\{ \ t<\tilde{T}: \ \left| \int _0^t \hbox {d}\tilde{B}(t')\right| > M \right\} . \end{aligned}$$
(5.33)

Hence, the second term is bounded above by \( P\left( \inf _{ 0\le t\le \tau } \tilde{Y}_\varepsilon (t)>0; \ \tau <a\varepsilon \ \right) \). Observe now that for any \(M_1>0\),

$$\begin{aligned}&P\left( \inf _{ 0\le t\le \tau } \tilde{Y}_\varepsilon (t)>0; \ \tau <a\varepsilon \ \right) \nonumber \\&\quad =\sum _{n=1}^\infty P\left( \inf _{ 0\le t\le \tau } \tilde{Y}_\varepsilon (t)>0; \ \tau <a\varepsilon , \ (n-1)M_1\le \sup _{\tau \le t\le \tau +\tilde{T}}\left| \int _\tau ^t \hbox {d}\tilde{B}(t')\right| < nM_1 \ \right) \nonumber \\&\quad \le \sum _{n=1}^\infty P\left( \inf _{0\le t\le \tau } \tilde{Y}_{n,\varepsilon }(t)>0; \ \tau <a\varepsilon \ \right) P\left( \ (n-1)M_1\le \left| \sup _{\tau \le t\le \tau +\tilde{T}}\int _\tau ^t \hbox {d}\tilde{B}(t')\ \right| < nM_1 \ \right) \nonumber \\&\quad = \sum _{n=1}^\infty P\left( \inf _{0\le t\le \tau } \tilde{Y}_{n,\varepsilon }(t)>0; \ \tau <a\varepsilon \ \right) P\left( \ (n-1)M_1\le \left| \sup _{0\le t\le \tilde{T}}\int _0^t \hbox {d}\tilde{B}(t')\ \right| < nM_1 \ \right) ,\nonumber \\ \end{aligned}$$
(5.34)

where \(\tilde{Y}_{n,\varepsilon }\) is given by the formula

$$\begin{aligned} \tilde{Y}_{n,\varepsilon }(t) =\tilde{y}(t)+\frac{C\sqrt{\varepsilon }(M+nM_1)t}{T}- \sqrt{\varepsilon }\frac{m_{1,A}(s)}{m_{1,A}(T)} \int _0^t \hbox {d}\tilde{B}(t') , \end{aligned}$$
(5.35)

and the constant C depends only on \(A_\infty T\). Note that in (5.34) we are using the fact that the variables

$$\begin{aligned} \tau \ \ \mathrm{and \ } \{\tilde{B}(t): \ 0<t\le \tau \} \quad \mathrm{are \ independent \ of \ the \ variable \ } \left| \sup _{\tau \le t\le \tau +\tilde{T}}\int _\tau ^t \hbox {d}\tilde{B}(t')\ \right| . \end{aligned}$$
(5.36)

To estimate \( P\left( \inf _{0\le t\le a\varepsilon } \tilde{Y}_{0,\varepsilon }(t)>0\right) \), we compare \(\tilde{Y}_{0,\varepsilon }(\cdot )\) to Brownian motion with constant drift as in (5.5). It follows from (5.32) that

$$\begin{aligned}&P\left( \inf _{0\le t\le a\varepsilon } \tilde{Y}_{0,\varepsilon }(t)>0\right) \nonumber \\&\qquad \le E\left[ \ P\left( \inf _{0\le t\le a\varepsilon } Z_\varepsilon (t)>0 \ \Big | \ \mu =\mu _\mathrm{rand}, \ Z_\varepsilon (0)=\lambda \varepsilon [1+Ca\varepsilon /T] \ \right) \ \right] ,\nonumber \\ \end{aligned}$$
(5.37)

where \(\mu _\mathrm{rand}\) is the random variable

$$\begin{aligned} \mu _\mathrm{rand} = 1-\frac{m_{2,A}(T)}{\sigma _A^2(T)}+\frac{m_{1,A}(T)y}{\sigma _A^2(T)}+\frac{Ca\varepsilon }{T}\left[ 1+\frac{y}{T}\right] + \frac{C\lambda \varepsilon }{T}+ \frac{C\sqrt{\varepsilon }}{T}\left[ M+ \left| \int _{a\varepsilon }^{\tilde{T}} \hbox {d}\tilde{B}(t') \right| \right] , \end{aligned}$$
(5.38)

and \(C>0\) is a constant depending only on \(A_\infty T\). To bound the RHS of (5.37), we use an identity similar to (5.10),

$$\begin{aligned}&P\left( \inf _{0<t<a'\varepsilon } Z_\varepsilon (t)>0 \ \big | \ Z_\varepsilon (0)=\lambda '\varepsilon \ \right) \nonumber \\&\quad =\left\{ 1- e^{-2\mu \lambda '}\right\} \frac{1}{\sqrt{2\pi }}\int _{[\lambda '-\mu a']/\sqrt{a'}}^\infty e^{-z^2/2} \ \hbox {d}z \ + \ \frac{1}{\sqrt{2\pi }}\int ^{[\lambda '-\mu a']/\sqrt{a'}}_{[-\lambda '-\mu a']/\sqrt{a'}} e^{-z^2/2} \hbox {d}z.\nonumber \\ \end{aligned}$$
(5.39)

From (5.39), we obtain the upper bound

$$\begin{aligned} P\left( \inf _{0<t<a'\varepsilon } Z_\varepsilon (t)>0 \ \big | \ Z_\varepsilon (0)=\lambda '\varepsilon \ \right) \le 1- e^{-2\mu \lambda '} + \frac{2\lambda '}{\sqrt{2\pi a'}} . \end{aligned}$$
(5.40)

Using (5.40), we estimate the RHS of (5.37) when \(a=\min \left[ (T/\varepsilon )^\alpha , \ \tilde{T}/\varepsilon \right] \) for some \(\alpha \) satisfying \(0<\alpha <1\). In that case \(\lambda '=\lambda [1+Ca\varepsilon /T]\le \lambda [1+C]\) for some constant C depending only on \(A_\infty T\). Taking \(M=C_1\sqrt{T}\) in (5.38) where \(C_1\) depends only on \(A_\infty T\), we conclude from (5.37), (5.40) that for \(0<\lambda \le 1, \ 0<\varepsilon \le T, \)

$$\begin{aligned}&P\left( \inf _{0\le t\le a\varepsilon } \tilde{Y}_{0,\varepsilon }(t)>0\right) \nonumber \\&\quad \le 1- \exp \left[ -2\lambda \left\{ 1-\frac{m_{2,A}(T)}{\sigma _A^2(T)}+\frac{m_{1,A}(T)y}{\sigma _A^2(T)}\right\} \ \right] \nonumber \\&\qquad +\, C_2\lambda \left[ \left( \frac{\varepsilon }{T}\right) ^{1-\alpha }\left\{ 1+\frac{y}{T}\right\} +\left( \frac{\varepsilon }{T}\right) ^{\alpha /2}\right] , \end{aligned}$$
(5.41)

where \(C_2\) in (5.41) depends only on \(A_\infty T\).

Next we estimate the probabilities on the RHS of (5.34). Evidently, we have from (5.35) that

$$\begin{aligned} \tilde{Y}_{n,\varepsilon }(\tau ) =\tilde{y}(\tau )+\frac{C\sqrt{\varepsilon }(M+nM_1)\tau }{T}\pm M\sqrt{\varepsilon }\frac{m_{1,A}(s(\tau ))}{m_{1,A}(T)} . \end{aligned}$$
(5.42)

We choose \(M_1=\sqrt{T}\) in (5.42) and \(M=C_1\sqrt{T}\) for a constant \(C_1\) depending only on \(A_\infty T\) so that \(M\sqrt{\varepsilon }/m_1(T)>2\varepsilon \). Since \(\tilde{Y}_{n,\varepsilon }(\tau )>0\), it follows that if (5.42) holds with the \(-\) sign, then there is a constant \(c>0\) depending only on \(A_\infty T\) such that

$$\begin{aligned} \tau > \ \tau _n =cT\sqrt{\frac{\varepsilon }{T}}\left[ 1+\frac{y}{T}+n \sqrt{\frac{\varepsilon }{T}} \ \right] ^{-1} . \end{aligned}$$
(5.43)

Observe now from (5.43) that if \(\alpha <1/2\), then \(\tau _n>a\varepsilon \) provided

$$\begin{aligned} 1+\frac{y}{T} +n\sqrt{\frac{\varepsilon }{T}} \le 2c_1\left( \frac{T}{\varepsilon }\right) ^{1/2-\alpha } \quad \mathrm{for \ } c_1>0 \ \mathrm{depending \ only \ on \ } A_\infty T. \end{aligned}$$
(5.44)

Since \(\tau <a\varepsilon \), it follows that (5.42) can hold with the minus sign only if

$$\begin{aligned} 1+\frac{y}{T} \ge c_1\left( \frac{T}{\varepsilon }\right) ^{1/2-\alpha } \quad \mathrm{or \ } n \ge c_1\left( \frac{T}{\varepsilon }\right) ^{1-\alpha } . \end{aligned}$$
(5.45)

In the case when \(\tau _n<a\varepsilon \), we see from (5.35) that there is a constant C depending only on \(A_\infty T\) and

$$\begin{aligned} P\left( \inf _{ 0\le t\le \tau _n} \tilde{Y}_{n,\varepsilon }(t)>0 \ \right) \le P\left( \inf _{0\le t\le \tau _n} Z_\varepsilon (t)>0 \ \Big | \ \mu =\mu _n, \ Z_\varepsilon (0)=\lambda \varepsilon [1+Ca\varepsilon /T] \ \right) , \end{aligned}$$
(5.46)

where \(Z_\varepsilon (\cdot )\) is the solution to the SDE (5.5). The drift \(\mu _n\) is given by the formula

$$\begin{aligned} \mu _n = C\left[ 1+\frac{y}{T} +n\sqrt{\frac{\varepsilon }{T}} \ \right] \quad \mathrm{where \ }C \ \mathrm{depends \ only \ on \ } A_\infty T. \end{aligned}$$
(5.47)

It follows then from (5.40), (5.46), (5.47) that

$$\begin{aligned} P\left( \inf _{ 0\le t\le \tau _n} \tilde{Y}_{n,\varepsilon }(t)>0 \ \right)\le & {} C_1\lambda \left[ 1+\frac{y}{T} +n\sqrt{\frac{\varepsilon }{T}} + \left( \frac{\varepsilon }{\tau _n}\right) ^{1/2}\ \right] \nonumber \\\le & {} C_2\lambda \left[ 1+\frac{y}{T} +n\sqrt{\frac{\varepsilon }{T}} \ \right] \end{aligned}$$
(5.48)

for some constants \(C_1,C_2\) depending only on \(A_\infty T\). We conclude from (5.48) that

$$\begin{aligned}&\sum _{n\ge c(T/\varepsilon )^{1-\alpha }} P\left( \inf _{0\le t\le \tau _n} \tilde{Y}_{n,\varepsilon }(t)>0 \ \right) P\left( \ (n-1)M_1\le \left| \sup _{0\le t\le \tilde{T}}\int _0^t \hbox {d}\tilde{B}(t')\ \right| < nM_1 \ \right) \nonumber \\&\quad \le C\lambda \sum _{n\ge c(T/\varepsilon )^{1-\alpha }} \left[ 1+\frac{y}{T} +n\sqrt{\frac{\varepsilon }{T}} \ \right] e^{-n^2/2} \le C_1\lambda \left( 1+\frac{y}{T}\right) \exp \left[ \ -c_1\left( \frac{T}{\varepsilon }\right) ^{2(1-\alpha )} \right] ,\nonumber \\ \end{aligned}$$
(5.49)

where the constants \(C_1,c_1\) depend only on \(A_\infty T\).

We consider next the situation where (5.42) holds with the plus sign. One sees that

$$\begin{aligned}&P\left( \inf _{ 0\le t\le \tau } \tilde{Y}_{n,\varepsilon }(t)>0; \ \tau <a\varepsilon , \ \int _0^\tau \hbox {d}\tilde{B}(t') \ \hbox {d}t'=-M \ \right) \nonumber \\&\quad \le P\left( \inf _{0\le t\le \tau } Z_\varepsilon (t)>0, \ \tau <a\varepsilon , \ \ Z_\varepsilon (\tau )\ge M\sqrt{\varepsilon } \ \Big | \ \mu =\mu _n, \ Z_\varepsilon (0)=\lambda \varepsilon [1+Ca\varepsilon /T] \ \right) ,\nonumber \\ \end{aligned}$$
(5.50)

where \( \mu _n\) is given by (5.47). Observe that the RHS of (5.50) is bounded by the probability that the diffusion \(Z_\varepsilon (\cdot )\) started at \(\lambda \varepsilon \{1+O[(\varepsilon /T)^{1-\alpha }]\}\) exits the interval \([0,C_1T(\varepsilon /T)^{1/2}]\) through the rightmost boundary in time less than \(T(\varepsilon /T)^{1-\alpha }\). This probability is bounded by \(K(\varepsilon /T, n,y/T)\lambda \) for some function K which has the property that \(\lim _{\varepsilon \rightarrow 0} K(\varepsilon /T, n,y/T)=0\) provided \(\alpha <1/2\). To find an expression for K, we first choose \(C_1\) depending only on \(A_\infty T\) large enough so that \(Z_\varepsilon (0)<C_1T(\varepsilon /T)^{1/2}/2\) for any \(\varepsilon \) satisfying \(0<\varepsilon \le T\). It is easy to see that for \(0<\lambda '<\Lambda '\),

$$\begin{aligned} P\left( 0<Z_\varepsilon (t)<\Lambda '\varepsilon , \ t<\tau , \ Z_\varepsilon (\tau )=\Lambda '\varepsilon \ \big | \ Z_\varepsilon (0)=\lambda '\varepsilon \ \right) =\frac{1-e^{-2\mu \lambda '}}{1-e^{-2\mu \Lambda '}} . \end{aligned}$$
(5.51)

We apply (5.51) with \(\Lambda '=C_1(T/\varepsilon )^{1/2}, \ \mu =\mu _n\) and \(\lambda '=\lambda [1+Ca\varepsilon /T]\), whence \(\mu \Lambda '\ge c\) for some positive constant c depending only on \(A_\infty T\). We conclude from (5.51) that the function K satisfies the inequality

$$\begin{aligned} K(\varepsilon /T, n,y/T) \le C\left[ 1+\frac{y}{T} +n\sqrt{\frac{\varepsilon }{T}} \ \right] \end{aligned}$$
(5.52)

for some constant C depending only on \(A_\infty T\).

To show that \(\lim _{\varepsilon \rightarrow 0}K(\varepsilon /T, n,y/T)=0\), we assume \(\lambda '<\Lambda '/2\) and use the inequality

$$\begin{aligned}&P\left( \inf _{0<t<a\varepsilon } Z_\varepsilon (t)>0, \ \sup _{0<t<a\varepsilon } Z_\varepsilon (t)\ge \Lambda '\varepsilon \ \big | \ Z_\varepsilon (0)=\lambda '\varepsilon \ \right) \nonumber \\&\quad \le P\left( 0<Z_\varepsilon (t)<\Lambda '\varepsilon /2, \ t<\tau , \ Z_\varepsilon (\tau )=\Lambda '\varepsilon /2 \ \big | \ Z_\varepsilon (0)=\lambda '\varepsilon \ \right) \nonumber \\&\qquad P\left( \sup _{0<t<a\varepsilon } Z_\varepsilon (t)\ge \Lambda '\varepsilon \ \big | \ Z_\varepsilon (0)=\Lambda '\varepsilon /2 \ \right) . \end{aligned}$$
(5.53)

The second probability on the RHS of (5.53) can be bounded as

$$\begin{aligned} P\left( \sup _{0<t<a\varepsilon } Z_\varepsilon (t)\ge \Lambda '\varepsilon \ \big | \ Z_\varepsilon (0)=\Lambda '\varepsilon /2 \ \right) \le C\exp \left[ -\frac{\Lambda '^2}{32a}\right] \end{aligned}$$
(5.54)

for some universal constant C provided \(\mu a<\Lambda '/4\). Observe now from (5.47) that the condition \(\mu _n a<\Lambda '/4\) is implied by (5.44). We conclude from (5.52), (5.54) that if (5.44) holds, then

$$\begin{aligned} K(\varepsilon /T, n,y/T) \le C_2\left[ 1+\frac{y}{T} +n\sqrt{\frac{\varepsilon }{T}} \ \right] \exp \left[ -c_2\left( \frac{T}{\varepsilon }\right) ^{1-\alpha } \right] \end{aligned}$$
(5.55)

for some positive constants \(C_2,c_2\) depending only on \(A_\infty T\). If (5.44) does not hold we can argue as before using (5.45), (5.52) to obtain an inequality similar to (5.49). The inequality (5.27) follows now from (5.41), (5.49), (5.52) on choosing \(\alpha =1/4\). \(\square \)

Lemma 5.2

Assume the function \(A(\cdot )\) is nonnegative and that \(0<\lambda \le 1, \ 0<\varepsilon \le T,\ y>0\). Then, there are positive constants Cc depending only on \(A_\infty T\) such that if \(\gamma =c(T/\varepsilon )^{1/8}(y/T)\ge 5\), then

$$\begin{aligned}&\frac{G_{\varepsilon ,D}(\lambda \varepsilon ,y,0,T)}{G_{\varepsilon }(\lambda \varepsilon ,y,0,T)} \nonumber \\&\quad \ge [1+e^{-\gamma ^2/4}]^{-2} \left( 1- \exp \left[ -\frac{2\lambda }{1+C(\varepsilon /T)^{1/8}}\left\{ 1-\frac{m_{2,A}(T)}{\sigma _A^2(T)}+\frac{m_{1,A}(T)y}{\sigma _A^2(T)}\right\} \ \right] \right) .\nonumber \\ \end{aligned}$$
(5.56)

Proof

We choose \(a=\min \left[ (T/\varepsilon )^\alpha , \tilde{T}/\varepsilon \right] \) with \(0<\alpha <1\) as in Lemma 5.1 and observe from (4.29), (5.28) that there is a constant \(c>0\) depending only on \(A_\infty T\) such that \(\tilde{y}(t)\ge cty/T\) for \(0\le t\le \tilde{T}\). Hence, there exists a constant \(c_1>0\) depending only on \(A_\infty T\) such that the process \(\tilde{Y}_\varepsilon (\cdot )\) of (5.30) satisfies:

$$\begin{aligned}&\tilde{Y}_\varepsilon (t)>0 \quad \mathrm{for \ } a\varepsilon \le t\le \tilde{T} \ \mathrm{\ if \ for \ } k=1,2,\ldots , \nonumber \\&\qquad \left| \int _0^{a\varepsilon } \hbox {d}\tilde{B}(t')\right| < c_1\sqrt{T}\left( \frac{\varepsilon }{T}\right) ^{1/2-\alpha } \frac{y}{T} \quad \mathrm{and \ } \sup _{a\varepsilon \le t\le (k+1)a\varepsilon }\left| \int _{a\varepsilon }^t \hbox {d}\tilde{B}(t')\right| \nonumber \\&\qquad \qquad \le c_1k\sqrt{T}\left( \frac{\varepsilon }{T}\right) ^{1/2-\alpha } \frac{y}{T} . \end{aligned}$$
(5.57)

It follows from (5.57) that

$$\begin{aligned} P\left( \inf _{0\le s\le T} Y_\varepsilon (s)>0\right) =P\left( \inf _{0\le t\le \tilde{T}} \tilde{Y}_\varepsilon (t)>0\right) \ge P\left( \inf _{0\le t\le a\varepsilon } \tilde{Y}_\varepsilon (t)>0 \ ; \mathcal {E}\right) , \end{aligned}$$
(5.58)

where \(\mathcal {E}\) is the event defined by the second line of (5.57). It is easy to see from (5.30) that for \(0<t\le a\varepsilon \) on the event \(\mathcal {E}\) there is a constant \(C>0\) depending only on \(A_\infty T\) such that

$$\begin{aligned} \tilde{Y}_\varepsilon (t)>0 \ \mathrm{if \ } \ \ \tilde{Z}_\varepsilon (t)=\frac{\tilde{y}(t)}{1+Ca\varepsilon /T}-Cc_1t \frac{y}{T} -\sqrt{\varepsilon }\int _0^t \hbox {d}\tilde{B}(t') > 0, \end{aligned}$$
(5.59)

where \(c_1\) is the constant of (5.57). We conclude from (5.57), (5.59) that

$$\begin{aligned}&P\left( \inf _{0\le s\le T} Y_\varepsilon (s)>0\right) \nonumber \\&\qquad \ge P\left( \ \tilde{Z}_\varepsilon (t)>0 , \ 0<t\le a\varepsilon ; \ \left| \int _0^{a\varepsilon } \hbox {d}\tilde{B}(t')\right| < c_1\sqrt{T}\left( \frac{\varepsilon }{T}\right) ^{1/2-\alpha } \frac{y}{T} \ \right) \ P(\mathcal {E}).\nonumber \\ \end{aligned}$$
(5.60)

In order to bound \(P(\mathcal {E})\) from below, we consider for \(\gamma >0\) the event \(\mathcal {E}_\gamma \) defined by

$$\begin{aligned} \left| \int _0^{1} \hbox {d}\tilde{B}(t')\right| < \gamma \ \quad \mathrm{and \ } \sup _{1\le t\le (k+1)}\left| \int _{1}^t \hbox {d}\tilde{B}(t')\right| < k\gamma \quad \mathrm{for \ } k=1,2,\ldots \end{aligned}$$
(5.61)

Then, we have that

$$\begin{aligned} P(\mathcal {E}) =P(\mathcal {E}_\gamma ) \quad \mathrm{where \ } \gamma =c_1\left( \frac{T}{\varepsilon }\right) ^{\alpha /2}\left( \frac{y}{T}\right) . \end{aligned}$$
(5.62)

Using the fact that

$$\begin{aligned} P\left( \sup _{1\le t\le (k+1)}\left| \int _{1}^t \hbox {d}\tilde{B}(t')\right| > k\gamma \right) \le 4e^{-k\gamma ^2/2} , \end{aligned}$$
(5.63)

we conclude that

$$\begin{aligned} P(\mathcal {E}_\gamma ) \ge [1+e^{-\gamma ^2/4} \ ]^{-1} \quad \mathrm{if \ } \gamma \ge 5. \end{aligned}$$
(5.64)

We bound from below the first probability on the RHS of (5.60) by comparing it to the constant drift Brownian motion (5.5). To do this, we use the inequality

$$\begin{aligned} \sigma _A^2(T)y(s)\ge & {} xm_{1,A}(s,T)\sigma _A^2(0,s)+ym_{1,A}(0,s)\sigma _A^2(s,T)\nonumber \\&+\,[\sigma _A^2(0,s)-m_{2,A}(0,s)]\sigma _A^2(s,T), \end{aligned}$$
(5.65)

which follows from (4.29) and the assumption that \(A(s)\ge 0, \ 0\le s\le T\). Since the function \(s\rightarrow \sigma _A^2(0,s)-m_{2,A}(0,s)\) is increasing, we conclude from (5.28), (5.65) that there is a constant \(C_1>0\) depending only on \(A_\infty T\) such that for \(0\le t\le a\varepsilon \),

$$\begin{aligned} \tilde{y}(t) \ge \frac{\lambda \varepsilon +\mu _\varepsilon t}{1+C_1a\varepsilon /T} \quad \mathrm{where \ } \mu _\varepsilon =\frac{m_{1,A}(T)y}{\sigma _A^2(T)}+\frac{\sigma _A^2(0,s(a\varepsilon ))-m_{2,A}(0,s(a\varepsilon ))}{\sigma _A^2(T)} . \end{aligned}$$
(5.66)

It follows from (5.59), (5.66) that for \(0\le t\le a\varepsilon \) there is a constant \(C_2>0\) depending only on \(A_\infty T\) such that

$$\begin{aligned} \tilde{Z}_\varepsilon (t) \ge Z_\varepsilon (t) \quad \mathrm{with \ } Z_\varepsilon (0)=\frac{\lambda \varepsilon }{1+C_2a\varepsilon /T} , \ \mu =\frac{\mu _\varepsilon }{1+C_2a\varepsilon /T} -C_2c_1 \frac{y}{T} . \end{aligned}$$
(5.67)

Hence, the first probability on the RHS of (5.60) is bounded below by

$$\begin{aligned} P\left( \ Z_\varepsilon (t)>0, \ 0<t\le a\varepsilon ; |Z_\varepsilon (a\varepsilon )-Z_\varepsilon (0)-a\varepsilon \mu |<\gamma \varepsilon \sqrt{a} \ \Big | \ Z_\varepsilon (0)=\frac{\lambda \varepsilon }{1+C_2a\varepsilon /T} \ \right) , \end{aligned}$$
(5.68)

where \(\gamma \) is as in (5.62).

To bound the probability in (5.68), we assume that the constant \(c_1\) in (5.62) is small enough so that \(\mu >0\) and \(\gamma <\mu \sqrt{a}\). Then, similarly to (5.10), (5.39) we have that

$$\begin{aligned}&P\left( \inf _{0<t<a\varepsilon } Z_\varepsilon (t)>0; \ |Z_\varepsilon (a\varepsilon )-\lambda '\varepsilon -a\varepsilon \mu |<\gamma \varepsilon \sqrt{a} \ \big | \ Z_\varepsilon (0)=\lambda '\varepsilon \ \right) \nonumber \\&\quad = \left\{ 1- e^{-2\mu \lambda '}\right\} \frac{1}{\sqrt{2\pi }}\int _{2\lambda '/\sqrt{a}-\gamma }^{\gamma } e^{-z^2/2} \ \hbox {d}z \ + \ \frac{1}{\sqrt{2\pi }}\int ^{2\lambda '/\sqrt{a}-\gamma }_{-\gamma } e^{-z^2/2} \ \hbox {d}z \nonumber \\&\qquad -\,e^{-2\mu \lambda '}\frac{1}{\sqrt{2\pi }}\int ^{2\lambda '/\sqrt{a}+\gamma }_{\gamma } e^{-z^2/2} \ \hbox {d}z \ge \left\{ 1- e^{-2\mu \lambda '}\right\} \frac{1}{\sqrt{2\pi }}\int _{-\gamma }^{\gamma } e^{-z^2/2} \ \hbox {d}z.\nonumber \\ \end{aligned}$$
(5.69)

We take \(\lambda '=\lambda /[1+C_2a\varepsilon /T]\) in (5.69) and choose \(\alpha =1/2\). Hence, the drift \(\mu \) of (5.67) satisfies the inequality

$$\begin{aligned} \mu \ge \frac{1}{1+C_3\sqrt{\varepsilon /T}}\left\{ 1-\frac{m_{2,A}(T)}{\sigma _A^2(T)}+\frac{m_{1,A}(T)y}{\sigma _A^2(T)}\right\} -C_3c_1 \frac{y}{T}-C_3\left( \frac{\varepsilon }{T}\right) ^{1/2} , \end{aligned}$$
(5.70)

for some constant \(C_3>0\) depending only on \(A_\infty T\). We choose now \(c_1=c(\varepsilon /T)^{1/8}\) where \(c>0\) depends only on \(A_\infty T\). It is clear that if \(\gamma \ge 5\), then the RHS of (5.70) is bounded below by

$$\begin{aligned} \frac{1}{1+C_4(\varepsilon /T)^{1/8}}\left\{ 1-\frac{m_{2,A}(T)}{\sigma _A^2(T)}+\frac{m_{1,A}(T)y}{\sigma _A^2(T)}\right\} \quad \mathrm{where \ } C_4 \ \mathrm{depends \ only \ on \ } A_\infty T. \end{aligned}$$
(5.71)

The inequality (5.56) follows now from (5.64), (5.69), (5.71). \(\square \)

6 Convergence as \(\varepsilon \rightarrow 0\) of Solutions to the Diffusive CP Model

Lemma 6.1

Let \(c_\varepsilon (x,t),\Lambda _\varepsilon (t), \ 0<x,t<\infty ,\) be the solution to the diffusive CP system (1.7), (1.8) with nonnegative initial data \(c_0(x), \ 0<x<\infty ,\) which is a locally integrable function satisfying

$$\begin{aligned} \int _0^\infty (1+x)c_0(x) \ {\mathrm{d}}x<\infty , \quad \int _0^\infty xc_0(x) \ {\mathrm{d}}x = 1. \end{aligned}$$
(6.1)

Then, for any \(T>0\) there are positive constants \(C_1,C_2\) depending only on T and \(c_0(\cdot )\) such that

$$\begin{aligned} C_1\le \Lambda _\varepsilon (t)\le C_2\quad \mathrm{for \ } 0<\varepsilon \le 1, \ 0\le t\le T. \end{aligned}$$
(6.2)

In addition, the set of functions \(\{\Lambda _\varepsilon :[0,T]\rightarrow \mathbf {R}: \ 0<\varepsilon \le 1\}\) form an equicontinuous family. Denote by \(c_0(x,t), \Lambda _0(t), \ 0<x,t<\infty ,\) the solution to the CP system (1.1), (1.2) with \(\varepsilon =0\) and initial data \(c_0(x), \ 0<x<\infty \). Then, for all \(x,t \ge 0\)

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}w_\varepsilon (x,t)= & {} w_0(x,t) , \end{aligned}$$
(6.3)
$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\ \Lambda _\varepsilon (t)= & {} \Lambda _0(t) , \end{aligned}$$
(6.4)

where \(w_\varepsilon \) is given in terms of \(c_\varepsilon \) by (2.49). The limit in (6.3), (6.4) is uniform for (xt) in any finite rectangle \(0<x\le x_0, \ 0<t\le T\).

Proof

It follows from (1.10) that \(\Lambda _\varepsilon (t)\) is an increasing function of t, whence the lower bound in (6.2) follows. We first prove the upper bound for the CP model (1.1), (1.2) corresponding to \(\varepsilon =0\). We see from (1.2), (1.3) that

$$\begin{aligned} \int _{\Lambda _0(0)/2}^\infty xc_0(x) \ \hbox {d}x\ge \frac{1}{2}\int _0^\infty xc_0(x) \ \hbox {d}x =\frac{1}{2} . \end{aligned}$$
(6.5)

Hence, from (2.3) there is a positive constant \(1/C_2\) depending only on \(c_0(\cdot )\) such that \(w_0(\Lambda _0(0)/2,0)\ge 1/C_2\). It follows then from (2.2), (2.4) that \(w_0(0,t) \ge 1/ C_2\) for \(0\le t\le \Lambda _0(0)/2\), whence (1.2), (1.3) implies that \(\Lambda _0(t)\le C_2\) for \(0\le t\le \Lambda _0(0)/2\). Furthermore, we see from (2.3) that \(\Lambda _0(t)\) is continuous in the interval \(0\le t\le \Lambda _0(0)/2\). Since \(\Lambda _0(t)\) is an increasing function of t, we can extend this argument in a finite number of steps to any interval \(0\le t\le T\). We have proven (6.2) in the case \(\varepsilon =0\).

To prove the upper bound in (6.2) for \(0<\varepsilon \le 1\), we use the representation

$$\begin{aligned} w_\varepsilon (x,t) =\int _0^\infty P\left( Y_\varepsilon (t)>x; \inf _{0\le s\le t}Y_\varepsilon (s) >0 \ \big | \ Y_\varepsilon (0)=y\right) c_0(y) \hbox {d}y , \end{aligned}$$
(6.6)

where \(Y_\varepsilon (s)\) is the solution to the SDE (4.8) with \(b(y,s)=y/\Lambda _\varepsilon (s)-1\). Since \(\Lambda _\varepsilon (s)\ge \Lambda _\varepsilon (0)=\Lambda _0(0)\) it follows from (4.9) that for any \(\delta >0, t>0\) there is a positive constant \(p_1\) depending only on \(\delta ,t,\Lambda _0(0)\) such that

$$\begin{aligned} P\left( \inf _{0\le s\le t} \{ \ Y_\varepsilon (s)-E[Y_\varepsilon (s)] \ \} \ge -\delta \right) \ge p_1 \quad \mathrm{for \ } 0<\varepsilon \le 1. \end{aligned}$$
(6.7)

We conclude from (6.7) by choosing \(\delta \) appropriately that there is a positive constant \(p_2\) depending only on \(\Lambda _0(0)\) such that if \(0<\varepsilon \le 1\), then

$$\begin{aligned} P\left( Y_\varepsilon (t)>0; \inf _{0\le s\le t}Y_\varepsilon (s) >0 \ \bigg | \ Y_\varepsilon (0)=y\right) \ge p_2 \quad \mathrm{for \ } t=\Lambda _0(0)/2, \ y\ge \Lambda _0(0)/2. \end{aligned}$$
(6.8)

It follows now from (6.5), (6.6), (6.8) that there is a positive constant \(C_2\) depending only on the initial data \(c_0(\cdot )\) such that \(w_\varepsilon (0,t)\ge 1/C_2\) for \(0<\varepsilon \le 1\) if \(t=\Lambda _0(0)/2\). The upper bound in (6.2) for all T then follows as in the previous paragraph.

To prove that \(\Lambda _\varepsilon (\cdot )\) is continuous, we first note that for any fixed \(t>0\) the function \(w_\varepsilon (x,t), \ x\ge 0,\) is continuous by virtue of the representation (6.6), the fact that \(\Lambda _\varepsilon (s)\ge \Lambda _0(0)\) for \(0\le s\le t\) and (4.11). The continuity is uniform for \(\varepsilon \) in the interval \(0<\varepsilon \le 1\) since \(c_0(\cdot )\) is a locally integrable function. Next we observe from (4.9) that for \(\Delta t>0\) there exists \(x(\Delta t)\) independent of \(\varepsilon \) in the interval \(0<\varepsilon \le 1\) such that \(\lim _{\Delta t\rightarrow 0}x(\Delta t)=0\) and

$$\begin{aligned}&P\left( Y_\varepsilon (t+\Delta t)>0; \inf _{0\le s\le t+\Delta t}Y_\varepsilon (s) >0 \ \bigg | \ Y_\varepsilon (0)=y\right) \nonumber \\&\qquad \ge [1-\Delta t]P\left( Y_\varepsilon (t)>x(\Delta t); \inf _{0\le s\le t}Y_\varepsilon (s) >0 \ \bigg | \ Y_\varepsilon (0)=y\right) \quad \mathrm{for \ } y\ge 0, \ 0<\varepsilon \le 1.\nonumber \\ \end{aligned}$$
(6.9)

It follows from (6.6), (6.9) that \(w_\varepsilon (0,t+\Delta t)\ge [1-\Delta t]w_\varepsilon (x(\Delta t),t)\) for \(0<\varepsilon \le 1\). Using the continuity of the function \(w_\varepsilon (x,t), \ x\ge 0,\) we conclude that \(\lim _{\Delta t\rightarrow 0}w_\varepsilon (0,t+\Delta t)=w_\varepsilon (0,t)\) and the limit is uniform for \(0<\varepsilon \le 1\). Hence, the function \(\Lambda _\varepsilon (\cdot )\) is continuous, and in fact, the family of functions \(\Lambda _\varepsilon (\cdot ), \ 0<\varepsilon \le 1,\) is equicontinuous.

To prove (6.3), (6.4), we first observe from the Ascoli–Arzela theorem that since the family of functions \(\Lambda _\varepsilon (\cdot ), \ 0<\varepsilon \le 1,\) is equicontinuous, the limit (6.4) holds uniformly on the interval \(0\le t\le T\) for a subsequence of \(\varepsilon \rightarrow 0\). For such a sequence, it follows from (2.2), (4.9), (6.6) that (6.3) holds with \(w_0(x,t)=w_0(F_{1/\Lambda _0}(x,t),0)\) and the conservation law (1.2) continues to hold for \(\varepsilon =0\). Hence, the limits on the RHS of (6.3), (6.4) are the solution to the CP model (1.1), (1.2) and are therefore unique. Consequently, (6.3), (6.4) hold for all \(\varepsilon \rightarrow 0\). The uniformity of the limits follows by similar argument. \(\square \)

To show that the coarsening rate (1.10) for the diffusive model (1.7), (1.8) converges as \(\varepsilon \rightarrow 0\) to the coarsening rate (1.4) for the CP model (1.1), (1.2), it will be sufficient to prove the following:

Lemma 6.2

Let \(c_\varepsilon (x,t),\Lambda _\varepsilon (t), \ 0<x,t<\infty ,\) and \(c_0(x), \ 0<x<\infty ,\) be as in Lemma 6.1 and satisfy (6.1). If \(c_0(\cdot )\) is a continuous function, then

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\frac{\varepsilon }{2}\frac{\partial c_\varepsilon (0,T)}{\partial x} =c_0(0,T) \quad \mathrm{for \ any \ }T>0. \end{aligned}$$
(6.10)

Proof

We use the identity

$$\begin{aligned} \frac{\varepsilon }{2}\frac{\partial c_\varepsilon (0,T)}{\partial x} = \lim _{\lambda \rightarrow 0} \frac{c_\varepsilon (\lambda \varepsilon ,T)}{2\lambda } \end{aligned}$$
(6.11)

and the representation for \(c_\varepsilon (\lambda \varepsilon ,T)\) from (4.15),

$$\begin{aligned} c_\varepsilon (\lambda \varepsilon ,T) =\int _{0}^\infty G_{\varepsilon ,D}(\lambda \varepsilon ,y,0,T) c_0(y) \ \hbox {d}y, \end{aligned}$$
(6.12)

where \(G_{\varepsilon ,D}\) is the Dirichlet Green’s function corresponding to the drift \(b(y,t)=y/\Lambda _\varepsilon (t)-1\). From (6.11), (6.12) and Lemma 5.1, we have that

$$\begin{aligned}&\frac{\varepsilon }{2}\frac{\partial c_\varepsilon (0,T)}{\partial x} \nonumber \\&\quad \le \int _{0}^\infty \left\{ 1-\frac{m_{2,1/\Lambda _\varepsilon }(T)}{\sigma _{1/\Lambda _\varepsilon }^2(T)}+\frac{m_{1,1/\Lambda _\varepsilon }(T)y}{\sigma _{1/\Lambda _\varepsilon }^2(T)}+C\Gamma \left( \frac{\varepsilon }{T},\frac{y}{T}\right) \left[ 1+\frac{y}{T}\right] \ \right\} \nonumber \\&\qquad G_{\varepsilon }(0,y,0,T) c_0(y) \ \hbox {d}y, \end{aligned}$$
(6.13)

where the constant C depends only on \(T/\Lambda _0(0)\). We conclude from Lemma 6.1, (4.11) and (6.13) that

$$\begin{aligned} \limsup _{\varepsilon \rightarrow 0} \frac{\varepsilon }{2}\frac{\partial c_\varepsilon (0,T)}{\partial x} \le \frac{1}{m_{1,1/\Lambda _0}(T)} c_0\left( \frac{m_{2,1/\Lambda _0}(T)}{m_{1,1/\Lambda _0}(T)}\right) , \end{aligned}$$
(6.14)

provided the function \(c_0(y), \ y>0,\) is continuous at \(y=m_{2,1/\Lambda _0}(T)/m_{1,1/\Lambda _0}(T)\).

We can obtain a lower bound on the LHS of (6.10) by using Lemma 5.2. Thus, we have that

$$\begin{aligned}&\left[ 1+C(\varepsilon /T)^{1/8}\right] \frac{\varepsilon }{2}\frac{\partial c_\varepsilon (0,T)}{\partial x}\nonumber \\&\quad \ge \int _{5(\varepsilon /T)^{1/8}T/c}^\infty \left[ 1+\exp \left( -c^2(T/\varepsilon )^{1/4}(y^2/4T^2\right) \ \right] ^{-2} \ \nonumber \\&\qquad \times \left\{ 1-\frac{m_{2,1/\Lambda _\varepsilon }(T)}{\sigma _{1/\Lambda _\varepsilon }^2(T)}+\frac{m_{1,1/\Lambda _\varepsilon }(T)y}{\sigma _{1/\Lambda _\varepsilon }^2(T)} \ \right\} G_{\varepsilon }(0,y,0,T) c_0(y) \ \hbox {d}y, \end{aligned}$$
(6.15)

where the constants \(C,c>0\) depend only on \(T/\Lambda _0(0)\). We conclude from Lemma 6.1, (4.11) and (6.15) that

$$\begin{aligned} \liminf _{\varepsilon \rightarrow 0} \frac{\varepsilon }{2}\frac{\partial c_\varepsilon (0,T)}{\partial x} \ge \frac{1}{m_{1,1/\Lambda _0}(T)} c_0\left( \frac{m_{2,1/\Lambda _0}(T)}{m_{1,1/\Lambda _0}(T)}\right) , \end{aligned}$$
(6.16)

provided the function \(c_0(y), \ y>0,\) is continuous at \(y=m_{2,1/\Lambda _0}(T)/m_{1,1/\Lambda _0}(T)\). Finally, we observe that the RHS of (6.14), (6.16) is the same as \(c_0(0,T)\). This follows by differentiating the function \(w(x,t)=w_0(F_{1/\Lambda _0}(x,t),0)\) with respect to x at \(x=0\), and using the formula (2.2) for the function \(F_{1/\Lambda _0}\). \(\square \)

7 Upper Bound on the Coarsening Rate of Diffusive CP Model

In this section, we prove Theorem 1.2. First, we show that \(\lim _{t\rightarrow \infty }\langle X_t\rangle =\infty \).

Lemma 7.1

Let \(c_\varepsilon (x,t), \Lambda _\varepsilon (t), \ 0<x,t<\infty ,\) be the solution to (1.7), (1.8) with \(\varepsilon >0\) and nonnegative initial data \(c_0(x), \ 0<x<\infty ,\) which is a locally integrable function satisfying (6.1). Then, \(\lim _{t\rightarrow \infty }\Lambda _\varepsilon (t)=\infty \).

Proof

We have already noted that \(\Lambda _\varepsilon (t)\) is an increasing function of t. It will therefore be sufficient to show that if for some finite \(\Lambda _\infty \) we have \(\Lambda _\varepsilon (t)\le \Lambda _\infty \) for all \(t\ge 0\), then there is a contradiction. To see this, we use the identity

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty xc_\varepsilon (x,t) \ \hbox {d}x = \frac{1}{\Lambda _\varepsilon (t)}\int _0^\infty xc_\varepsilon (x,t) \ \hbox {d}x -\int _0^\infty c_\varepsilon (x,t) \ \hbox {d}x , \end{aligned}$$
(7.1)

which follows from (1.7). Using the conservation law (1.8) and (7.1), we see that

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty xc_\varepsilon (x,t) \ \hbox {d}x \ge \frac{1}{2\Lambda _\infty }\int _0^\infty xc_\varepsilon (x,t) \ \hbox {d}x -\int _0^{2\Lambda _\infty } c_\varepsilon (x,t) \ \hbox {d}x \\&\quad =\frac{1}{2\Lambda _\infty } -\int _0^{2\Lambda _\infty } c_\varepsilon (x,t) \ \hbox {d}x \ge \ \frac{1}{2\Lambda _\infty } -\int _0^{2\Lambda _\infty } \hbox {d}x\int _0^\infty \hbox {d}y \ G_\varepsilon (x,y,0,t)c_0(y),\nonumber \end{aligned}$$
(7.2)

where \(G_\varepsilon \) is the function (4.11) with \(A(s)=1/\Lambda _\varepsilon (s), \ s\ge 0\). Hence, we conclude that

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty xc_\varepsilon (x,t) \ \hbox {d}x \ge \frac{1}{2\Lambda _\infty }- \frac{2\Lambda _\infty }{\Lambda _\varepsilon (0)\sqrt{2\pi \varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)}} . \end{aligned}$$
(7.3)

From (4.10), we see that \(\sigma _{1/\Lambda _\varepsilon }^2(t)\ge t\), and hence, (7.3) implies that

$$\begin{aligned} \lim _{t\rightarrow \infty }\int _0^\infty xc_\varepsilon (x,t) \ \hbox {d}x =\infty , \end{aligned}$$
(7.4)

but this is a contradiction to the conservation law (1.8). \(\square \)

We begin the proof of the inequality (1.16):

Lemma 7.2

Suppose \(c_0:[0,\infty )\rightarrow \mathbf {R}^+\) satisfies (6.1) and \(c_\varepsilon (x,t),\ x \ge 0, t>0\) is the solution to (1.7), (1.8) with initial data \(c_0(\cdot )\) and Dirichlet boundary condition \(c_\varepsilon (0,t)=0, \ t>0\). Assume that \(\Lambda _\varepsilon (0)=1\) and that the function \(h_\varepsilon (x,t)\) defined by (2.49) is log-concave in x at \(t=0\). Then, there exist positive universal constants \(C,\varepsilon _0\) with \(0<\varepsilon _0\le 1\) such that

$$\begin{aligned} c_\varepsilon (\lambda \varepsilon ,1) \le C\lambda c_\varepsilon (\varepsilon ,1) \quad \mathrm{for \ } 0<\varepsilon \le \varepsilon _0,\ 0<\lambda \le 1. \end{aligned}$$
(7.5)

Proof

Let \(X_0\) be the positive random variable with pdf \(c_0(x)/\int _0^\infty c_0(x') \ \hbox {d}x', \ x>0\). Then from (1.9) and the assumption \(\Lambda _\varepsilon (0)=1\) we see that \(\langle X_0\rangle =1\). Since the beta function (2.35) for \(X_0\) is also bounded by 1 it follows from the Chebshev inequality and the identity (29) of Conlon (2011) that for \(\delta \) with \(0<\delta <1\), there exists a constant \(\nu (\delta )>0\) depending only on \(\delta \) such that

$$\begin{aligned} P(X_0<\nu (\delta ))+P(X_0>1/[1-\delta ]) \ \le 1-\delta /2. \end{aligned}$$
(7.6)

Now recall that the Green’s function (4.11) has the form (4.24) where the function \(y\rightarrow q(x,y,0,t)\) takes its minimum at \(y=F_A(x,t)\) where \(F_A\) is defined by (2.2). In the case of \(A(\cdot )=1/\Lambda _\varepsilon (\cdot ), \ x=\lambda \varepsilon \) and \(t=1\), we see that \( 1-(1-\lambda \varepsilon )/e \le F_{1/\Lambda _\varepsilon }(\lambda \varepsilon ,1)\le 1+ \lambda \varepsilon \). This follows from the fact that the function \(\Lambda _\varepsilon (\cdot )\) is increasing and \(\Lambda _\varepsilon (0)=1\). We choose now \(\delta ,\varepsilon _0>0\) sufficiently small so that \(F_{1/\Lambda _\varepsilon }(\lambda \varepsilon ,1)-\nu (\delta )>1/(1-\delta )-F_{1/\Lambda _\varepsilon }(\lambda \varepsilon ,1)>0\) for \( 0<\lambda \le 1, \ \varepsilon \le \varepsilon _0\). It follows then from Lemma 5.1 that

$$\begin{aligned} c_\varepsilon (\lambda \varepsilon ,1) \le C_1\lambda \int _{\nu (\delta )}^{2F_{1/\Lambda _\varepsilon }(\lambda \varepsilon ,1)-\nu (\delta )} G_\varepsilon (\varepsilon ,y,0,1) c_0(y) \ \hbox {d}y \quad \mathrm{for \ } 0<\varepsilon \le \varepsilon _0, \ 0<\lambda \le 1, \end{aligned}$$
(7.7)

where \(C_1>0\) depends only on \(\delta ,\varepsilon _0\). Next we apply Lemma 5.2 to conclude that

$$\begin{aligned}&\int _{\nu (\delta )}^{2F_{1/\Lambda _\varepsilon }(\lambda \varepsilon ,1)-\nu (\delta )} G_\varepsilon (\varepsilon ,y,0,1) c_0(y) \ \hbox {d}y \nonumber \\&\quad \le C_2\int _{\nu (\delta )}^{2F_{1/\Lambda _\varepsilon }(\lambda \varepsilon ,1)-\nu (\delta )} G_{\varepsilon ,D}(\varepsilon ,y,0,1) c_0(y) \ \hbox {d}y \le C_2 c_\varepsilon (\varepsilon ,1) , \end{aligned}$$
(7.8)

for some constant \(C_2>0\) depending only on \(\delta ,\varepsilon _0\). Actually, a strict application of Lemma 5.2 requires us to impose an additional restriction on \(\varepsilon _0>0\) so that the condition \(\gamma \ge 5\) of Lemma 5.2 holds. Now (7.5) follows from (7.7), (7.8). \(\square \)

Lemma 7.3

Suppose the initial data \(c_0(\cdot )\) for (1.7), (1.8) satisfies the conditions of Lemma 7.2, the function \(h_\varepsilon (x,t)\) is log-concave in x for all \(t\ge 0\), and \(0<\varepsilon \le \varepsilon _0\). Then, there is a universal constant C such that \({\mathrm{d}}\Lambda _\varepsilon (t)/{\mathrm{d}}t\le C\) for \(t\ge 1\).

Proof

From (1.8), (1.10), (2.35), (2.49) and Lemma 7.2, we see there is a universal constant \(C_1\) such that

$$\begin{aligned} \frac{\hbox {d}\Lambda _\varepsilon (1)}{\hbox {d}t} \le \frac{C_1c_\varepsilon (\varepsilon ,1)h_\varepsilon (0,1)}{w_\varepsilon (0,1)^2} \le \frac{C_1\beta _{X_1}(\varepsilon )h_\varepsilon (0,1)}{h_\varepsilon (\varepsilon ,1)} , \end{aligned}$$
(7.9)

where \(X_t\) is the random variable with pdf proportional to \(c_\varepsilon (\cdot ,t)\). We can bound \(h_\varepsilon (\varepsilon ,1)\) below by a constant times \(h_\varepsilon (0,1)\). To see this, consider a positive random variable X and observe that

$$\begin{aligned} E[ \ X-\langle X\rangle /2; \ X>\langle X\rangle /2 \ ]\ge & {} E[ \ X-\langle X\rangle /2; \ X>3\langle X\rangle /4 \ ]\nonumber \\\ge & {} \frac{1}{3} E[ \ X; \ X>3\langle X\rangle /4 \ ] \ge \frac{1}{12} \langle X\rangle . \end{aligned}$$
(7.10)

Since the function \(\Lambda _\varepsilon (\cdot )\) is increasing, it follows that \(\langle X_1\rangle \ge 1\). We conclude then from (7.10) that

$$\begin{aligned} h_\varepsilon (\varepsilon ,1) =\int _\varepsilon ^\infty (x-\varepsilon ) c_\varepsilon (x,1) \ \hbox {d}x \ge \frac{1}{12} \int _0^\infty x c_\varepsilon (x,1) \ \hbox {d}x =\frac{1}{12}h_\varepsilon (0,1) , \end{aligned}$$
(7.11)

provided \(\varepsilon <1/2\). Since the log-concavity of \(h_\varepsilon (\cdot ,1)\) implies that \(\beta _{X_1}(\varepsilon )\le 1\), we obtain from (7.9), (7.11) an upper bound on \(\hbox {d}\Lambda _\varepsilon (t)/\hbox {d}t\) when \(t=1\).

The upper bound for \(t>1\) now follows from the scaling property of (1.7), (1.8) mentioned in the discussion following the statement of Theorem 1.2. To see this, we define a function \(\tau (\lambda ), \ \lambda \ge 1, \) as the solution to the equation \(\Lambda _\varepsilon (\lambda \tau (\lambda ))=\lambda \). Observe from (1.10) and the Hopf maximum principle (Protter and Weinberger 1984) that the function \(\Lambda _\varepsilon (\cdot )\) is strictly increasing. Hence, \(\tau (\lambda )\) is uniquely determined. Furthermore, the function \(\tau (\cdot )\) is continuous. Rescaling (1.7), (1.8) by \(\lambda \), we conclude from the result of the previous paragraph that

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t} \ \Lambda _\varepsilon (\lambda [\tau (\lambda )+t]) \le C\lambda \quad \mathrm{at \ } t=1. \end{aligned}$$
(7.12)

We have shown then that \(\hbox {d}\Lambda _\varepsilon (t)/\hbox {d}t\le C\) at \(t=\lambda [\tau (\lambda )+1]\). Since the function \(\lambda \rightarrow \lambda \tau (\lambda )\) is monotonically increasing with range \([0,\infty )\) the result follows. \(\square \)

To complete the proof of the inequality (1.16), we first observe that by Lemma 7.1 there exists \(T_\varepsilon \ge 0\) such that \(\varepsilon /\Lambda _\varepsilon (T_\varepsilon )\le \varepsilon _0\), where \(\varepsilon _0\) is the universal constant of Lemma 7.2. We now rescale (1.7), (1.8) with \(\lambda =\Lambda _\varepsilon (T_\varepsilon )\), which puts us into the situation of Lemma 7.3. The result follows on taking \(T=T_\varepsilon +\Lambda _\varepsilon (T_\varepsilon )\), provided we have the log-concavity property of the function \(h_\varepsilon \) in the statement of Lemma 7.3. The assumption of Theorem 1.2, that the function \(x\rightarrow E[X_0-x \ | \ X_0>x], \ 0\le x<\Vert X_0\Vert _\infty \), decreases, is equivalent to the assumption that the function \(h_\varepsilon (\cdot ,0)\) is log-concave. Hence, it remains to be shown that if \(h_\varepsilon (\cdot ,0)\) is log-concave, then \(h_\varepsilon (\cdot ,t)\) is also log-concave for all \(t>0\).

If we make the approximation (2.51) for \(h_\varepsilon \), the log-concavity of \(h_\varepsilon (\cdot ,t)\) follows from the Prékopa–Leindler theorem (Theorem 6.4 of Villani 2003). In our situation, we follow the approach of Korevaar (1983) and differentiate the PDE (2.67) which \(h_\varepsilon \) satisfies. Thus, \(v_\varepsilon (x,t)=-\frac{\partial }{\partial x} \log h_\varepsilon (x,t)\) is a solution of the PDE (2.69), whence \(u_\varepsilon (x,t)=\partial v_\varepsilon (x,t)/\partial x\) is a solution to the PDE

$$\begin{aligned} \frac{\partial u_\varepsilon (x,t)}{\partial t}+\left[ \frac{x}{\Lambda _\varepsilon (t)}-1+\varepsilon v_\varepsilon (x,t)\right] \frac{\partial u_\varepsilon (x,t)}{\partial x} +\frac{2u_\varepsilon (x,t)}{\Lambda _\varepsilon (t)} +\varepsilon u_\varepsilon (x,t)^2= \frac{\varepsilon }{2} \frac{\partial ^2 u_\varepsilon (x,t)}{\partial x^2} . \end{aligned}$$
(7.13)

Observe now that

$$\begin{aligned} u_\varepsilon (x,t) \ = \ v_\varepsilon (x,t)^2[1-c_\varepsilon (x,t)h_\varepsilon (x,t)/w_\varepsilon (x,t)^2] . \end{aligned}$$
(7.14)

Since \(\lim _{x\rightarrow 0} c_\varepsilon (x,t)=0\) for \(t>0\), it follows from (7.14) that \( \liminf _{x\rightarrow 0} u_\varepsilon (x,t)\ge 0\) for \(t>0\). If \(h_\varepsilon (\cdot ,0)\) is log-concave, then the initial data \(u_\varepsilon (x,0), \ x>0,\) for (7.13) is also nonnegative. We expect then from the maximum principle that \(u_\varepsilon (x,t)\) is nonnegative for all \(x,t>0\), and hence, \(h_\varepsilon (\cdot ,t)\) is a log-concave function for all \(t>0\).

Although Korevaar’s argument is simple in concept, the rigorous implementation requires that certain technical difficulties be overcome. Our first step toward rigorous implementation is to approximate an arbitrary nonnegative random variable X satisfying \(\langle X\rangle <\infty \), and having log-concave function \(h_X(\cdot )\) as defined by (2.33), by random variables with some regularity. The approximating random variables Y have the properties:

$$\begin{aligned} Y \ \mathrm{is \ nonnegative \ with \ continuous \ pdf \ } c_Y(y), \ y\ge 0, \quad \mathrm{and \ } c_Y(0)=0. \end{aligned}$$
(7.15)
$$\begin{aligned}&\mathrm{There \ exists \ } K,a,L,y_0>0 \ \mathrm{and \ } c_Y(y)\\&\quad =K\exp \left[ -a(y-y_0)-\{a(y-y_0)\}^2/2L\right] \ \mathrm{for \ } y\ge y_0.\\&\mathrm{The \ beta \ function \ (2.35) \ of \ } Y \ \mathrm{satisfies \ } \beta _Y(y)<1 \quad \mathrm{for \ all \ } y\ge 0. \end{aligned}$$

We assume that the function \(\Lambda _\varepsilon :[0,\infty )\rightarrow \mathbf {R}^+\) of (1.7) is positive and continuous, and consider solutions \(c_\varepsilon \) to (1.7) with Dirichlet boundary condition \(c_\varepsilon (0,t)=0, \ t>0\), and initial condition given by the integrable pdf \(c_{X_0}(\cdot )\) of a nonnegative random variable \(X_0\) satisfying \(\langle X_0\rangle <\infty \). We denote by \(X_t\) the random variable with pdf \(c_\varepsilon (\cdot ,t), \ t>0\).

Lemma 7.4

Assume that the function \(h_{X_0}(\cdot )\) for the initial data random variable \(X_0\) of (1.7) is log-concave. Then, there is a sequence of random variables \(Y_0^k, \ k=1,2,\ldots ,\) satisfying (7.15) such that the functions \((x,t)\rightarrow h_{Y^k_t}(x), \ k=1,2,\ldots ,\) converge as \(k\rightarrow \infty \), uniformly in any rectangle \(\{(x,t) \ : \ 0\le x\le x_0, \ 0\le t\le T\}\), to the function \((x,t)\rightarrow h_{X_t}(x)\).

Proof

We first assume that \(\Vert X_0\Vert _\infty <\infty \) and define the beta function for \(Y_0^k\) in the interval \(0\le x\le (1-2/k)\Vert X_0\Vert _\infty \) in terms of the beta function for \(X_0\) as follows:

$$\begin{aligned} \beta _{Y^k_0}(x)= & {} \left( 1-\frac{1}{k}\right) \frac{k}{\Vert X_0\Vert _\infty }\int _0^x \beta _{X_0}(z) \ \hbox {d}z \quad \mathrm{for \ } 0\le x\le \frac{\Vert X_0\Vert _\infty }{k} , \nonumber \\ \beta _{Y^k_0}(x)= & {} \left( 1-\frac{1}{k}\right) \frac{k}{\Vert X_0\Vert _\infty }\int _{x-\Vert X_0\Vert _\infty /k}^x \beta _{X_0}(z) \ \hbox {d}z \quad \mathrm{for \ } \frac{\Vert X_0\Vert _\infty }{k}\le x\le \left( 1-\frac{2}{k}\right) \Vert X_0\Vert _\infty .\nonumber \\ \end{aligned}$$
(7.16)

It follows from (2.35), (7.16) that \(c_{Y^k_0}(x)\) is continuous in the interval \(0\le x\le (1-2/k)\Vert X_0\Vert _\infty \) and \(c_{Y^k_0}(0)=0\). To continue the definition of \(\beta _{Y^k_0}(\cdot )\), we choose \(L_k>0\) sufficiently large so that the function \(\beta _{L_k}\) of Lemma 2.1 satisfies \(\beta _L(0)\ge 1-1/k\) and \(\beta _L(z)\le 1-1/2L(1+z/L)^2\) for \(z\ge 0, \ L\ge L_k\). We then define \(\beta _{Y^k_0}(x)\) for \((1-2/k)\Vert X_0\Vert _\infty \le x\le (1-1/k)\Vert X_0\Vert _\infty \) by linear interpolation, taking the value \(\beta _{Y^k_0}\left( (1-2/k)\Vert X_0\Vert _\infty \right) \) at the left end of the interval and the value \(\beta _{L_k}(0)\) at the right end. Finally, we extend \(c_{Y^k_0}(x)\) to \(x\ge (1-1/k)\Vert X_0\Vert _\infty \) by setting it equal to the Gaussian in (7.15) with \(y_0=(1-1/k)\Vert X_0\Vert _\infty \), choosing K so that \(c_{Y^k_0}(\cdot )\) is continuous and a so that \(E[Y^k_0-x \ | \ Y^k_0>x]=E[X_0-x \ | \ X_0>x]\) when \(x=(1-1/k)\Vert X_0\Vert _\infty \). The random variable \(Y^k_0\) satisfies (7.15). In particular, since \(h_{X_0}(\cdot )\) is log-concave it follows that \(\beta _{Y^k_0}(x)\le 1-1/k\) for \(x\le (1-2/k)\Vert X_0\Vert _\infty \).

To construct the pdf \(c_{Y^k_0}(\cdot )\) from the function \(\beta _{Y^k_0}(\cdot )\) defined in the previous paragraph, we first observe from (2.35) that

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}y} E[Y^k_0-y \ | \ Y^k_0>y] =\beta _{Y^k_0}(y)-1 , \quad E[Y^k_0-y^k \ | \ Y^k_0>y^k]=E[X_0-y^k \ | \ X_0>y^k] , \end{aligned}$$
(7.17)

where \(y^k=(1-1/k)\Vert X_0\Vert _\infty \). The function \(v^k(y)=E[Y^k_0-y \ | \ Y^k_0>y]^{-1}, \ y\ge 0,\) is uniquely determined by (7.17). From (2.33), (2.34), we see that

$$\begin{aligned} h_{Y^k_0}(x) =A^k \exp \left[ -\int _0^x v^k(y) \ \hbox {d}y\right] , \quad x\ge 0, \end{aligned}$$
(7.18)

for some constant \(A^k\). If we define the function \(f_k(\cdot )\) by

$$\begin{aligned} f_k(x) =\beta _{Y^k_0}(x)v^k(x)^2 \exp \left[ -\int _0^x v^k(y) \ \hbox {d}y\right] , \quad x\ge 0, \end{aligned}$$
(7.19)

then (2.35) implies that \(c_{Y^k_0}(x)=A^kf_k(x), \ x\ge 0\). Using the normalization condition for the probability measure \(c_{Y^k_0}(\cdot )\), we conclude that the constant K in (7.15) is given by the formula

$$\begin{aligned} K =f_k(y^k)\big / \int _0^\infty f_k(x) \ \hbox {d}x . \end{aligned}$$
(7.20)

We show that the functions \(w_{Y^k_0}(\cdot )\) converge as \(k\rightarrow \infty \) to \(w_{X_0}(\cdot )\), uniformly in \([0,\infty )\). To do this, we use the identity

$$\begin{aligned} \int _x^\infty f_k(y) \ \hbox {d}y = v^k(x) \exp \left[ -\int _0^x v^k(y) \ \hbox {d}y\right] , \quad x\ge 0, \end{aligned}$$
(7.21)

obtained from (7.17), (7.19). From (7.17), we have that

$$\begin{aligned} E[Y^k_0-y \ | \ Y^k_0>y] =E[X_0-y^k \ | \ X_0>y^k]+\int _y^{y^k} \left[ 1-\beta _{Y^k_0}(y')\right] \ \hbox {d}y' , \quad 0\le y\le y^k . \end{aligned}$$
(7.22)

It follows from (7.16), (7.22) that

$$\begin{aligned} \lim _{k\rightarrow \infty } E[Y^k_0-y \ | \ Y^k_0>y] =\int _y^{\Vert X_0\Vert _\infty } \left[ 1-\beta _{X_0}(y')\right] \ \hbox {d}y' , \quad 0\le y<\Vert X_0\Vert _\infty , \end{aligned}$$
(7.23)

and the convergence is uniform in any interval \(\{y \ : \ 0\le y<\Vert X_0\Vert _\infty (1-\delta )\}\) for which \(\delta >0\). We conclude from (7.21), (7.23) upon setting \(v^\infty (x)=E[X_0-x \ | \ X>x]^{-1}\) that

$$\begin{aligned} \lim _{k\rightarrow \infty }\int _x^\infty f_k(y) \ \hbox {d}y = v^\infty (x) \exp \left[ -\int _0^x v^\infty (y) \ \hbox {d}y\right] , \quad 0\le x<\Vert X_0\Vert _\infty , \end{aligned}$$
(7.24)

and the convergence is uniform in any interval \(\{x \ : \ 0\le x<\Vert X_0\Vert _\infty (1-\delta )\}\) for which \(\delta >0\). Taking \(x=0\) in (7.24), we have that \(\lim _{k\rightarrow \infty } A_k=E[X_0]\). Hence, (7.24) implies that

$$\begin{aligned} \lim _{k\rightarrow \infty }\int _x^\infty c_{Y^k_0}(y) \ \hbox {d}y =\int _x^\infty c_{X_0}(y) \ \hbox {d}y , \quad 0\le x<\Vert X_0\Vert _\infty , \end{aligned}$$
(7.25)

and the convergence is uniform in any interval \(\{x \ : \ 0\le x<\Vert X_0\Vert _\infty (1-\delta )\}\) for which \(\delta >0\). In view of the integrability of \(c_{X_0}(\cdot )\), we conclude that the convergence (7.25) is uniform for \(0\le x<\infty \).

We can easily estimate \(w_{Y^k_0}(x)\) for \(x\ge \Vert X_0\Vert _\infty \) since the pdf of \(Y^k_0\) is Gaussian when \(x\ge \Vert X_0\Vert _\infty \). To do this, we define a function \(g:[0,\infty )\rightarrow \mathbf {R}\) by

$$\begin{aligned} g(z) = E[Z \ | \ Z>z]^{-1} =e^{z^2/2}\int _z^\infty e^{-z'^2/2} \ \hbox {d}z' =\int _0^\infty e^{-z'z-z'^2/2} \ \hbox {d}z' , \end{aligned}$$
(7.26)

where Z is the standard normal variable. We conclude from (7.26) that \(g(\cdot )\) is a decreasing function and \(\lim _{z\rightarrow \infty } g(z)=0\). We can estimate g(z) for large z from the final integral on the RHS of (7.26) to obtain the inequality

$$\begin{aligned} 0 < g(z) < \frac{1}{z}\left[ 1-\frac{1}{z^2}+\frac{3}{z^4} \right] , \end{aligned}$$
(7.27)

whence it follows that

$$\begin{aligned} E[Z \ | \ Z>z] > z\left[ 1+\frac{1}{2z^2}\right] \quad \mathrm{for \ } z> \sqrt{6}. \end{aligned}$$
(7.28)

We conclude from (7.28) that there exists \(\gamma _0>0\) and

$$\begin{aligned} E[Z \ | \ Z>z] -z\ge \min \{\gamma _0, 1/2z\} , \quad z\ge 0. \end{aligned}$$
(7.29)

The random variable Y of (7.15) has for \(y\ge y_0\) the pdf of a normal variable with mean \(y_0-L/a\) and variance \(L/a^2\). We can estimate the value of a when \(Y=Y^k_0\) by using the equality \(E[Y^k_0-y^k \ | \ Y^k_0>y^k]=E[X_0-y^k \ | \ X_0>y^k]\le \Vert X_0\Vert _\infty /k\). Observe now that

$$\begin{aligned} E[Y^k_0-y^k \ | \ Y^k_0>y^k] =\frac{\sqrt{L_k}}{a}\left( E[Z \ | \ Z>\sqrt{L_k}]-\sqrt{L_k}\right) . \end{aligned}$$
(7.30)

Hence, using the upper bound on the LHS of (7.30), we obtain from (7.29), (7.30) a lower bound

$$\begin{aligned} a \ge \frac{k\sqrt{L_k}}{\Vert X_0\Vert _\infty } \min \{\gamma _0,1/2\sqrt{L_k}\} \end{aligned}$$
(7.31)

for a. Since \(\lim _{k\rightarrow \infty } L_k=\infty \), it follows from (7.31) that \(\lim _{k\rightarrow \infty } a_k=\infty ,\) where \(a_k\) is the value of a in (7.15) when \(Y=Y^k_0\).

We have from (7.15) that

$$\begin{aligned} w_{Y^k_0}(x) =\int _{x-y_k}^\infty K_k\exp [-a_ky-(a_ky)^2/2L_k] \ \hbox {d}y \quad \mathrm{for \ } x\ge y_k . \end{aligned}$$
(7.32)

Furthermore, from (7.25) it follows that for any \(\eta >0\) there exists an integer \(k_\eta \) such that \(w_{Y^k_0}(y_k)<\eta \) for \(k\ge k_\eta \). This implies a bound on \(K_k\) in (7.32) of the form \(K_k\le \eta a_k, \ k\ge k_\eta \). Hence, from (7.32) it follows that \( h_{Y^k_0}(y_k)\le \eta / a_k\) and hence \(\lim _{k\rightarrow \infty } h_{Y^k_0}(y_k)=0\). We conclude from this and (7.25) that the functions \(x\rightarrow h_{Y^k_0}(x), \ k=1,2,\ldots ,\) converge as \(k\rightarrow \infty \), uniformly in any interval \(\{x \ : \ 0\le x\le x_0 \}\), to the function \(x\rightarrow h_{X_0}(x)\).

To prove that \(h_{Y^k_t}(\cdot ), \ k=1,2,\ldots ,\) converges to \(h_{X_t}(\cdot )\) for \(t>0\) we use the fact that the function \(w_\varepsilon \) defined by (2.49) is a solution to the PDE (2.66). Furthermore, the Dirichlet boundary condition \(c_\varepsilon (0,t)=0\) for (1.7) becomes a Neumann boundary condition \(\partial w_\varepsilon (0,t)/\partial x=0\) for (2.66). By the Hopf maximum principle (Protter and Weinberger 1984), we then have that

$$\begin{aligned} \sup \left| w_{Y^k_t}(\cdot )-w_{X_t}(\cdot )\right| \le \sup \left| w_{Y^k_0}(\cdot )-w_{X_0}(\cdot )\right| , \quad t\ge 0. \end{aligned}$$
(7.33)

From (7.25), (7.33), we see that \(w_{Y^k_t}(\cdot ), \ k=1,2,\ldots ,\) converges to \(w_{X_t}(\cdot )\) for any \(t\ge 0\). This implies convergence of \(h_{Y^k_t}(\cdot ), \ k=1,2,\ldots ,\) to \(h_{X_t}(\cdot )\) provided we can obtain a suitable uniform bound on \(w_{Y^k_t}(x), \ k=1,2,\ldots ,\) for large x. To carry this out, we use the representation (6.6) for \(w_\varepsilon \). Thus, we have that

$$\begin{aligned} w_{Y^k_t}(x)\le & {} \int _x^\infty \hbox {d}x'\int _0^\infty G_\varepsilon (x',y,0,t) \ c_{Y^k_0}(y) \ \hbox {d}y \nonumber \\= & {} \int _x^\infty \hbox {d}x' \ G_\varepsilon (x',0,0,t )w_{Y^k_0}(0)+m_{1,1/\Lambda _\varepsilon }(t)\int _0^\infty G_\varepsilon (x,y,0,t) w_{Y^k_0}(y) \ \hbox {d}y ,\nonumber \\ \end{aligned}$$
(7.34)

where \(G_\varepsilon \) is given by (4.11) with \(A(\cdot )=1/\Lambda _\varepsilon (\cdot )\). Evidently, (7.25) and (7.34) yield a uniform upper bound on \(w_{Y^k_t}(x), \ k=1,2,\ldots ,\) for large x, which decays rapidly as \(x\rightarrow \infty \) to 0.

We have therefore proven the result for random variables \(X_0\) which satisfy \(\Vert X_0\Vert _\infty <\infty \). In the case when \(\Vert X_0\Vert _\infty =\infty \) we proceed similarly, approximating \(X_0\) with variables \(Y^k_0\) satisfying (7.15) by averaging the beta function of \(X_0\) over intervals of length 1 / k as in (7.16) for \(0\le x\le k, \ k=1,2,\ldots \) \(\square \)

Lemma 7.5

Assume that the function \(h_{X_0}(\cdot )\) for the initial data random variable \(X_0\) of (1.7) satisfies (7.15) and \(T>0\). Then, there exists \(x_T>0\) such that \(\beta _{X_t}(x)<1\) for all \(x\ge x_T, \ 0\le t\le T\).

Proof

We first obtain a lower bound on the ratio of the half-line Dirichlet Green’s function defined by (4.14) to the full line Green’s function (4.11). Letting \(A_\infty \) be given by (4.1), we show that for any \(\gamma >0\), there are positive constants \(C_1,C_2\) depending only on \(\gamma , \ A_\infty T\) such that

$$\begin{aligned} \frac{G_{\varepsilon ,D}(x,y,0,T)}{G_{\varepsilon }(x,y,0,T)} \ge 1-\exp \left[ -\frac{x^2}{C_1\varepsilon T}\right] \quad \mathrm{for \ } y\ge \gamma x, \ x\ge C_2[T+\sqrt{\varepsilon T}]. \end{aligned}$$
(7.35)

We see from (4.29), (4.41) that in order to establish (7.35) it is sufficient to show that there are constants \(C_1,C_2\) depending only on \(A_\infty T\) such that

$$\begin{aligned} P\left( \sup _{0\le s\le T}\left| \int _0^T k(s,s') \ \hbox {d}B(s')\right| > z\right) \le \exp \left[ -\frac{z^2}{C_1 T}\right] \quad \mathrm{for \ } z\ge C_2\sqrt{T}. \end{aligned}$$
(7.36)

The inequality (7.36) follows from Doob’s Martingale inequality as in the proof of Proposition 5.1.

We have now that

$$\begin{aligned} c_{X_t}(x)= & {} \int _0^\infty G_{\varepsilon ,D}(x,y,0,t)c_{X_0}(y) \ \hbox {d}y \nonumber \\\le & {} \int _0^{y_0} G_{\varepsilon }(x,y,0,t)c_{X_0}(y) \ \hbox {d}y+\int _{y_0}^\infty G_{\varepsilon }(x,y,0,t)c_{X_0}(y) \ \hbox {d}y,\qquad \qquad \end{aligned}$$
(7.37)

where \(A(\cdot )=1/\Lambda _\varepsilon (\cdot )\) in (4.11) and \(y_0\) is given in (7.15). We can bound the first integral on the RHS of (7.37) using integration by parts to obtain the inequality

$$\begin{aligned} \int _0^{y_0} G_{\varepsilon }(x,y,0,t)c_{X_0}(y) \ \hbox {d}y\le & {} G_{\varepsilon }(x,0,0,t)w_{X_0}(0)\nonumber \\&-m_{1,1/\Lambda _\varepsilon }(t)\frac{\partial }{\partial x} \int _0^{y_0} G_{\varepsilon }(x,y,0,t)w_{X_0}(y) \ \hbox {d}y.\nonumber \\ \end{aligned}$$
(7.38)

Since \(c_{X_0}(y)\) is Gaussian for \(y\ge y_0\) as given in (7.15), the second integral on the RHS of (7.37) is bounded by a Gaussian,

$$\begin{aligned}&\int _{y_0}^\infty G_{\varepsilon }(x,y,0,t)c_{X_0}(y) \ \hbox {d}y \nonumber \\&\quad \le K\int _{-\infty }^\infty G_{\varepsilon }(x,y,0,t)\exp [-a(y-y_0)-\{a(y-y_0)\}^2/2L] \ \hbox {d}y\nonumber \\&\quad =K\exp [L/2]\left( \frac{L}{a^2\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)+Lm_{1,1/\Lambda _\varepsilon }(t)^2}\right) ^{1/2} \ \nonumber \\&\qquad \times \exp \left[ -\frac{\left( x+m_{2,1/\Lambda _\varepsilon }(t)-m_{1,1/\Lambda _\varepsilon }(t)y_0+m_{1,1/\Lambda _\varepsilon }(t)L/a\right) ^2}{2\left\{ \varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)+Lm_{1,1/\Lambda _\varepsilon }(t)^2/a^2\right\} }\right] .\quad \qquad \end{aligned}$$
(7.39)

We conclude from (4.11), (7.37)–(7.39) that there exist positive constants \(x_T,M_T\) such that for \(0<t\le T, \ x\ge x_T\),

$$\begin{aligned} c_{X_t}(x) \le \left[ 1+\exp [-x^2/M_T\right] K_t\exp \left[ -\frac{\left( x+ \bar{x}_t\right) ^2}{2\left\{ \varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)+Lm_{1,1/\Lambda _\varepsilon }(t)^2/a^2\right\} }\right] , \end{aligned}$$
(7.40)

where \(K_t,\bar{x}_t\) are given by the formulas

$$\begin{aligned} K_t= & {} K\exp [L/2]\left( \frac{L}{a^2\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)+Lm_{1,1/\Lambda _\varepsilon }(t)^2}\right) ^{1/2} , \\ \bar{x}_t= & {} m_{2,1/\Lambda _\varepsilon }(t)-m_{1,1/\Lambda _\varepsilon }(t)y_0+m_{1,1/\Lambda _\varepsilon }(t)L/a . \nonumber \end{aligned}$$
(7.41)

We can use (7.35) to obtain a lower bound on \(c_{X_t}(\cdot )\). Thus, we have that

$$\begin{aligned} c_{X_t}(x) \ge \left\{ 1-\exp \left[ -\frac{x^2}{C_1\varepsilon T}\right] \right\} \int _{\gamma x}^\infty G_\varepsilon (x,y,0,t)c_{X_0}(y) \ \hbox {d}y . \end{aligned}$$
(7.42)

Assuming x sufficiently large so that \(\gamma x\ge y_0\) then from (7.39) we see that

$$\begin{aligned}&\int _{\gamma x}^\infty G_\varepsilon (x,y,0,t)c_{X_0}(y) \ \hbox {d}y =K_t\exp \left[ -\frac{\left( x+ \bar{x}_t\right) ^2}{2\left\{ \varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)+Lm_{1,1/\Lambda _\varepsilon }(t)^2/a^2\right\} }\right] \nonumber \\&\quad - K\int _{-\infty }^{\gamma x} G_{\varepsilon }(x,y,0,t)\exp [-a(y-y_0)-\{a(y-y_0)\}^2/2L] \ \hbox {d}y . \end{aligned}$$
(7.43)

If \(\gamma >0\) is sufficiently small, then the second term on the RHS of (7.43) is much smaller than the first term. Hence, we conclude that there exist positive constants \(x_T,M_T\) such that for \(0<t\le T, \ x\ge x_T\),

$$\begin{aligned} c_{X_t}(x) \ge \left[ 1-\exp [-x^2/M_T\right] K_t\exp \left[ -\frac{\left( x+ \bar{x}_t\right) ^2}{2\left\{ \varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)+Lm_{1,1/\Lambda _\varepsilon }(t)^2/a^2\right\} }\right] . \end{aligned}$$
(7.44)

We can use (7.40), (7.44) to obtain an upper bound for \(\beta _{X_t}(x)\) when \(0<t\le T, \ x\ge x_T\). In fact, from the formula (2.35) we immediately conclude that

$$\begin{aligned} \beta _{X_t}(x) \le \left[ 1+\exp [-x^2/M_T\right] ^2\left[ 1-\exp [-x^2/M_T\right] ^{-2} \beta _Y(x) , \quad \mathrm{for \ } x\ge x_T, \end{aligned}$$
(7.45)

where \(\beta _Y(\cdot )\) is the beta function for a Gaussian variable Y with mean m and variance \(\sigma ^2\) given by

$$\begin{aligned} m =-\bar{x}_t , \quad \sigma ^2 =\varepsilon \sigma _{1/\Lambda _\varepsilon }^2(t)+Lm_{1,1/\Lambda _\varepsilon }(t)^2/a^2 . \end{aligned}$$
(7.46)

Now from Lemma 2.1 we see that if Y is Gaussian with mean m and variance \(\sigma ^2\), then there exists a universal constant C such that

$$\begin{aligned} \beta _Y(x) \le 1-\frac{\sigma ^2}{2(x-m)^2} \quad \mathrm{for \ } x\ge C\sigma +m. \end{aligned}$$
(7.47)

The result follows from (7.45), (7.47) upon choosing \(x_T\) sufficiently large. \(\square \)

Remark 4

The inequality (7.35) was easy to obtain from the explicit representation for the stochastic process \(Y_\varepsilon (s), \ 0\le s\le T,\) of (4.8) conditioned on \(Y_\varepsilon (0)=y, \ Y_\varepsilon (T)=x\), when the drift \(b(\cdot ,\cdot )\) is linear. It is much more difficult to obtain estimates on probabilities for the conditioned process in the case of nonlinear \(b(\cdot ,\cdot )\). Theorem 1.2 of Conlon and Guha (2014) proves a result for the cdf of \(Y_\varepsilon (t), \ 0<t<T,\) in the case of \(b(\cdot ,\cdot )\) satisfying the uniform Lipschitz condition (4.1), which is analagous to (7.35).

Proof of log-concavity of the function \(h_\varepsilon \). We first assume that the initial condition random variable \(X_0\) for (1.7) satisfies (7.15). Then by standard regularity theorems (Friedman 1964) for solutions to (1.7), the function \(u_\varepsilon \) of (7.13) is continuous on the closed set \(\{(x,t) \ : \ x\ge 0, t\ge 0\}\). Furthermore, Lemma 7.5 implies that for any \(T>0\) there exists \(x_T>0\) such that \(u_\varepsilon (x,t)>0\) for \(0\le t\le T, \ x\ge x_T\). In addition, (7.14) implies that \(u_\varepsilon (0,t)>0, \ 0\le t\le T,\) and (7.15) that \(u_\varepsilon (x,0)>0, \ x\ge 0\). Since \(u_\varepsilon \) is a classical solution to (7.13), the maximum principle (Protter and Weinberger 1984) implies that \(u_\varepsilon (x,t)>0\) for \(0\le t\le T, \ 0\le x\le x_T\). We have therefore proven that the function \(h_\varepsilon (\cdot ,t)\) is log-concave for \(0 \le t\le T\) when the initial data random variable \(X_0\) satisfies (7.15). The log-concavity of \(h_\varepsilon (\cdot ,t), \ t>0,\) for general log-concave initial data random variable \(X_0\) now follows from Lemma 7.4. \(\square \)

Remark 5

The main difficulty in implementing Korevaar’s argument is to show that the solution of the PDE is log-concave on the boundary of the region. We accomplished this here by taking advantage of the fact that the full line Green’s function is Gaussian when the drift \(b(\cdot ,\cdot )\) for (4.2) is linear. In the case of nonlinear \(b(\cdot ,\cdot )\), it is not possible to argue in this way. An alternative approach is to use Korevaar’s observation (Korevaar 1983) that a Dirichlet boundary condition implies log-concavity close to the boundary on account of the Hopf maximum principle (Protter and Weinberger 1984). Some log-concavity theorems for nonlinear \(b(\cdot ,\cdot )\) are proved in the appendix of Conlon and Guha (2014) using this method.