# The Cauchy-Dirichlet Problem for a Class of Linear Parabolic Differential Equations with Unbounded Coefficients in an Unbounded Domain

Authors: Gerardo Rubio
Journal: International Journal of Stochastic Analysis
 (2011)

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## Abstract

We consider the Cauchy-Dirichlet problem in[0,∞)×D for a class of linear parabolic partial differential equations. We assume that D⊂ℝd is an
unbounded, open, connected set with regular boundary. Our hypotheses are unbounded and locally Lipschitz coefficients, not necessarily differentiable, with continuous data and local uniform ellipticity. We construct a classical solution to the nonhomogeneous Cauchy-Dirichlet problem using stochastic differential equations and parabolic differential equations in bounded domains.


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## Body

## 1. Introduction

In this paper, we study the existence and uniqueness of a classical solution to the Cauchy-Dirichlet problem for a linear parabolic differential equation in a general unbounded domain. Letℒ be the differential operator (1.1)L[u](t,x):=∑i,j=1daij(t,x)Diju(t,x)+∑i=1dbi(t,x)Diu(t,x),
where {aij}=a=σσ′, Di=∂/∂xi, and Dij=∂2/∂xi∂xj. The Cauchy-Dirichlet problem is(1.2)-ut(t,x)+L[u](t,x)+c(t,x)u(t,x)=-f(t,x),(t,x)∈(0,∞)×D,u(0,x)=h(x),x∈D,u(t,x)=g(t,x),(t,x)∈(0,∞)×∂D,
where D⊂ℝd is an unbounded, open, connected set with regular boundary.In the case of bounded domains, the Cauchy-Dirichlet problem is well understood (see [1, 2] for a detailed description of this problem). Moreover, when the domain is unbounded and the coefficients are bounded, the existence of a classical solution to (1.2) is well known. For a survey of this theory see [3, 4] where the problem is studied with analytical methods and [5] for a probabilistic approach.In the last years, parabolic equations with unbounded coefficients in unbounded domains have been studied in great detail. For the particular case whenD=ℝd, there exist many papers in which the existence, uniqueness, and regularity of the solution is studied under different hypotheses on the coefficients; see for example, [6–17].In the case of general unbounded domains, Fornaro et al. in [18] studied the homogeneous, autonomous Cauchy-Dirichlet problem. They proved, using analytical methods in semigroups, the existence and uniqueness of a solution to the Cauchy-Dirichlet problem when the coefficients are locally C1,α, with aij bounded, b and c functions with a Lyapunov type growth; that is, there exists a function φ∈C1,2((0,T)×ℝd) such that (1.3)lim⁡|x|→∞inf⁡0≤t≤Tφ(t,x)=∞
and for some λ>0,(1.4)sup⁡[0,T]×Rd{(-∂∂t+L)φ(t,x)-λφ(t,x)}<∞.
It is also assumed that D has a C2 boundary. Schauder-type estimates were obtained for the gradient of the solution in terms of the data. Bertoldi and Fornaro in [19] obtained analogous results for the Cauchy-Neumann problem for an unbounded convex domain. Later, in [20] Bertoldi et al. generalized the method to nonconvex sets with C2 boundary. They studied the existence, uniqueness, and gradient estimates for the Cauchy-Neumann problem. For a survey of this results, see [21].Using the theory of semigroups, Da Prato and Lunardi studied, in [22, 23], the realization of the elliptic operator 𝒜=(1/2)Δ-〈DU,D·〉, in the functions spaces L2(D), L∞(D) and Cb(D), when U is an unbounded convex function defined in a convex set D. They proved existence and uniqueness for the elliptic and parabolic equations associated with 𝒜 and studied the regularity of the semigroup generated by 𝒜. Geissert et al. in [24], made a similar approach for the Ornstein-Uhlenbeck operator.In the paper of Hieber et al. [25], the existence and uniqueness of a classical solution for the autonomous, nonhomogeneous Cauchy-Dirichlet and Cauchy-Neumann problems is proved. The domain is considered to be an exterior domain with C3 boundary. The coefficients are assumed to be C3,α continuous functions with Lyapunov type growth. The continuity properties of the semigroup generated by the solution of the parabolic problem are studied in the spaces Cb(D¯) and Lp(D).In all the papers cited above, the uniformly elliptic condition is assumed; that is, there existsλ>0 such that for all (t,x)∈[0,∞)×D¯ we have that ∑aij(t,x)ηiηj≥λ∥η∥2 for all η∈ℝd.In this paper, we prove the existence and uniqueness of a classical solution to (1.2), when the coefficients are locally Lipschitz continuous in x and locally Hölder continuous in t, aij has a quadratic growth, bi has linear growth, and c is bounded from above. We allow f, g, and h to have a polynomial growth of any order. We also consider the elliptic condition to be local; that is, for any [0,T]×A⊂[0,∞)×D¯, there exists λ(T,A) such that ∑aij(t,x)ηiηj≥λ(T,A)∥η∥2 for all t∈[0,T], x∈A and η∈ℝd. We assume that D is an unbounded, connected, open set with regular boundary (see [1] Chapter III, Section 4, for a definition of regular boundary). Furthermore, we prove that the solution is locally Hölder continuous up to the second space derivative and the first time derivative.Our approach is using stochastic differential equations and parabolic differential equations in bounded domains. For proving existence, many analytical methods construct the solution by solving the problem in nested bounded domains that approximate the domainD. In these cases, the convergence of the approximating solutions is always a very difficult task. Unlike these methods, first we propose, as a solution to (1.2), a functional of the solution to a SDE, (1.5)v(t,x)=Ex[∫0t∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds]+Ex[e∫0tc(t-r,X(r))drh(X(t))1τD≥t]+Ex[e∫0τDc(t-r,X(r))drg(t-τD,X(τD))1τD<t],
where (1.6)dX(s)=b(t-s,X(s))ds+σ(t-s,X(s))dW(s),X(0)=x,τD:=inf⁡⁡{s>0∣X(s)∉D¯}.
Using the continuity of the paths of the SDE, we prove that this function is continuous in [0,∞)×D¯. Then, using the theory of parabolic equations in bounded domains, we study locally the regularity of the function v and prove that it is a C1,2 function. Finally, with some standard arguments, we prove that it solves the Cauchy-Dirichlet problem. This kind of idea has been used for several partial differential problems (see [5, 26, 27]).In Section2, we introduce the notation and the hypotheses used throughout the paper. Section 3 presents the main result. In this section we prove that if the function v is smooth, then it has to be the solution to the Cauchy-Dirichlet problem. Section 4 is devoted to prove the required differentiability for the candidate function. Finally, in Section 5, the reader will find some of the results used in the proof of our main theorem.


## 2. Preliminaries and Notation

In this section, we present the hypotheses and the notation used in this paper.


### 2.1. Domain

LetD⊂ℝd be an unbounded, open, connected set with boundary ∂D and closure D¯. We assume that D has a regular boundary; that is, for any x∈∂D, x is a regular point (see [1] Chapter III, Section 4 or [28] Chapter 2, Section 4, for a detailed discussion of regular points). We denote the hypotheses on D by H0.


### 2.2. Stochastic Differential Equation

Let(Ω,ℱ,ℙ,{ℱs}s≥0) be a complete filtered probability space and let {W}={Wi}i=1d be a d-dimensional brownian motion defined in it. For t≥0 and x∈D¯, consider the stochastic differential equation(2.1)dX(s)=b(t-s,X(s))ds+σ(t-s,X(s))dW(s),X(0)=x,
where b={bi}i=1d and σ={σij}i,j=1d. Although this process is the natural one for solving equation (1.2), it does not posses many good properties. The continuity of the flow process does not imply the continuity with respect to t. Furthermore, although this process is a strong Markov process, it is not homogeneous in time, a very useful property for proving the results in this paper.To overcome these difficulties, we augment the dimension considering the following process(2.2)dξ(s)=-ds,ξ(0)=t.
Then the process {ξ(s),X(s)} is solution to(2.3)dξ(s)=-ds,dX(s)=b(ξ(s),X(s))ds+σ(ξ(s),X(s))dW(s),
with (ξ(0),X(0))=(t,x). Throughout this paper we will use both processes, X(s) and (ξ(s),X(s)), in order to simplify the exposition. For the expectation, we use the notation(2.4)Ex[⋅]:=E[⋅∣X(0)=x]
when considering the process X defined as in (2.1) and the notation(2.5)Et,x[⋅]:=E[⋅∣(ξ(0),X(0))=(t,x)]
when working with the joint process (ξ,X) defined in (2.3).We need to define the following stopping times(2.6)τD:=inf⁡⁡{s>0∣X(s)∉D¯},τ:=τD∧t.Remark 2.1.
Observe thatτ is the exit time of the process (ξ(s),X(s)) from the set [0,∞)×D¯, that is,
(2.7)τ=inf⁡⁡{s>0∣(ξ(s),X(s))∉[0,∞)×D¯}.
We cannot guarantee that the process X(s) leaves the set D in a finite time however, the process ξ(s) reaches the boundary s=0 at time t. Thus, the joint process (ξ(s),X(s)) leaves the set [0,∞)×D¯ in a bounded time.We assume the following hypotheses on the coefficientsb and σ. We denote them by H1. The matrix norm considered is ∥σ∥2:=trσσ′=∑i,jσij2.H1.Let(2.8)σ:R×Rd⟶M(Rd×Rd),b:R×Rd⟶Rd
be continuous functions such that  (1)
Continuity. Let λ∈(0,1). For all T>0, n≥1, there exists L1(T,n) such that
(2.9)‖σ(r,x)-σ(s,y)‖2+‖b(r,x)-b(s,y)‖2≤L1(T,n)2(|r-s|2λ+‖x-y‖2),
for all |r|,|s|≤T, ∥x∥≤n, ∥y∥≤n.(2)
Linear-Growth. For each T>0, there exists a constant K1(T) such that
(2.10)‖σ(r,x)‖2+‖b(r,x)‖2≤K1(T)2(1+‖x‖2),
for all |r|≤T, x∈ℝd.(3)
Local Ellipticity. Let A⊂D¯ be any bounded, open, connected set and T>0. There exists λ(T,A)>0 such that, for all (r,x)∈[0,T]×A¯ and η∈ℝd,
(2.11)∑i,jaij(r,x)ηiηj≥λ(T,A)‖η‖2,
where {aij}=a=σσ′.Remark 2.2.
Observe that the local ellipticity is only assumed on[0,∞)×D¯. This condition is used to prove the existence of a classical solution to (1.2) and so is only needed in that set. The local Lipschitz condition and the linear growth are assumed on ℝ×ℝd to ensure the existence of a strong solution to (2.3) for s∈[0,∞).Remark 2.3.
If we assume, in the more natural case, that(2.12)σ:[0,∞)×Rd⟶M(Rd×Rd),b:[0,∞)×Rd⟶Rd
are continuous functions satisfying the hypotheses in H1 restricted to the set [0,∞)×ℝd, then we can extend them to be defined for negative values of r as follows: let b̂ and σ̂ be defined as
(2.13)b̂(r,x)={b(r,x)ifr≥0,b(0,x)ifr<0,σ̂(r,x)={σ(r,x)ifr≥0,σ(0,x)ifr<0.
It is easy to see that these functions satisfy H1 with the same constants L1 and K1.Remark 2.4.
It follows, from the nondegeneracy (the local ellipticity) of the processX(s), the regular boundary of the set D and Lemma 4.2, Chapter 2 in [28], that, for any t,x∈[0,∞)×D¯,
(2.14)Pt,x[τ=τ′]=1,
where τ′:=inf{s>0∣(ξ(s),X(s))∉(0,∞)×D} (see Remark 2.1).The next proposition presents some of the properties of the process(ξ,X) required in this work.Proposition 2.5.
As a consequence ofH1, (ξ,X) has the following properties.  (i)
For all(t,x)∈[0,∞)×ℝd, there exists a unique strong solution to (2.3).(ii)
The process{ξ(s),X(s)}s≥0 is a strong homogeneous Markov process.(iii)
The process{ξ(s),X(s)}s≥0 does not explode in finite time a.s..(iv)
For allx∈ℝd, T>0, and r≥1,
(2.15)Ex[sup⁡0≤s≤T‖X(s)‖2r]≤C(T,K1,r)(1+‖x‖2r).Proof.
See(i)
Theorem 1.1 ChapterV in [26] or Theorem 3.2, Chapter 6 in [29].(ii)
Theorem 4.6, Chapter 5 in [27] or Proposition 3.15, Chapter 6 in [29].(iii)
Theorem10.2.2 in [30].(iv)
Theorem 2.3, Chapter 5 in [31] or Corollary 3.3, Chapter 6 in [29].


### 2.3. The Cauchy-Dirichlet Problem

Consider the following differential operator:(2.16)L[u](t,x):=∑i,j=1daij(t,x)Diju(t,x)+∑i=1dbi(t,x)Diu(t,x),
where Di=∂/∂xi, Dij=∂2/∂xi∂xj, and {aij}i,j=1d=a=σσ′. For the rest of the paper, we assume that the coefficients of ℒ satisfy H1.The Cauchy-Dirichlet problem for a linear parabolic equation is(2.17)-ut(t,x)+L[u](t,x)+c(t,x)u(t,x)=-f(t,x),(t,x)∈(0,∞)×D,u(0,x)=h(x),x∈D,u(t,x)=g(t,x),(t,x)∈(0,∞)×∂D.
We assume the following hypotheses for the functions c, f, h, and g. We denote them by H2.H2.
(1)
Let(2.18)c:[0,∞)×D¯⟶R,f:[0,∞)×D¯⟶R
be continuous functions such that
(i)
Continuity. Let λ∈(0,1). For all T>0, n≥1, there exists a constant L2(T,n) such that
(2.19)‖f(r,x)-f(s,y)‖2+‖c(r,x)-c(s,y)‖2≤L2(T,n)2(|r-s|2λ+‖x-y‖2),
for all 0≤s, r≤T, x,y∈D¯ with ∥x∥≤n, ∥y∥≤n.(ii)
Growth. There exists c0≥0 such that
(2.20)c(r,x)≤c0∀(r,x)∈[0,∞)×D¯.
There exists k>0, such that for all T>0, a constant K2(T) exists such that
(2.21)|f(r,x)|≤K2(T)(1+‖x‖k),
for all 0≤r≤T, x∈D¯(2)
Let(2.22)h:D¯⟶R,g:[0,∞)×∂D⟶R
be continuous functions such that
(i)
Growth. There exists k>0, such that, for all T>0, there exists a constant K3(T) such that
(2.23)|h(x)|+|g(r,x)|≤K3(T)(1+‖x‖k)
for all (r,x)∈[0,T]×D¯.(ii)
Consistency. There exists consistency in the intersection of the space and the time boundaries, that is,
(2.24)h(x)=g(0,x)
for x∈∂D.


### 2.4. Additional Notation

Ifμ is a locally Lipschitz function defined in some set R, then, for any bounded open set A for which A¯⊂R, we denote, by Kμ(A) and Lμ(A), the constants (2.25)Kμ(A):=sup⁡x∈A‖μ(x)‖<∞,Lμ(A):=sup⁡x,y∈A,x≠y‖μ(x)-μ(y)‖‖x-y‖<∞.
If ν:[0,∞)→ℝd, then, for all T>0,(2.26)‖ν‖T:=sup⁡0≤s≤T‖ν(s)‖.
The space Cloc1,2,λ((0,∞)×D) is the space of all functions such that they and all their derivatives up to the second order in x and first order in t, are locally Hölder continuous of order λ.


## 2.1. Domain

LetD⊂ℝd be an unbounded, open, connected set with boundary ∂D and closure D¯. We assume that D has a regular boundary; that is, for any x∈∂D, x is a regular point (see [1] Chapter III, Section 4 or [28] Chapter 2, Section 4, for a detailed discussion of regular points). We denote the hypotheses on D by H0.


## 2.2. Stochastic Differential Equation

Let(Ω,ℱ,ℙ,{ℱs}s≥0) be a complete filtered probability space and let {W}={Wi}i=1d be a d-dimensional brownian motion defined in it. For t≥0 and x∈D¯, consider the stochastic differential equation(2.1)dX(s)=b(t-s,X(s))ds+σ(t-s,X(s))dW(s),X(0)=x,
where b={bi}i=1d and σ={σij}i,j=1d. Although this process is the natural one for solving equation (1.2), it does not posses many good properties. The continuity of the flow process does not imply the continuity with respect to t. Furthermore, although this process is a strong Markov process, it is not homogeneous in time, a very useful property for proving the results in this paper.To overcome these difficulties, we augment the dimension considering the following process(2.2)dξ(s)=-ds,ξ(0)=t.
Then the process {ξ(s),X(s)} is solution to(2.3)dξ(s)=-ds,dX(s)=b(ξ(s),X(s))ds+σ(ξ(s),X(s))dW(s),
with (ξ(0),X(0))=(t,x). Throughout this paper we will use both processes, X(s) and (ξ(s),X(s)), in order to simplify the exposition. For the expectation, we use the notation(2.4)Ex[⋅]:=E[⋅∣X(0)=x]
when considering the process X defined as in (2.1) and the notation(2.5)Et,x[⋅]:=E[⋅∣(ξ(0),X(0))=(t,x)]
when working with the joint process (ξ,X) defined in (2.3).We need to define the following stopping times(2.6)τD:=inf⁡⁡{s>0∣X(s)∉D¯},τ:=τD∧t.Remark 2.1.
Observe thatτ is the exit time of the process (ξ(s),X(s)) from the set [0,∞)×D¯, that is,
(2.7)τ=inf⁡⁡{s>0∣(ξ(s),X(s))∉[0,∞)×D¯}.
We cannot guarantee that the process X(s) leaves the set D in a finite time however, the process ξ(s) reaches the boundary s=0 at time t. Thus, the joint process (ξ(s),X(s)) leaves the set [0,∞)×D¯ in a bounded time.We assume the following hypotheses on the coefficientsb and σ. We denote them by H1. The matrix norm considered is ∥σ∥2:=trσσ′=∑i,jσij2.H1.Let(2.8)σ:R×Rd⟶M(Rd×Rd),b:R×Rd⟶Rd
be continuous functions such that  (1)
Continuity. Let λ∈(0,1). For all T>0, n≥1, there exists L1(T,n) such that
(2.9)‖σ(r,x)-σ(s,y)‖2+‖b(r,x)-b(s,y)‖2≤L1(T,n)2(|r-s|2λ+‖x-y‖2),
for all |r|,|s|≤T, ∥x∥≤n, ∥y∥≤n.(2)
Linear-Growth. For each T>0, there exists a constant K1(T) such that
(2.10)‖σ(r,x)‖2+‖b(r,x)‖2≤K1(T)2(1+‖x‖2),
for all |r|≤T, x∈ℝd.(3)
Local Ellipticity. Let A⊂D¯ be any bounded, open, connected set and T>0. There exists λ(T,A)>0 such that, for all (r,x)∈[0,T]×A¯ and η∈ℝd,
(2.11)∑i,jaij(r,x)ηiηj≥λ(T,A)‖η‖2,
where {aij}=a=σσ′.Remark 2.2.
Observe that the local ellipticity is only assumed on[0,∞)×D¯. This condition is used to prove the existence of a classical solution to (1.2) and so is only needed in that set. The local Lipschitz condition and the linear growth are assumed on ℝ×ℝd to ensure the existence of a strong solution to (2.3) for s∈[0,∞).Remark 2.3.
If we assume, in the more natural case, that(2.12)σ:[0,∞)×Rd⟶M(Rd×Rd),b:[0,∞)×Rd⟶Rd
are continuous functions satisfying the hypotheses in H1 restricted to the set [0,∞)×ℝd, then we can extend them to be defined for negative values of r as follows: let b̂ and σ̂ be defined as
(2.13)b̂(r,x)={b(r,x)ifr≥0,b(0,x)ifr<0,σ̂(r,x)={σ(r,x)ifr≥0,σ(0,x)ifr<0.
It is easy to see that these functions satisfy H1 with the same constants L1 and K1.Remark 2.4.
It follows, from the nondegeneracy (the local ellipticity) of the processX(s), the regular boundary of the set D and Lemma 4.2, Chapter 2 in [28], that, for any t,x∈[0,∞)×D¯,
(2.14)Pt,x[τ=τ′]=1,
where τ′:=inf{s>0∣(ξ(s),X(s))∉(0,∞)×D} (see Remark 2.1).The next proposition presents some of the properties of the process(ξ,X) required in this work.Proposition 2.5.
As a consequence ofH1, (ξ,X) has the following properties.  (i)
For all(t,x)∈[0,∞)×ℝd, there exists a unique strong solution to (2.3).(ii)
The process{ξ(s),X(s)}s≥0 is a strong homogeneous Markov process.(iii)
The process{ξ(s),X(s)}s≥0 does not explode in finite time a.s..(iv)
For allx∈ℝd, T>0, and r≥1,
(2.15)Ex[sup⁡0≤s≤T‖X(s)‖2r]≤C(T,K1,r)(1+‖x‖2r).Proof.
See(i)
Theorem 1.1 ChapterV in [26] or Theorem 3.2, Chapter 6 in [29].(ii)
Theorem 4.6, Chapter 5 in [27] or Proposition 3.15, Chapter 6 in [29].(iii)
Theorem10.2.2 in [30].(iv)
Theorem 2.3, Chapter 5 in [31] or Corollary 3.3, Chapter 6 in [29].


## 2.3. The Cauchy-Dirichlet Problem

Consider the following differential operator:(2.16)L[u](t,x):=∑i,j=1daij(t,x)Diju(t,x)+∑i=1dbi(t,x)Diu(t,x),
where Di=∂/∂xi, Dij=∂2/∂xi∂xj, and {aij}i,j=1d=a=σσ′. For the rest of the paper, we assume that the coefficients of ℒ satisfy H1.The Cauchy-Dirichlet problem for a linear parabolic equation is(2.17)-ut(t,x)+L[u](t,x)+c(t,x)u(t,x)=-f(t,x),(t,x)∈(0,∞)×D,u(0,x)=h(x),x∈D,u(t,x)=g(t,x),(t,x)∈(0,∞)×∂D.
We assume the following hypotheses for the functions c, f, h, and g. We denote them by H2.H2.
(1)
Let(2.18)c:[0,∞)×D¯⟶R,f:[0,∞)×D¯⟶R
be continuous functions such that
(i)
Continuity. Let λ∈(0,1). For all T>0, n≥1, there exists a constant L2(T,n) such that
(2.19)‖f(r,x)-f(s,y)‖2+‖c(r,x)-c(s,y)‖2≤L2(T,n)2(|r-s|2λ+‖x-y‖2),
for all 0≤s, r≤T, x,y∈D¯ with ∥x∥≤n, ∥y∥≤n.(ii)
Growth. There exists c0≥0 such that
(2.20)c(r,x)≤c0∀(r,x)∈[0,∞)×D¯.
There exists k>0, such that for all T>0, a constant K2(T) exists such that
(2.21)|f(r,x)|≤K2(T)(1+‖x‖k),
for all 0≤r≤T, x∈D¯(2)
Let(2.22)h:D¯⟶R,g:[0,∞)×∂D⟶R
be continuous functions such that
(i)
Growth. There exists k>0, such that, for all T>0, there exists a constant K3(T) such that
(2.23)|h(x)|+|g(r,x)|≤K3(T)(1+‖x‖k)
for all (r,x)∈[0,T]×D¯.(ii)
Consistency. There exists consistency in the intersection of the space and the time boundaries, that is,
(2.24)h(x)=g(0,x)
for x∈∂D.


## 2.4. Additional Notation

Ifμ is a locally Lipschitz function defined in some set R, then, for any bounded open set A for which A¯⊂R, we denote, by Kμ(A) and Lμ(A), the constants (2.25)Kμ(A):=sup⁡x∈A‖μ(x)‖<∞,Lμ(A):=sup⁡x,y∈A,x≠y‖μ(x)-μ(y)‖‖x-y‖<∞.
If ν:[0,∞)→ℝd, then, for all T>0,(2.26)‖ν‖T:=sup⁡0≤s≤T‖ν(s)‖.
The space Cloc1,2,λ((0,∞)×D) is the space of all functions such that they and all their derivatives up to the second order in x and first order in t, are locally Hölder continuous of order λ.


## 3. Main Result

In this section, we present the main result of this work and some parts of the proof.Theorem 3.1.
AssumeH0,  H1, and H2. Then, there exists a unique solution u∈C([0,∞)×D¯)∩Cloc1,2,λ((0,∞)×D) to (2.17). The solution has the representation
(3.1)u(t,x)=Ex[∫0t∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds]+Ex[e∫0tc(t-r,X(r))drh(X(t))1τD≥t]+Ex[e∫0τDc(t-r,X(r))drg(t-τD,X(τD))1τD<t],
where X is the solution to the stochastic differential equation
(3.2)dX(s)=b(t-s,X(s))ds+σ(t-s,X(s))dW(s),X(0)=x,τD:=inf⁡⁡{s>0∣X(s)∉D¯}.
Furthermore, for all T>0,
(3.3)sup⁡0≤t≤T|u(t,x)|≤C(T,c0,K1,K2,K3,k)(1+‖x‖k),x∈D¯,
where c0, K1, K2, K3, and k are the constants defined in H1 and H2.The proof of this theorem is given by several lemmas. The method we will use has the following steps: first we define a functional of the processX as a candidate solution. Let v:[0,∞)×D¯→ℝ be defined as(3.4)v(t,x):=Ex[∫0t∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds]+Ex[e∫0tc(t-r,X(r))drh(X(t))1τD≥t]+Ex[e∫0τDc(t-r,X(r))drg(t-τD,X(τD))1τD<t].
If v∈C([0,∞)×D¯)∩C1,2((0,∞)×D), then there exist some standard arguments (see [27, chapter 4]) to prove that v is the unique solution to (2.17). The rest of this section is devoted to proving Theorem 3.1 in the case when v is a “regular” function. The proof is divided into two lemmas: the first one proves that if v∈C([0,∞)×D¯)∩C1,2((0,∞)×D), then v is a solution to equation. The second one proves that in case of existence of a classical solution, u, to problem (2.17), then it is unique and has the form given by v in (3.4). The regularity of v is proved in Section 4 below.The next proposition gives an extension of the boundary data to all the space[0,∞)×ℝd. This extension is given to simplify the notation and is required in the proofs of Lemmas 4.2 and 4.3.Proposition 3.2.
AssumeH2. Then, there exists a continuous function G:[0,∞)×ℝd→ℝ such that
(3.5)G(t,x)=g(t,x),(t,x)∈[0,∞)×∂D,G(0,x)=h(x),x∈D¯.Proof.
Thanks to the consistency condition inH2 and the continuity of g and h, we can extend by Tietze's extension theorem (see [32, Section 2.6]) the functions g, h from the closed set {0}×D¯∪[0,∞)×∂D to a continuous function G defined in [0,∞)×ℝd.As a consequence of Proposition3.2, we can write v in (3.4) as follows:
(3.6)v(t,x)=Ex[∫0t∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds]+Ex[e∫0t∧τDc(t-r,X(r))drG(t-t∧τD,X(t∧τD))].
We are ready to prove both lemmas explained above.Lemma 3.3.
AssumeH0,  H1, and H2. Let v be defined as in (3.6) and assume that v∈C([0,∞)×D¯)∩C1,2((0,∞)×D). Then, v fulfils the following equation:
(3.7)-ut(t,x)+L[u](t,x)+c(t,x)u(t,x)=-f(t,x)(t,x)∈(0,∞)×D,u(t,x)=G(t,x)(t,x)∈∂((0,∞)×D).
Furthermore, for all T>0, there exists C such that
(3.8)sup⁡0≤t≤T|v(t,x)|≤C(T,c0,K1,K2,K3,k)(1+‖x‖k),x∈D¯,
where c0, K1, K2, K3, and k are the constants defined in H1 and H2.Proof.
Let0≤α≤t, then, following the same argument used to prove (4.80) in the proof of Theorem 4.4 in Section 4 we have that
(3.9)Ex[∫0t∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds+e∫0t∧τDc(t-r,X(r))drG(t-t∧τD,X(t∧τD))∣Fα]=∫0α∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds+e∫0α∧τDc(t-r,X(r))v(t-α∧τD,X(α∧τD)).
Because of H1 and H2, we have that the random variable inside the conditional expectation is integrable and so the left-hand side of (3.9) is a ℱα-martingale, for α∈[0,t]. Since v∈C1,2, we can apply Ito's formula to e∫0αcdrv() to get
(3.10)e∫0α∧τDc(t-r,X(r))drv(t-α∧τD,X(α∧τD))=v(t,x)+∫0α∧τDe∫0sc(t-r,X(r))dr(-vt+L[v]+cv)(t-s,X(s))ds+∫0α∧τDDv(t-s,X(s))⋅σ(t-s,X(s))dW(s).
It follows from the continuity of Dv, σ and X(·) that
(3.11)sup⁡0≤s≤α‖Dv(t-s,X(s))‖‖σ(t-s,X(s))‖
is a.s. finite and then
(3.12)∫0α∧τDDv(t-s,X(s))⋅σ(t-s,X(s))dW(s)
is a local martingale for 0≤α≤t. So, combining (3.9) and (3.10), we get that
(3.13)M(α):=∫0α∧τDe∫0sc(t-r,X(r))dr(-vt+L[v]+cv+f)(t-s,X(s))ds
is a continuous local martingale for α∈[0,t]. Since M is locally of bounded variation, then M(α)≡0. This implies that -vt+ℒ[v]+cv+f=0 for all (t,x)∈(0,∞)×D.
For the boundary condition, it follows from the regularity of the setD and the local ellipticity (see Remark 2.4) that
(3.14)Px[t∧τD=0]=1,for(t,x)∈∂([0,∞)×D¯).
From this, it is clear that the first addend of the right-hand side of (3.6) is zero. For the second addend, we get that the exponential term is equal to one and that 𝔼x[G(t-t∧τD,X(t∧τD))]=G(t,X(0))=G(t,x), and so we conclude that v satisfies the boundary condition.
The second statement of the theorem is proved with the same argument used to prove (4.9) and (4.42) in the proofs of Lemmas 4.2 and 4.3 in Section 4.The next Lemma proves the uniqueness of the solution.Lemma 3.4.
AssumeH0,  H1, and H2. Assume that there exists a classical solution u∈C([0,∞)×D¯)∩C1,2((0,∞)∩D) to equations
(3.15)-ut(t,x)+L[u](t,x)+c(t,x)u(t,x)=-f(t,x)(t,x)∈(0,∞)×D,u(t,x)=G(t,x)(t,x)∈∂((0,∞)×D),
such that, for all T>0, there exists C for which
(3.16)sup⁡0≤t≤T|u(t,x)|≤C(T)(1+‖x‖μ)
for some μ>0. Then, u has the following representation:
(3.17)u(t,x)=Ex[∫0t∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds]+Ex[e∫0t∧τDc(t-r,X(r))drG(t-t∧τD,X(t∧τD))],
and hence the solution is unique.Proof.
Consider, forα∈[0,t], the process
(3.18)e∫0α∧τDc(t-r,X(r))dru(t-α∧τD,X(α∧τD)).
Applying Ito's rule, we get
(3.19)e∫0α∧τDc(t-r,X(r))dru(t-α∧τD,X(α∧τD))=u(t,x)+∫0α∧τDe∫0sc(t-r,X(r))dr(-ut+L[u]+cu)(t-s,X(s))ds+∫0α∧τDDu(t-s,X(s))⋅σ(t-s,X(s))dW(s).
A similar argument as the one used in the proof of Lemma 3.3 shows that
(3.20)∫0α∧τDDu(t-s,X(s))⋅σ(t-s,X(s))dW(s)
is a local martingale. Due to (3.15), we conclude that
(3.21)M(α):=e∫0α∧τDc(t-r,X(r))dru(t-α∧τD,X(α∧τD))+∫0α∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds
is a local martingale for α∈[0,t]. Let {θn}n≥1 be a sequence of localization times for M(α); that is, θn↑∞ a.s. as n→∞ and M(α∧θn) is a martingale for all n≥1. Then, for all n≥1,
(3.22)u(t,x)=Ex[e∫0α∧τD∧θnc(t-r,X(r))dru(t-α∧τD∧θn,X(α∧τD∧θn))]+Ex[∫0α∧τD∧θne∫0sc(t-r,X(r))drf(t-s,X(s))ds].
Since 0≤α∧τD∧θn≤t, using estimate (3.16), we get
(3.23)e∫0α∧τD∧θnc(t-r,X(r))dr|u(t-α∧τD∧θn,X(α∧τD∧θn))|≤ec0tC(t)(1+‖X(α∧τD∧θn)‖μ)≤ec0tC(t)(1+sup⁡0≤s≤t‖X(s)‖μ),|∫0α∧τD∧θne∫0sc(t-r,X(r))drf(t-s,X(s))ds|≤∫0α∧τD∧θnec0sK2(t)(1+‖X(s)‖k)ds≤ec0ttK2(t)(1+sup⁡0≤s≤t‖X(s)‖k).
By (2.15) and the dominated convergence theorem, letting n→∞, we get
(3.24)u(t,x)=Ex[e∫0α∧τDc(t-r,X(r))dru(t-α∧τD,X(α∧τD))]+Ex[∫0α∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds].
Letting α↑t, a similar argument and the boundary condition proof that
(3.25)u(t,x)=Ex[e∫0t∧τDc(t-r,X(r))drG(t-t∧τD,X(t∧τD))]+Ex[∫0t∧τDe∫0sc(t-r,X(r))drf(t-s,X(s))ds],
and the proof is complete.


## 4. Regularity ofv

In this section, we prove thatv∈C([0,∞)×D¯)∩Cloc1,2,λ((0,∞)×D). First, we prove, using the continuity of the flow process X, that v is a continuous function in [0,∞)×D¯. Since we are only assuming the continuity of the coefficients, then the flow is not necessarily differentiable and so we can not prove the regularity of v in terms of the regularity of the flow. To prove that v∈C1,2, we show that v is the solution to a parabolic differential equation in a bounded domain, for which we have the existence of a classical solution and hence v∈C1,2.


### 4.1. Continuity ofv

Let(ξ,X) denote the solution to (2.3) and let G be defined as in Proposition 3.2, then v has the following form:(4.1)v(t,x)=Et,x[∫0t∧τDe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds]+Et,x[e∫0t∧τDc(ξ(r),X(r))drG(ξ(t∧τD),X(t∧τD))].
For simplicity, we write v=v1+v2, where(4.2)v1(t,x):=Et,x[∫0t∧τDe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds],(4.3)v2(t,x):=Et,x[e∫0t∧τDc(ξ(r),X(r))drG(ξ(t∧τD),X(t∧τD))].Theorem 4.1.
AssumeH0,  H1, and H2. Let v be defined as in (4.1). Then, v is continuous on [0,∞)×D¯.The proof of this theorem is divided into two lemmas.Lemma 4.2.
AssumeH0,  H1, and H2. Let v1 be defined as in (4.2). Then, v1 is continuous on [0,∞)×D¯.Proof.
First, we prove the continuity on(0,∞)×D. For that, let
(4.4)(tn,xn)⟶n→∞(t,x)
in (0,∞)×D and ϵ>0. We need to prove that there exists N(ϵ)∈ℕ such that, for all n≥N(ϵ)(4.5)|v1(tn,xn)-v1(t,x)|<ϵ.
Denote by (ξ,X) and (ξn,Xn) the solutions to (2.3) with initial conditions (t,x) and (tn,xn), respectively. To simplify the notation in the proof, let τ:=t∧τD and τn:=tn∧τDn denote their corresponding exit times from [0,∞)×D¯.
Letα>0, then there exists N1∈ℕ such that, for all n≥N1(4.6)‖(tn,xn)-(t,x)‖<α.
Observe that, for all n≥N1, we get
(4.7)τn≤tn≤t+α,τ≤t≤t+α.
Define the random variables Yn as
(4.8)Yn:=|∫0τne∫0sc(ξn(r),Xn(r))drf(ξn(s),Xn(s))ds-∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|.
The sequence {Yn}n≥N1 is uniformly integrable. Thanks to Theorem 4.2, in Chapter 5 of [33], it is sufficient to prove that supn𝔼[Yn2]<∞. So,
(4.9)E[Yn2]≤2E[|∫0τne∫0sc(ξn(r),Xn(r))drf(ξn(s),Xn(s))ds|2]+2E[|∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|2]≤2E[|∫0τnec0(t+α)K2(t+α)(1+‖Xn(s)‖k)ds|2]+2E[|∫0τec0(t+α)K2(t+α)(1+‖X(s)‖k)ds|2]≤2e2c0(t+α)K22(t+α)E[|∫0τn(1+sup⁡0≤r≤t+α‖Xn(r)‖k)ds|2]+2e2c0(t+α)K22(t+α)E[|∫0τ(1+sup⁡0≤r≤t+α‖X(r)‖k)ds|2]≤4e2c0(t+α)K22(t+α)(t+α)2(1+E[sup⁡0≤r≤t+α‖Xn(r)‖2k])+4e2c0(t+α)K22(t+α)(t+α)2(1+E[sup⁡0≤r≤t+α‖X(r)‖2k])≤C(1+K(1+‖xn‖2k))+C(1+K(1+‖x‖2k))≤C(1+K(1+(‖x‖+α)2k))+C(1+K(1+‖x‖2k))<∞,
where C=C(t,α,c0) and K=K(t,α,k). We use (4.6), (4.7), (2.15) and the polynomial growth of f.
LetM>0, 0<η<1, and β>0 and define the set(4.10)EM,n,η,β:={‖X‖t+α≤M}∩{‖Xn-X‖t+α≤η}∩{|τn-τ|≤β}.
Then,
(4.11)|v1(tn,xn)-v1(t,x)|≤∫ΩYndP=∫EM,n,η,βYndP+∫Ω∖EM,n,η,βYndP.
Since the sequence {Yn} is uniformly integrable, there exists δ(ϵ) such that for any E∈ℱ that satisfies ℙ[E]<δ, we have
(4.12)sup⁡n≥N1∫EYndP<ϵ2.
It follows, from Remark 2.4, Proposition 2.5, and Theorems 5.3 and 5.4 in Section 5, the existence of M and N2 such that
(4.13)P[Ω∖EM,n,η,β]<δ(ϵ)
for all n≥N2. Then for n≥N1∨N2, we get that
(4.14)|v1(tn,xn)-v1(t,x)|≤∫EM,n,η,βYndP+ϵ2.
For simplicity of notation, we write the set EM,n,η,β as E and define
(4.15)A:=[0,t+α]×[-M-1,M+1]d,Dt:=[0,t+α]×D¯
and let
(4.16)BA:=(-m,m)d+1
be an open set such that A⊂BA.
On the setE, for all n≥N1 and 0≤s≤t+α, it is satisfied that(4.17)(ξn(s),Xn(s)),(ξ(s),X(s))∈A.
We have that
(4.18)∫EYndP≤∫E∫0τn∧τ|e∫0sc(ξn(r),Xn(r))drf(ξn(s),Xn(s))-e∫0sc(ξ(r),X(r))drf(ξ(s),X(s))|dsdP(4.19)+∫E∫τn∧ττn∨τ(e∫0sc(ξn(r),Xn(r))dr|f(ξn(s),Xn(s))|1τn>τ+e∫0sc(ξ(r),X(r))dr|f(ξ(s),X(s))|1τn≤τ)dsdP.
We first analyse (4.19):
(4.20)(4.19)≤∫E∫τn∧ττn∨τec0(t+α)Kf(A∩Dt)(1τn≤τ+1τn>τ)dsdP=ec0(t+α)Kf(A∩Dt)∫E|τn-τ|dP≤ec0(t+α)Kf(A∩Dt)β.
For (4.18), we get
(4.21)(4.18)≤∫E∫0τn∧τe∫0sc(ξn(r),Xn(r))dr|f(ξn(s),Xn(s))-f(ξ(s),X(s))|dsdP(4.22)+∫E∫0τn∧τ|f(ξ(s),X(s))||e∫0sc(ξn(r),Xn(r))dr-e∫0sc(ξ(r),X(r))dr|dsdP.
Now,
(4.23)(4.21)≤∫Eec0(t+α)∫0τn∧τLf(BA)(|tn-t|λ+‖Xn(s)-X(s)‖)dsdP≤ec0(t+α)(t+α)Lf(BA)(|tn-t|λ+η).
For (4.22), we need the following bound:
(4.24)|e∫0sc(ξn(r),Xn(r))dr-e∫0sc(ξ(r),X(r))dr|=e∫0sc(ξ(r),X(r))dr|exp⁡{∫0s(c(ξn(r),Xn(r))-c(ξ(r),X(r)))dr}-1|≤ec0s(exp⁡{∫0s|c(ξn(r),Xn(r))-c(ξ(r),X(r))|dr}-1)≤ec0s(exp⁡{∫0sLc(BA)(|tn-t|λ+‖Xn(r)-X(r)‖)dr}-1)≤ec0s(exp⁡{Lc(BA)s(|tn-t|λ+η)}-1),
since |ex-1|≤e|x|-1. If we choose N3∈ℕ such that |tn-t|λ≤1/(2Lc(BA)(t+α)) for all n≥N3 and η≤1/(2Lc(BA)(t+α)), then we get by the mean value theorem that
(4.25)|e∫0sc(ξn(r),Xn(r))dr-e∫0sc(ξ(r),X(r))dr|≤ec0seLc(BA)s(|tn-t|λ+η).
Then,
(4.26)(4.22)≤Kf(A∩Dt)ec0(t+α)eLc(BA)(t+α)2(|tn-t|λ+η).
To summarize, we get with the above estimations that
(4.27)∫EYndP≤ec0(t+α)(t+α)Lf(BA)(|tn-t|λ+η)+Kf(A∩Dt)ec0(t+α)eLc(BA)(t+α)2(|tn-t|λ+η)+ec0(t+α)Kf(A∩Dt)β.
Hence, to prove continuity, we proceed as follows,  (i)
Letϵ>0 and 0<α≪1.(ii)
LetN1≥ℕ such that, for all n≥N1,
(4.28)∥(tn,xn)-(t,x)∥<α.(iii)
Letδ(ϵ)>0 fulfil the uniformly integrability condition (4.12).(iv)
TakeM>0 such that ℙ[∥X∥t+α>M]<δ(ϵ)/3.(v)
DefineA:=[0,t+α]×[-M-1,M+1]d and Dt:=[0,t+α]×D¯.(vi)
Let(4.29)η<min{1,12Lc(BA)(t+α),ϵ16ec0(t+α)(t+α)Lf(BA),ϵ16Kf(A∩Dt)ec0(t+α)eLc(BA)(t+α)2}.(vii)
ChooseN2∈ℕ such that ℙ[∥Xn-X∥t+α>η]≤δ(ϵ)/3 for all n≥N2(viii)
LetN3∈ℕ be such that
(4.30)|tn-t|λ<min{12Lc(BA)(t+α),ϵ16ec0(t+α)(t+α)Lf(BA),ϵ16Kf(A∩Dt)ec0(t+α)eLc(BA)(t+α)},
for all n≥N3.(ix)
Let(4.31)β<ϵ4ec0(t+α)Kf(A∩Dt)
and choose N4∈ℕ such that, for all n≥N4, ℙ[|τn-τ|>β]≤δ(ϵ)/3.
Thus, ifN=N1∨N2∨N3∨N4, then, for all n≥N,(4.32)|v1(tn,xn)-v1(t,x)|<ϵ.
Therefore, v1 is continuous in (0,∞)×D.
For the continuity at the boundary we make a similar argument. Let(tn,xn)→n→∞(t,x), where (tn,xn)∈(0,∞)×D and (t,x)∈∂((0,∞)×D), that is, either t=0 or x∈∂D. In both cases we get that τ=0 a.s. and so v1(t,x)=0. Then, we need to prove that(4.33)|v1(tn,xn)|⟶n→∞0.
Let 0<α≪1 and N1∈ℕ be such that
(4.34)‖(tn,xn)-(t,x)‖<α.
We get
(4.35)τn≤tn<t+α
for all n≥N1. For the continuity, we have
(4.36)|v1(tn,xn)|≤E[∫0τne∫0sc(ξn(r),Xn(r))dr|f(ξn(s),Xn(s))|ds]≤E[∫0τnec0sK2(t+α)(1+‖Xn(s)‖k)ds]≤ec0(t+α)K2(t+α)E[τn(1+sup⁡0≤r≤t+α‖Xn(r)‖k)]⟶n→∞0.
The convergence follows from the uniform integrability of
(4.37)τn(1+sup⁡0≤r≤t+α‖Xn(r)‖k)
and the fact that τn→n→∞ℙ0 (see Theorem 5.2 in Chapter 5 of [33] and Theorem 5.3 in Section 5). This completes the proof.Lemma 4.3.
AssumeH0,  H1, and H2. Let v2 be defined as in (4.3). Then, v2 is continuous on [0,∞)×D¯.Proof.
We use an analogous argument to the one in the proof of Lemma4.2. First, we prove the continuity in (0,∞)×D. Let
(4.38)(tn,xn)⟶n→∞(t,x),
with (tn,xn),(t,x)∈(0,∞)×D. Denote by (ξn,Xn) and (ξ,X) the solutions to (2.3) with initial conditions (tn,xn) and (t,x), respectively, and let τn:=tn∧τDn and τ:=t∧τD be their corresponding exit times from [0,∞)×D¯. Let 0<α≪1 and N1 be such that, for all n≥N1,
(4.39)∥(tn,xn)-(t,x)∥<α.
This implies that
(4.40)τn≤t+α,τ≤t+α.
First, we prove that the sequence of random variables
(4.41)Yn:=|e∫0τnc(ξn(r),Xn(r))drG(ξn(τn),Xn(τn))-e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))|
is uniformly integrable for all n≥N1. As in (4.9),
(4.42)E[Yn2]≤2E[|e∫0τnc(ξn(r),Xn(r))drG(ξn(τn),Xn(τn))|2]+2E[|e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))|2]≤2E[e2c0(t+α)K32(t+α)(1+‖Xn(τn)‖k)2]+2E[e2c0(t+α)K32(t+α)(1+‖X(τ)‖k)2]≤2e2c0(t+α)K32(t+α)E[(1+sup⁡0≤r≤t+α‖Xn(r)‖k)2]+2e2c0(t+α)K32(t+α)E[(1+sup⁡0≤r≤t+α‖X(r)‖k)2]≤4e2c0(t+α)K32(t+α)(1+E[sup⁡0≤r≤t+α‖Xn(r)‖2k])+4e2c0(t+α)K32(t+α)(1+E[sup⁡0≤r≤t+α‖X(r)‖2k])≤C(1+K(1+‖xn‖2k))+C(1+K(1+‖x‖2k))≤C(1+K(1+(‖x‖+α)2k))+C(1+K(1+‖x‖2k))<∞,
where C=C(t,α,c0) and K=K(t,α,k). We use (4.39), (4.40), (2.15), and the polynomial growth of G in ∂((0,∞)×D). As in Lemma 4.2, let ϵ>0, then there exists δ(ϵ)>0 such that
(4.43)sup⁡n≥N1∫EYndP<ϵ2
for all E∈ℱ, with ℙ[E]<δ(ϵ).
LetEM,n,η,β be defined as in (4.10), and choose M>0 and N2∈ℕ such that(4.44)P[Ω∖EM,n,η,β]<δ(ϵ),
for all n≥N2.
For simplicity of notation, denoteEM,n,η,β as E. Then,(4.45)|v2(tn,xn)-v2(t,x)|≤∫EYndP+∫Ω∖EYndP≤∫EYndP+ϵ2.
Let A, Dt, and BA be defined as in Lemma 4.2 (see (4.15) and (4.16)). Then, on the set E, we get that, for all n≥N1 and 0≤s≤t+α,
(4.46)(ξn(s),Xn(s)),(ξ(s),X(s))∈A.
So,
(4.47)∫EYndP≤∫Ee∫0τnc(ξn(r),Xn(r))dr×|G(ξn(τn),Xn(τn))-G(ξ(τ),X(τ))|dP(4.48)+∫E|G(ξ(τ),X(τ))||e∫0τnc(ξn(r),Xn(r))dr-e∫0τc(ξ(r),X(r))dr|dP.
We study each addend of the right-hand side separately:
(4.49)(4.47)≤ec0(t+α)∫E|G(tn-τn,Xn(τn))-G(tn-τn,X(τn))|dP(4.50)+ec0(t+α)∫E|G(tn-τn,X(τn))-G(t-τ,X(τ))|dP.
First, we get a bound for (4.49). Since G is continuous, then it is uniformly continuous on A. Then, for ϵ>0, there exists γ(c0,t,α,ϵ,M) such that
(4.51)|G(t1,x1)-G(t2,x2)|<ϵ8ec0(t+α)
for all (t1,x1),(t2,x2)∈A with ∥(t1,x1)-(t2,x2)∥<γ(c0,t,α,ϵ,M). On the set E, we have (tn-τn,Xn(τn)),(tn-τn,X(τn))∈A and
(4.52)‖(tn-τn,Xn(τn))-(tn-τn,X(τn))‖<η.
So, if we choose η<γ, then we get
(4.53)(4.49)<ϵ8.
Next, we study (4.50). Thanks to Theorem 5.3, we know that τn→n→∞a.sτ. This and the continuity of X(·) and G imply that
(4.54)G(tn-τn,X(τn))⟶n→∞a.s.G(t-τ,X(τ)).
On the set E, we have that (tn-τn,X(τn)),(t-τ,X(τ))∈A and so
(4.55)|G(tn-τn,X(τn))-G(t-τ,X(τ))|1E≤2KG(A).
By the dominated convergence theorem, there exists N3∈ℕ such that
(4.56)(4.50)<ϵ8
for all n≥N3.
To give a bound for (4.48), we observe that, on the set E,(4.57)|∫0τnc(ξn(r),Xn(r))dr-∫0τc(ξ(r),X(r))dr|≤∫0τn∧τ|c(ξn(r),Xn(r))-c(ξ(r),X(r))|dr+∫τn∧ττn∨τ(|c(ξn(r),Xn(r))|1τn≥τ+|c(ξ(r),X(r))|1τn<τ)dr≤∫0τn∧τLc(BA)(|tn-t|λ+‖Xn(r)-X(r)‖)dr+Kc(A∩Dt)|τn-τ|≤Lc(BA)(t+α)(|tn-t|λ+η)+Kc(A∩Dt)β.
Making a similar argument as the one made in (4.24) and (4.25), we get
(4.58)|e∫0τnc(ξn(r),Xn(r))dr-e∫0τc(ξ(r),X(r))dr|≤ec0(t+α)e[Lc(BA)(t+α)(|tn-t|λ+η)+Kc(A∩Dt)β]
if |tn-t|λ<1/3Lc(BA)(t+α), η<1/3Lc(BA)(t+α), and β<1/3Kc(A∩Dt). Then,
(4.59)(4.48)≤KG(A)ec0(t+α)e[Lc(BA)(t+α)(|tn-t|λ+η)+Kc(A∩Dt)β].
To summarize, we get with the above estimations that
(4.60)∫EYndP≤ϵ4+KG(A)ec0(t+α)e[Lc(BA)(t+α)(|tn-t|λ+η)+Kc(A∩Dt)β].
Hence, to prove continuity, we proceed as follows. (i)
Letϵ>0 and 0<α≪1.(ii)
LetN1∈ℕ be such that, for all n≥N1,
(4.61)‖(tn,xn)-(t,x)‖<α.(iii)
Letδ(ϵ)>0 fulfil the uniformly integrability condition (4.43).(iv)
TakeM>0 such that ℙ[∥X∥t+α>M]<δ(ϵ)/3.(v)
DefineA:=[0,t+α]×[-M-1,M+1]d and Dt:=[0,t+α]×D¯.(vi)
Let(4.62)η<min{1,γ(c0,t,α,ϵ,M),13Lc(BA)(t+α),ϵ12KG(A)ec0(t+α)eLc(BA)(t+α)}.(vii)
ChooseN2∈ℕ such that ℙ[∥Xn-X∥t+α>η]<δ(ϵ)/3 for all n≥N2(viii)
Let(4.63)β<min{13Kc(A∩Dt),ϵ12KG(A)ec0(t+α)eKc(A∩Dt)}.(ix)
ChooseN3∈ℕ such that ℙ[|τn-τ|>β]<δ(ϵ)/3 for all n≥N3.(x)
LetN4∈ℕ be such that
(4.64)|tn-t|λ<min{13Lc(BA)(t+α),ϵ12KG(A)ec0(t+α)eLc(BA)(t+α)},
for all n≥N4.(xi)
LetN5∈ℕ to get
(4.65)∫E|G(tn-τn,X(τn))-G(t-τ,X(τ))|dP<ϵ8ec0(t+α),
for all n≥N5.
So, forN=N1∨N2∨N3∨N4∨N5, we have that, if n≥N, then(4.66)|v2(tn,xn)-v2(t,x)|<ϵ
and we conclude that v2 is continuous over (0,∞)×D.
Next, we prove the continuity in the boundary. Let(tn,xn)→n→∞(t,x), where (tn,xn)∈(0,∞)×D and (t,x)∈∂((0,∞)×D), that is, either t=0 or x∈∂D. In both cases, we get that τ=0 a.s., We need to prove that(4.67)|v2(tn,xn)-G(t,x)|⟶n→∞0.
Let 0<α≪1 and N1∈ℕ such that
(4.68)‖(tn,xn)-(t,x)‖<α.
So, for all n≥N1,
(4.69)τn≤tn<t+α.
We have that
(4.70)|v2(tn,xn)-G(t,x)|≤E[e∫0τnc(ξn(r),Xn(r))dr|G(tn-τn,Xn(τn))-G(t,x)|](4.71)+E[|G(t,x)||e∫0τnc(ξn(r),Xn(r))dr-1|].
Because |e∫0τnc(ξn,Xn)dr-1|≤ec0(t+α)+1, we have that (4.71) is uniformly integrable and repeating the same argument made with (4.48), we can prove that
(4.72)E[|G(t,x)||e∫0τnc(ξn(r),Xn(r))dr-1|]⟶n→∞0.
Next, we work with (4.70). As in estimate (4.42), we can prove that the sequence
(4.73){e∫0τnc(ξn(r),Xn(r))dr|G(tn-τn,Xn(τn))-G(t,x)|}n≥N1
is uniformly integrable. We have that
(4.74)(4.70)≤ec0(t+α)E[|G(tn-τn,Xn(τn))-G(tn-τn,X(τn))|](4.75)+ec0(t+α)E[|G(tn-τn,X(τn))-G(t,x))|].
We repeat the same arguments made for the estimates to (4.49) and (4.50) with (4.74) and (4.75), respectively. Then, we can prove that
(4.76)E[|G(tn-τn,Xn(τn))-G(tn-τn,X(τn))|]⟶n→∞0,E[|G(tn-τn,X(τn))-G(t,x))|]⟶n→∞0.
So, v2∈C([0,∞)×D¯) and the proof is complete.


### 4.2. Differentiability ofv

Let0≤T0<T1 and A⊂D be a bounded, open, connected set with C2 boundary. Consider the following Cauchy-Dirichlet problem:(4.77)-ut(t,x)+L[u](t,x)+c(t,x)u(t,x)=-f(t,x)(t,x)∈(T0,T1]×A,u(T0,x)=v(T0,x)forx∈A,u(t,x)=v(t,x)for(t,x)∈(T0,T1]×∂A,
where the boundary data is v. If we assume H0, H1, and H2, then by the continuity of v (Theorem 4.1) and Theorem 5.5, we can guarantee the existence of a unique classical solution to problem (4.77). To prove the regularity of v, we show that it coincides with the solution to (4.77) in the set (T0,T1)×A and so v∈C1,2((T0,T1)×A). Since T0, T1, and A are arbitrary, we get the desired regularity. We are ready to prove the next theorem.Theorem 4.4.
AssumeH0,  H1, and H2. Let v be defined as in (4.1). Then v∈Cloc1,2,λ((0,∞)×ℝd).Proof.
Letw be the solution to (4.77). Define the following stopping times
(4.78)θT:=inf⁡⁡{s>0∣ξ(s)<T0}θA:=inf⁡⁡{s>0∣X(s)∉A¯},θ:=θT∧θA.
Following the same arguments of Section 5 in Chapter 6 of [31], we can prove that w has the following representation:
(4.79)w(t,x)=Ex[∫0θe∫0sc(t-r,X(r))drf(t-s,X(s))ds]+Ex[e∫0θc(t-r,X(r))dsv(t-θ,X(θ))].
Next, we prove that v satisfies the following equality:
(4.80)v(t,x)=Et,x[∫0θe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds]+Et,x[e∫0θc(ξ(r),X(r))dsv(ξ(θ),X(θ))].
Let v1 and v2 be defined as in (4.2) and (4.3), where τ:=t∧τD is introduced to simplify the notation. We will use the following representation of v1 and v2:
(4.81)v1(t,x)=Et,x[∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds],v2(t,x)=Et,x[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))].
First, we work with v1(4.82)v1(t,x)=Et,x[E[∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|Fθ]]=Et,x[E[∫0θe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|Fθ]](4.83)+Et,x[E[∫θτe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|Fθ]].
We study the addends of the right-hand side separately
(4.84)(4.82)=Et,x[∫0θe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds].
For (4.83), we make a couple of changes of variables to get
(4.85)(4.83)=Et,x[E[∫0τ-θe∫0s+θc(ξ(r),X(r))drf(ξ(s+θ),X(s+θ))ds|Fθ]]=Et,x[e∫0θc(ξ(r),X(r))drE[∫0τ-θe∫0sc(ξ(r+θ)),X(r+θ))drf(ξ(s+θ),X(s+θ))ds|Fθ]].
Since θ<τD and θ is bounded, we get that  (see Remark 2.1)
(4.86)τ=inf⁡⁡{s>0∣(ξ(s),X(s))∉[0,∞)×D¯}=θ+inf⁡⁡{s>0∣(ξ(s+θ),X(s+θ))∉[0,∞)×D¯},
so
(4.87)τ-θ=Θθ∘τ,
where Θ· denotes the shift operator. Since the process (ξ,X) is a homogeneous strong Markov process, we get that
(4.88)E[∫0Θθ∘τe∫0sc(Θθ∘(ξ,X)(r))drf(Θθ∘(ξ,X)(s))ds∣Fθ]=Eξ(θ),X(θ)[∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds]=v1(ξ(θ),X(θ)).
The last equality follows from a general form of the strong Markov property (see Theorem 4.18 in Chapter 5 of [29] Theorem 4.6 in Chapter 5 of [27], or Theorem 5.1 in Chapter V of [26]) that states that if F:Ω→ℝ is a bounded measurable functional, then
(4.89)E[(F(ξ(⋅),X(⋅))∘Θθ)1{θ<∞}∣Fθ]=Eξ(θ),X(θ)[F(ξ(⋅),X(⋅))].
In both sides of (4.88), we consider the sequence (F∨-n)∧n and apply the conditional dominated convergence Theorem.
So,(4.90)v1(t,x)=Et,x[∫0θe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds]+Et,x[e∫0θc(ξ(r),X(r))drv1(ξ(θ),X(θ))].
Next, we study v2. Again, for the integral, we use a couple of changes of variables to get
(4.91)v2(t,x)=Et,x[E[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))∣Fθ]]=Et,x[e∫0θc(ξ(r),X(r))drE[e∫0τ-θc(ξ(r+θ),X(r+θ))drG(ξ(τ),X(τ))∣Fθ]].
Using (4.87), we write
(4.92)G(ξ(τ),X(τ))=G(ξ(τ-θ+θ),X(τ-θ+θ))=G(ξ(Θθ∘τ+θ),X(Θθ∘τ+θ)).
Then, the expression inside the conditional expectation can be written as
(4.93)e∫0Θθ∘τc(Θθ∘(ξ,X)(r))drG(Θθ∘(ξ,X)(Θθ∘τ)).
Repeating the same argument used for (4.88), we have that the conditional expectation is
(4.94)E[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))∣Fθ]=E[e∫0Θθ∘τc(Θθ∘(ξ,X)(r))drG(Θθ∘(ξ,X)(Θθ∘τ))∣Fθ]=Eξ(θ),X(θ)[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))].
Finally we get(4.95)v2(t,x)=Et,x[e∫0θc(ξ(r),X(r))drEξ(θ),X(θ)[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))]]=Et,x[e∫0θc(ξ(r),X(r))drv2(ξ(θ),X(θ))].
Combining equations (4.90) and (4.95) we prove that (4.80) holds.
So due to equations (4.79) and (4.80) we have that v=w. Since w∈C1,2,λ((T0,T1)×A) (see Theorem 5.5 below) and T0T1 and A are arbitrary we get that v∈C1,2((0,∞)×ℝd)∩Cloc1,2,λ((0,∞)×ℝd) and the proof is complete.We are ready to proof the Main TheoremProof of Theorem3.1.
The proof follows from Theorems4.1 and 4.4 and Lemmas 3.3 and 3.4.


## 4.1. Continuity ofv

Let(ξ,X) denote the solution to (2.3) and let G be defined as in Proposition 3.2, then v has the following form:(4.1)v(t,x)=Et,x[∫0t∧τDe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds]+Et,x[e∫0t∧τDc(ξ(r),X(r))drG(ξ(t∧τD),X(t∧τD))].
For simplicity, we write v=v1+v2, where(4.2)v1(t,x):=Et,x[∫0t∧τDe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds],(4.3)v2(t,x):=Et,x[e∫0t∧τDc(ξ(r),X(r))drG(ξ(t∧τD),X(t∧τD))].Theorem 4.1.
AssumeH0,  H1, and H2. Let v be defined as in (4.1). Then, v is continuous on [0,∞)×D¯.The proof of this theorem is divided into two lemmas.Lemma 4.2.
AssumeH0,  H1, and H2. Let v1 be defined as in (4.2). Then, v1 is continuous on [0,∞)×D¯.Proof.
First, we prove the continuity on(0,∞)×D. For that, let
(4.4)(tn,xn)⟶n→∞(t,x)
in (0,∞)×D and ϵ>0. We need to prove that there exists N(ϵ)∈ℕ such that, for all n≥N(ϵ)(4.5)|v1(tn,xn)-v1(t,x)|<ϵ.
Denote by (ξ,X) and (ξn,Xn) the solutions to (2.3) with initial conditions (t,x) and (tn,xn), respectively. To simplify the notation in the proof, let τ:=t∧τD and τn:=tn∧τDn denote their corresponding exit times from [0,∞)×D¯.
Letα>0, then there exists N1∈ℕ such that, for all n≥N1(4.6)‖(tn,xn)-(t,x)‖<α.
Observe that, for all n≥N1, we get
(4.7)τn≤tn≤t+α,τ≤t≤t+α.
Define the random variables Yn as
(4.8)Yn:=|∫0τne∫0sc(ξn(r),Xn(r))drf(ξn(s),Xn(s))ds-∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|.
The sequence {Yn}n≥N1 is uniformly integrable. Thanks to Theorem 4.2, in Chapter 5 of [33], it is sufficient to prove that supn𝔼[Yn2]<∞. So,
(4.9)E[Yn2]≤2E[|∫0τne∫0sc(ξn(r),Xn(r))drf(ξn(s),Xn(s))ds|2]+2E[|∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|2]≤2E[|∫0τnec0(t+α)K2(t+α)(1+‖Xn(s)‖k)ds|2]+2E[|∫0τec0(t+α)K2(t+α)(1+‖X(s)‖k)ds|2]≤2e2c0(t+α)K22(t+α)E[|∫0τn(1+sup⁡0≤r≤t+α‖Xn(r)‖k)ds|2]+2e2c0(t+α)K22(t+α)E[|∫0τ(1+sup⁡0≤r≤t+α‖X(r)‖k)ds|2]≤4e2c0(t+α)K22(t+α)(t+α)2(1+E[sup⁡0≤r≤t+α‖Xn(r)‖2k])+4e2c0(t+α)K22(t+α)(t+α)2(1+E[sup⁡0≤r≤t+α‖X(r)‖2k])≤C(1+K(1+‖xn‖2k))+C(1+K(1+‖x‖2k))≤C(1+K(1+(‖x‖+α)2k))+C(1+K(1+‖x‖2k))<∞,
where C=C(t,α,c0) and K=K(t,α,k). We use (4.6), (4.7), (2.15) and the polynomial growth of f.
LetM>0, 0<η<1, and β>0 and define the set(4.10)EM,n,η,β:={‖X‖t+α≤M}∩{‖Xn-X‖t+α≤η}∩{|τn-τ|≤β}.
Then,
(4.11)|v1(tn,xn)-v1(t,x)|≤∫ΩYndP=∫EM,n,η,βYndP+∫Ω∖EM,n,η,βYndP.
Since the sequence {Yn} is uniformly integrable, there exists δ(ϵ) such that for any E∈ℱ that satisfies ℙ[E]<δ, we have
(4.12)sup⁡n≥N1∫EYndP<ϵ2.
It follows, from Remark 2.4, Proposition 2.5, and Theorems 5.3 and 5.4 in Section 5, the existence of M and N2 such that
(4.13)P[Ω∖EM,n,η,β]<δ(ϵ)
for all n≥N2. Then for n≥N1∨N2, we get that
(4.14)|v1(tn,xn)-v1(t,x)|≤∫EM,n,η,βYndP+ϵ2.
For simplicity of notation, we write the set EM,n,η,β as E and define
(4.15)A:=[0,t+α]×[-M-1,M+1]d,Dt:=[0,t+α]×D¯
and let
(4.16)BA:=(-m,m)d+1
be an open set such that A⊂BA.
On the setE, for all n≥N1 and 0≤s≤t+α, it is satisfied that(4.17)(ξn(s),Xn(s)),(ξ(s),X(s))∈A.
We have that
(4.18)∫EYndP≤∫E∫0τn∧τ|e∫0sc(ξn(r),Xn(r))drf(ξn(s),Xn(s))-e∫0sc(ξ(r),X(r))drf(ξ(s),X(s))|dsdP(4.19)+∫E∫τn∧ττn∨τ(e∫0sc(ξn(r),Xn(r))dr|f(ξn(s),Xn(s))|1τn>τ+e∫0sc(ξ(r),X(r))dr|f(ξ(s),X(s))|1τn≤τ)dsdP.
We first analyse (4.19):
(4.20)(4.19)≤∫E∫τn∧ττn∨τec0(t+α)Kf(A∩Dt)(1τn≤τ+1τn>τ)dsdP=ec0(t+α)Kf(A∩Dt)∫E|τn-τ|dP≤ec0(t+α)Kf(A∩Dt)β.
For (4.18), we get
(4.21)(4.18)≤∫E∫0τn∧τe∫0sc(ξn(r),Xn(r))dr|f(ξn(s),Xn(s))-f(ξ(s),X(s))|dsdP(4.22)+∫E∫0τn∧τ|f(ξ(s),X(s))||e∫0sc(ξn(r),Xn(r))dr-e∫0sc(ξ(r),X(r))dr|dsdP.
Now,
(4.23)(4.21)≤∫Eec0(t+α)∫0τn∧τLf(BA)(|tn-t|λ+‖Xn(s)-X(s)‖)dsdP≤ec0(t+α)(t+α)Lf(BA)(|tn-t|λ+η).
For (4.22), we need the following bound:
(4.24)|e∫0sc(ξn(r),Xn(r))dr-e∫0sc(ξ(r),X(r))dr|=e∫0sc(ξ(r),X(r))dr|exp⁡{∫0s(c(ξn(r),Xn(r))-c(ξ(r),X(r)))dr}-1|≤ec0s(exp⁡{∫0s|c(ξn(r),Xn(r))-c(ξ(r),X(r))|dr}-1)≤ec0s(exp⁡{∫0sLc(BA)(|tn-t|λ+‖Xn(r)-X(r)‖)dr}-1)≤ec0s(exp⁡{Lc(BA)s(|tn-t|λ+η)}-1),
since |ex-1|≤e|x|-1. If we choose N3∈ℕ such that |tn-t|λ≤1/(2Lc(BA)(t+α)) for all n≥N3 and η≤1/(2Lc(BA)(t+α)), then we get by the mean value theorem that
(4.25)|e∫0sc(ξn(r),Xn(r))dr-e∫0sc(ξ(r),X(r))dr|≤ec0seLc(BA)s(|tn-t|λ+η).
Then,
(4.26)(4.22)≤Kf(A∩Dt)ec0(t+α)eLc(BA)(t+α)2(|tn-t|λ+η).
To summarize, we get with the above estimations that
(4.27)∫EYndP≤ec0(t+α)(t+α)Lf(BA)(|tn-t|λ+η)+Kf(A∩Dt)ec0(t+α)eLc(BA)(t+α)2(|tn-t|λ+η)+ec0(t+α)Kf(A∩Dt)β.
Hence, to prove continuity, we proceed as follows,  (i)
Letϵ>0 and 0<α≪1.(ii)
LetN1≥ℕ such that, for all n≥N1,
(4.28)∥(tn,xn)-(t,x)∥<α.(iii)
Letδ(ϵ)>0 fulfil the uniformly integrability condition (4.12).(iv)
TakeM>0 such that ℙ[∥X∥t+α>M]<δ(ϵ)/3.(v)
DefineA:=[0,t+α]×[-M-1,M+1]d and Dt:=[0,t+α]×D¯.(vi)
Let(4.29)η<min{1,12Lc(BA)(t+α),ϵ16ec0(t+α)(t+α)Lf(BA),ϵ16Kf(A∩Dt)ec0(t+α)eLc(BA)(t+α)2}.(vii)
ChooseN2∈ℕ such that ℙ[∥Xn-X∥t+α>η]≤δ(ϵ)/3 for all n≥N2(viii)
LetN3∈ℕ be such that
(4.30)|tn-t|λ<min{12Lc(BA)(t+α),ϵ16ec0(t+α)(t+α)Lf(BA),ϵ16Kf(A∩Dt)ec0(t+α)eLc(BA)(t+α)},
for all n≥N3.(ix)
Let(4.31)β<ϵ4ec0(t+α)Kf(A∩Dt)
and choose N4∈ℕ such that, for all n≥N4, ℙ[|τn-τ|>β]≤δ(ϵ)/3.
Thus, ifN=N1∨N2∨N3∨N4, then, for all n≥N,(4.32)|v1(tn,xn)-v1(t,x)|<ϵ.
Therefore, v1 is continuous in (0,∞)×D.
For the continuity at the boundary we make a similar argument. Let(tn,xn)→n→∞(t,x), where (tn,xn)∈(0,∞)×D and (t,x)∈∂((0,∞)×D), that is, either t=0 or x∈∂D. In both cases we get that τ=0 a.s. and so v1(t,x)=0. Then, we need to prove that(4.33)|v1(tn,xn)|⟶n→∞0.
Let 0<α≪1 and N1∈ℕ be such that
(4.34)‖(tn,xn)-(t,x)‖<α.
We get
(4.35)τn≤tn<t+α
for all n≥N1. For the continuity, we have
(4.36)|v1(tn,xn)|≤E[∫0τne∫0sc(ξn(r),Xn(r))dr|f(ξn(s),Xn(s))|ds]≤E[∫0τnec0sK2(t+α)(1+‖Xn(s)‖k)ds]≤ec0(t+α)K2(t+α)E[τn(1+sup⁡0≤r≤t+α‖Xn(r)‖k)]⟶n→∞0.
The convergence follows from the uniform integrability of
(4.37)τn(1+sup⁡0≤r≤t+α‖Xn(r)‖k)
and the fact that τn→n→∞ℙ0 (see Theorem 5.2 in Chapter 5 of [33] and Theorem 5.3 in Section 5). This completes the proof.Lemma 4.3.
AssumeH0,  H1, and H2. Let v2 be defined as in (4.3). Then, v2 is continuous on [0,∞)×D¯.Proof.
We use an analogous argument to the one in the proof of Lemma4.2. First, we prove the continuity in (0,∞)×D. Let
(4.38)(tn,xn)⟶n→∞(t,x),
with (tn,xn),(t,x)∈(0,∞)×D. Denote by (ξn,Xn) and (ξ,X) the solutions to (2.3) with initial conditions (tn,xn) and (t,x), respectively, and let τn:=tn∧τDn and τ:=t∧τD be their corresponding exit times from [0,∞)×D¯. Let 0<α≪1 and N1 be such that, for all n≥N1,
(4.39)∥(tn,xn)-(t,x)∥<α.
This implies that
(4.40)τn≤t+α,τ≤t+α.
First, we prove that the sequence of random variables
(4.41)Yn:=|e∫0τnc(ξn(r),Xn(r))drG(ξn(τn),Xn(τn))-e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))|
is uniformly integrable for all n≥N1. As in (4.9),
(4.42)E[Yn2]≤2E[|e∫0τnc(ξn(r),Xn(r))drG(ξn(τn),Xn(τn))|2]+2E[|e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))|2]≤2E[e2c0(t+α)K32(t+α)(1+‖Xn(τn)‖k)2]+2E[e2c0(t+α)K32(t+α)(1+‖X(τ)‖k)2]≤2e2c0(t+α)K32(t+α)E[(1+sup⁡0≤r≤t+α‖Xn(r)‖k)2]+2e2c0(t+α)K32(t+α)E[(1+sup⁡0≤r≤t+α‖X(r)‖k)2]≤4e2c0(t+α)K32(t+α)(1+E[sup⁡0≤r≤t+α‖Xn(r)‖2k])+4e2c0(t+α)K32(t+α)(1+E[sup⁡0≤r≤t+α‖X(r)‖2k])≤C(1+K(1+‖xn‖2k))+C(1+K(1+‖x‖2k))≤C(1+K(1+(‖x‖+α)2k))+C(1+K(1+‖x‖2k))<∞,
where C=C(t,α,c0) and K=K(t,α,k). We use (4.39), (4.40), (2.15), and the polynomial growth of G in ∂((0,∞)×D). As in Lemma 4.2, let ϵ>0, then there exists δ(ϵ)>0 such that
(4.43)sup⁡n≥N1∫EYndP<ϵ2
for all E∈ℱ, with ℙ[E]<δ(ϵ).
LetEM,n,η,β be defined as in (4.10), and choose M>0 and N2∈ℕ such that(4.44)P[Ω∖EM,n,η,β]<δ(ϵ),
for all n≥N2.
For simplicity of notation, denoteEM,n,η,β as E. Then,(4.45)|v2(tn,xn)-v2(t,x)|≤∫EYndP+∫Ω∖EYndP≤∫EYndP+ϵ2.
Let A, Dt, and BA be defined as in Lemma 4.2 (see (4.15) and (4.16)). Then, on the set E, we get that, for all n≥N1 and 0≤s≤t+α,
(4.46)(ξn(s),Xn(s)),(ξ(s),X(s))∈A.
So,
(4.47)∫EYndP≤∫Ee∫0τnc(ξn(r),Xn(r))dr×|G(ξn(τn),Xn(τn))-G(ξ(τ),X(τ))|dP(4.48)+∫E|G(ξ(τ),X(τ))||e∫0τnc(ξn(r),Xn(r))dr-e∫0τc(ξ(r),X(r))dr|dP.
We study each addend of the right-hand side separately:
(4.49)(4.47)≤ec0(t+α)∫E|G(tn-τn,Xn(τn))-G(tn-τn,X(τn))|dP(4.50)+ec0(t+α)∫E|G(tn-τn,X(τn))-G(t-τ,X(τ))|dP.
First, we get a bound for (4.49). Since G is continuous, then it is uniformly continuous on A. Then, for ϵ>0, there exists γ(c0,t,α,ϵ,M) such that
(4.51)|G(t1,x1)-G(t2,x2)|<ϵ8ec0(t+α)
for all (t1,x1),(t2,x2)∈A with ∥(t1,x1)-(t2,x2)∥<γ(c0,t,α,ϵ,M). On the set E, we have (tn-τn,Xn(τn)),(tn-τn,X(τn))∈A and
(4.52)‖(tn-τn,Xn(τn))-(tn-τn,X(τn))‖<η.
So, if we choose η<γ, then we get
(4.53)(4.49)<ϵ8.
Next, we study (4.50). Thanks to Theorem 5.3, we know that τn→n→∞a.sτ. This and the continuity of X(·) and G imply that
(4.54)G(tn-τn,X(τn))⟶n→∞a.s.G(t-τ,X(τ)).
On the set E, we have that (tn-τn,X(τn)),(t-τ,X(τ))∈A and so
(4.55)|G(tn-τn,X(τn))-G(t-τ,X(τ))|1E≤2KG(A).
By the dominated convergence theorem, there exists N3∈ℕ such that
(4.56)(4.50)<ϵ8
for all n≥N3.
To give a bound for (4.48), we observe that, on the set E,(4.57)|∫0τnc(ξn(r),Xn(r))dr-∫0τc(ξ(r),X(r))dr|≤∫0τn∧τ|c(ξn(r),Xn(r))-c(ξ(r),X(r))|dr+∫τn∧ττn∨τ(|c(ξn(r),Xn(r))|1τn≥τ+|c(ξ(r),X(r))|1τn<τ)dr≤∫0τn∧τLc(BA)(|tn-t|λ+‖Xn(r)-X(r)‖)dr+Kc(A∩Dt)|τn-τ|≤Lc(BA)(t+α)(|tn-t|λ+η)+Kc(A∩Dt)β.
Making a similar argument as the one made in (4.24) and (4.25), we get
(4.58)|e∫0τnc(ξn(r),Xn(r))dr-e∫0τc(ξ(r),X(r))dr|≤ec0(t+α)e[Lc(BA)(t+α)(|tn-t|λ+η)+Kc(A∩Dt)β]
if |tn-t|λ<1/3Lc(BA)(t+α), η<1/3Lc(BA)(t+α), and β<1/3Kc(A∩Dt). Then,
(4.59)(4.48)≤KG(A)ec0(t+α)e[Lc(BA)(t+α)(|tn-t|λ+η)+Kc(A∩Dt)β].
To summarize, we get with the above estimations that
(4.60)∫EYndP≤ϵ4+KG(A)ec0(t+α)e[Lc(BA)(t+α)(|tn-t|λ+η)+Kc(A∩Dt)β].
Hence, to prove continuity, we proceed as follows. (i)
Letϵ>0 and 0<α≪1.(ii)
LetN1∈ℕ be such that, for all n≥N1,
(4.61)‖(tn,xn)-(t,x)‖<α.(iii)
Letδ(ϵ)>0 fulfil the uniformly integrability condition (4.43).(iv)
TakeM>0 such that ℙ[∥X∥t+α>M]<δ(ϵ)/3.(v)
DefineA:=[0,t+α]×[-M-1,M+1]d and Dt:=[0,t+α]×D¯.(vi)
Let(4.62)η<min{1,γ(c0,t,α,ϵ,M),13Lc(BA)(t+α),ϵ12KG(A)ec0(t+α)eLc(BA)(t+α)}.(vii)
ChooseN2∈ℕ such that ℙ[∥Xn-X∥t+α>η]<δ(ϵ)/3 for all n≥N2(viii)
Let(4.63)β<min{13Kc(A∩Dt),ϵ12KG(A)ec0(t+α)eKc(A∩Dt)}.(ix)
ChooseN3∈ℕ such that ℙ[|τn-τ|>β]<δ(ϵ)/3 for all n≥N3.(x)
LetN4∈ℕ be such that
(4.64)|tn-t|λ<min{13Lc(BA)(t+α),ϵ12KG(A)ec0(t+α)eLc(BA)(t+α)},
for all n≥N4.(xi)
LetN5∈ℕ to get
(4.65)∫E|G(tn-τn,X(τn))-G(t-τ,X(τ))|dP<ϵ8ec0(t+α),
for all n≥N5.
So, forN=N1∨N2∨N3∨N4∨N5, we have that, if n≥N, then(4.66)|v2(tn,xn)-v2(t,x)|<ϵ
and we conclude that v2 is continuous over (0,∞)×D.
Next, we prove the continuity in the boundary. Let(tn,xn)→n→∞(t,x), where (tn,xn)∈(0,∞)×D and (t,x)∈∂((0,∞)×D), that is, either t=0 or x∈∂D. In both cases, we get that τ=0 a.s., We need to prove that(4.67)|v2(tn,xn)-G(t,x)|⟶n→∞0.
Let 0<α≪1 and N1∈ℕ such that
(4.68)‖(tn,xn)-(t,x)‖<α.
So, for all n≥N1,
(4.69)τn≤tn<t+α.
We have that
(4.70)|v2(tn,xn)-G(t,x)|≤E[e∫0τnc(ξn(r),Xn(r))dr|G(tn-τn,Xn(τn))-G(t,x)|](4.71)+E[|G(t,x)||e∫0τnc(ξn(r),Xn(r))dr-1|].
Because |e∫0τnc(ξn,Xn)dr-1|≤ec0(t+α)+1, we have that (4.71) is uniformly integrable and repeating the same argument made with (4.48), we can prove that
(4.72)E[|G(t,x)||e∫0τnc(ξn(r),Xn(r))dr-1|]⟶n→∞0.
Next, we work with (4.70). As in estimate (4.42), we can prove that the sequence
(4.73){e∫0τnc(ξn(r),Xn(r))dr|G(tn-τn,Xn(τn))-G(t,x)|}n≥N1
is uniformly integrable. We have that
(4.74)(4.70)≤ec0(t+α)E[|G(tn-τn,Xn(τn))-G(tn-τn,X(τn))|](4.75)+ec0(t+α)E[|G(tn-τn,X(τn))-G(t,x))|].
We repeat the same arguments made for the estimates to (4.49) and (4.50) with (4.74) and (4.75), respectively. Then, we can prove that
(4.76)E[|G(tn-τn,Xn(τn))-G(tn-τn,X(τn))|]⟶n→∞0,E[|G(tn-τn,X(τn))-G(t,x))|]⟶n→∞0.
So, v2∈C([0,∞)×D¯) and the proof is complete.


## 4.2. Differentiability ofv

Let0≤T0<T1 and A⊂D be a bounded, open, connected set with C2 boundary. Consider the following Cauchy-Dirichlet problem:(4.77)-ut(t,x)+L[u](t,x)+c(t,x)u(t,x)=-f(t,x)(t,x)∈(T0,T1]×A,u(T0,x)=v(T0,x)forx∈A,u(t,x)=v(t,x)for(t,x)∈(T0,T1]×∂A,
where the boundary data is v. If we assume H0, H1, and H2, then by the continuity of v (Theorem 4.1) and Theorem 5.5, we can guarantee the existence of a unique classical solution to problem (4.77). To prove the regularity of v, we show that it coincides with the solution to (4.77) in the set (T0,T1)×A and so v∈C1,2((T0,T1)×A). Since T0, T1, and A are arbitrary, we get the desired regularity. We are ready to prove the next theorem.Theorem 4.4.
AssumeH0,  H1, and H2. Let v be defined as in (4.1). Then v∈Cloc1,2,λ((0,∞)×ℝd).Proof.
Letw be the solution to (4.77). Define the following stopping times
(4.78)θT:=inf⁡⁡{s>0∣ξ(s)<T0}θA:=inf⁡⁡{s>0∣X(s)∉A¯},θ:=θT∧θA.
Following the same arguments of Section 5 in Chapter 6 of [31], we can prove that w has the following representation:
(4.79)w(t,x)=Ex[∫0θe∫0sc(t-r,X(r))drf(t-s,X(s))ds]+Ex[e∫0θc(t-r,X(r))dsv(t-θ,X(θ))].
Next, we prove that v satisfies the following equality:
(4.80)v(t,x)=Et,x[∫0θe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds]+Et,x[e∫0θc(ξ(r),X(r))dsv(ξ(θ),X(θ))].
Let v1 and v2 be defined as in (4.2) and (4.3), where τ:=t∧τD is introduced to simplify the notation. We will use the following representation of v1 and v2:
(4.81)v1(t,x)=Et,x[∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds],v2(t,x)=Et,x[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))].
First, we work with v1(4.82)v1(t,x)=Et,x[E[∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|Fθ]]=Et,x[E[∫0θe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|Fθ]](4.83)+Et,x[E[∫θτe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds|Fθ]].
We study the addends of the right-hand side separately
(4.84)(4.82)=Et,x[∫0θe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds].
For (4.83), we make a couple of changes of variables to get
(4.85)(4.83)=Et,x[E[∫0τ-θe∫0s+θc(ξ(r),X(r))drf(ξ(s+θ),X(s+θ))ds|Fθ]]=Et,x[e∫0θc(ξ(r),X(r))drE[∫0τ-θe∫0sc(ξ(r+θ)),X(r+θ))drf(ξ(s+θ),X(s+θ))ds|Fθ]].
Since θ<τD and θ is bounded, we get that  (see Remark 2.1)
(4.86)τ=inf⁡⁡{s>0∣(ξ(s),X(s))∉[0,∞)×D¯}=θ+inf⁡⁡{s>0∣(ξ(s+θ),X(s+θ))∉[0,∞)×D¯},
so
(4.87)τ-θ=Θθ∘τ,
where Θ· denotes the shift operator. Since the process (ξ,X) is a homogeneous strong Markov process, we get that
(4.88)E[∫0Θθ∘τe∫0sc(Θθ∘(ξ,X)(r))drf(Θθ∘(ξ,X)(s))ds∣Fθ]=Eξ(θ),X(θ)[∫0τe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds]=v1(ξ(θ),X(θ)).
The last equality follows from a general form of the strong Markov property (see Theorem 4.18 in Chapter 5 of [29] Theorem 4.6 in Chapter 5 of [27], or Theorem 5.1 in Chapter V of [26]) that states that if F:Ω→ℝ is a bounded measurable functional, then
(4.89)E[(F(ξ(⋅),X(⋅))∘Θθ)1{θ<∞}∣Fθ]=Eξ(θ),X(θ)[F(ξ(⋅),X(⋅))].
In both sides of (4.88), we consider the sequence (F∨-n)∧n and apply the conditional dominated convergence Theorem.
So,(4.90)v1(t,x)=Et,x[∫0θe∫0sc(ξ(r),X(r))drf(ξ(s),X(s))ds]+Et,x[e∫0θc(ξ(r),X(r))drv1(ξ(θ),X(θ))].
Next, we study v2. Again, for the integral, we use a couple of changes of variables to get
(4.91)v2(t,x)=Et,x[E[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))∣Fθ]]=Et,x[e∫0θc(ξ(r),X(r))drE[e∫0τ-θc(ξ(r+θ),X(r+θ))drG(ξ(τ),X(τ))∣Fθ]].
Using (4.87), we write
(4.92)G(ξ(τ),X(τ))=G(ξ(τ-θ+θ),X(τ-θ+θ))=G(ξ(Θθ∘τ+θ),X(Θθ∘τ+θ)).
Then, the expression inside the conditional expectation can be written as
(4.93)e∫0Θθ∘τc(Θθ∘(ξ,X)(r))drG(Θθ∘(ξ,X)(Θθ∘τ)).
Repeating the same argument used for (4.88), we have that the conditional expectation is
(4.94)E[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))∣Fθ]=E[e∫0Θθ∘τc(Θθ∘(ξ,X)(r))drG(Θθ∘(ξ,X)(Θθ∘τ))∣Fθ]=Eξ(θ),X(θ)[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))].
Finally we get(4.95)v2(t,x)=Et,x[e∫0θc(ξ(r),X(r))drEξ(θ),X(θ)[e∫0τc(ξ(r),X(r))drG(ξ(τ),X(τ))]]=Et,x[e∫0θc(ξ(r),X(r))drv2(ξ(θ),X(θ))].
Combining equations (4.90) and (4.95) we prove that (4.80) holds.
So due to equations (4.79) and (4.80) we have that v=w. Since w∈C1,2,λ((T0,T1)×A) (see Theorem 5.5 below) and T0T1 and A are arbitrary we get that v∈C1,2((0,∞)×ℝd)∩Cloc1,2,λ((0,∞)×ℝd) and the proof is complete.We are ready to proof the Main TheoremProof of Theorem3.1.
The proof follows from Theorems4.1 and 4.4 and Lemmas 3.3 and 3.4.


## 5. Auxiliary Results

### 5.1. Continuity of the Stopping Times

Theorem 5.1.
Let{Z(t)}t≥0 be a stochastic process with continuous paths a.s. and A⊂ℝd an open, connected set with regular boundary. Let
(5.1)τ:=inf⁡⁡{t>0∣Z(t)∉A¯}.
Assume that ℙ[τ<∞∣Z(0)=z]=1 and ℙ[τ=τ′∣Z(0)=z]=1 for all z∈A, where
(5.2)τ′:=inf⁡⁡{t>0∣Z(t)∉A}.
For a>0, define
(5.3)Aa:={x∈Rd∣d(x,∂A)<a},Aa+:=A∪Aa,Aa-:=A∖Aa
and the corresponding exit times
(5.4)τa+:=inf⁡⁡{t>0∣Z(t)∉Aa+¯},τa-:=inf⁡⁡{t>0∣Z(t)∉Aa-¯}.
Then, if Z(0)=z∈A, then
(5.5)τa+⟶a↓0a.s.τ,τa-⟶a↓0a.s.τ.Proof.
LetZ(0)=z∈A and
(5.6)Bz:={Z(t)iscontinuous}∩{τ=τ′}∩{τ<∞}.
By the hypotheses, we have that ℙ[Bz]=1. Observe that τa-≤τ≤τa+ for all a>0, then we need to prove that, for all ω∈Bz and α>0, there exists γ(ω,α)>0 such that, for all 0<a<γ,
(5.7)0≤τa+(ω)-τ(ω)<α,0≤τ(ω)-τa-(ω)<α.
Let ω∈Bz and α>0, then Z(t,ω) is continuous and τ(ω)<∞. We first prove the continuity for τa+. Define
(5.8)γ+(α,ω):=sup⁡{t∈[τ(ω),τ(ω)+α),Z(t,ω)∉A¯}{d(Z(t,ω),∂A)}.
Since Z(t,ω) is continuous, then γ+(α,ω)>0. So, there exists t+∈[τ(ω),τ(ω)+α) such that Z(t+,ω)∉Aa+ for all 0<a≤(γ+/2). Let γ:=γ+/2, then, for all 0<a≤γ, we get that τa+(ω)∈[τ(ω),τ(ω)+α) and so
(5.9)0≤τa+(ω)-τ(ω)<α.
For τa-, we proceed in a similar way. Let
(5.10)β(ω):=inf⁡t≤τ(ω)-α{d(Z(t,ω),∂A)}.
Since τ(ω)=τ′(ω) and Z(t,ω) is continuous, we get that β(ω)>0. Define
(5.11)γ-(α,ω):=sup⁡t∈(τ(ω)-α,τ(ω)]{d(Z(t,ω),∂A)}.
Again, it follows from the continuity of Z(t,ω) that γ-(α,ω)>0. So, there exists t-∈(τ(ω)-α,τ(ω)] such that Z(t-,ω)∉Aa- for all 0<a≤(β∧γ-)/2. Let γ:=(β∧γ-)/2, then, for all 0<a≤γ, we get that τa-(ω)∈(τ(ω)-α,τ(ω)] and so
(5.12)0≤τ(ω)-τa-(ω)<α
and the proof is complete.Remark 5.2.
For the continuity ofτa-, we need the extra constant β. If we only consider the constant γ-, then we can guarantee that the process Z is “close” to ∂A in the interval (τ-α,τ] but we cannot guarantee that it is the first time that it does it and so we do not necessarily have that τa- and τ are “close enough.”Theorem 5.3.
Let{Z(t)}t≥0 be a stochastic process with Z(0)=z∈ℝd and A⊂ℝd an open, connected set with regular boundary. Let {zn} be a sequence such that
(5.13)zn⟶n→∞z0
with zn,z0∈A. Denote by Z0 and Zn the stochastic processes with initial conditions z0 and zn, respectively. Define
(5.14)τ:=inf⁡⁡{t>0∣Z0(t)∉A¯},τn:=inf⁡⁡{t>0∣Zn(t)∉A¯}.
Assume that τ<∞ and τ=τ′ a.s., where
(5.15)τ′:=inf⁡⁡{t>0∣Z0(t)∉A}.
Assume also that, for all T>0,
(5.16)‖Zn-Z‖T⟶n→∞P0.
Then,
(5.17)τn⟶n→∞Pτ.Proof.
Letϵ>0 and η>0. We need to prove that there exists N(ϵ,η)∈ℕ such that, for all n≥N(ϵ,η), we have
(5.18)P[|τn-τ|≤ϵ]≥1-η.
Since τ<∞ a.s., then there exists M=M(η) such that
(5.19)P[τ≤M]≥1-η3.
Thanks to Theorem 5.1, we get the existence of a0=a0(ϵ,η) such that, for all a≤a0(ϵ,η), it is satisfied that
(5.20)P[τa+≤τ+ϵ,τa-≥τ-ϵ]≥1-η3.
Let a1:=a0/2. By hypothesis, we have the existence of N=N(M,a1,ϵ,η)=N(ϵ,η) such that for all n≥N(ϵ,η)(5.21)P[‖Zn-Z‖M+ϵ≤a1]≥1-η3.
So, over the intersection of the three sets, we get
(5.22){τ≤M}∩{τa1+≤τ+ϵ,τa1-≥τ-ϵ}∩{‖Zn-Z‖M+ϵ≤a1}⊂{τa1-≤τn≤τa1+}∩{τa1+≤τ+ϵ,τa1-≥τ-ϵ}⊂{τ-ϵ≤τn≤τ+ϵ}.
The first inclusion follows from the fact that, over the set {τ≤M}∩{∥Zn-Z∥M+ϵ≤a1}, we have that the process Zn cannot leave the set A before the process Z leaves the set Aa1- and so we get the τa1-≤τn. On the intersection of the three sets, we have that τa1+≤M+ϵ. Considering a similar argument, we obtain the remaining part, that is, τn≤τa1+.
Finally, we get from inequalities (5.19), (5.20), (5.21) and Bonferroni's inequality that for all n≥N(ϵ,η),(5.23)P[τ-ϵ≤τn≤τ+ϵ]≥1-η
and the proof is complete.


### 5.2. Additional Results

Theorem 5.4.
Let{(tn,xn)}n∈ℕ⊂[0,∞)×ℝd be a sequence such that (tn,xn)→n→∞(t,x). Denote by Xn and X the solutions of the following equations:
(5.24)dXn(s)=b(tn-s,Xn(s))ds+σ(tn-s,Xn(s))dW(s),Xn(0)=xn,dX(s)=b(t-s,X(s))ds+σ(t-s,X(s))dW(s),X(0)=x.
Then, for all T>0,
(5.25)‖Xn-X‖T⟶n→∞P0.Proof.
This theorem is consequence of Theorem 1.5 in ChapterV of [26].The following theorem is Theorem 9 of Chapter 3 in [3]. Let 0≤T0<T1 and let A⊂ℝd be a bounded open set with C2 boundary. Since σ, b, c, and f are locally Lipschitz, then they are locally Hölder of any order β∈(0,1)Theorem 5.5.
AssumeH1 and H2. Consider the following differential equation:
(5.26)-ut(t,x)+L[u](t,x)+c(t,x)u(t,x)=-f(t,x)(t,x)∈[T0,T1]×A,u(T0,x)=g(T0,x)forx∈A,u(t,x)=g(t,x)for(t,x)∈(T0,T1]×∂A.
If g is continuous, then there exists a classical solution w∈C([T0,T1)×A¯)∩C1,2,β((T0,T1)×A) of (5.26).Remark 5.6.
Letw be the solution of (5.26) and define z as w(t,x)=ec0tz(t,x) in [T0,T1]×A. Then, z fulfils (5.26) with c′=c-c0 and f′(t,x)=e-c0tf(t,x). And so the hypotheses of Theorem 9 of Chapter 3 in [3] are satisfied.


## 5.1. Continuity of the Stopping Times

Theorem 5.1.
Let{Z(t)}t≥0 be a stochastic process with continuous paths a.s. and A⊂ℝd an open, connected set with regular boundary. Let
(5.1)τ:=inf⁡⁡{t>0∣Z(t)∉A¯}.
Assume that ℙ[τ<∞∣Z(0)=z]=1 and ℙ[τ=τ′∣Z(0)=z]=1 for all z∈A, where
(5.2)τ′:=inf⁡⁡{t>0∣Z(t)∉A}.
For a>0, define
(5.3)Aa:={x∈Rd∣d(x,∂A)<a},Aa+:=A∪Aa,Aa-:=A∖Aa
and the corresponding exit times
(5.4)τa+:=inf⁡⁡{t>0∣Z(t)∉Aa+¯},τa-:=inf⁡⁡{t>0∣Z(t)∉Aa-¯}.
Then, if Z(0)=z∈A, then
(5.5)τa+⟶a↓0a.s.τ,τa-⟶a↓0a.s.τ.Proof.
LetZ(0)=z∈A and
(5.6)Bz:={Z(t)iscontinuous}∩{τ=τ′}∩{τ<∞}.
By the hypotheses, we have that ℙ[Bz]=1. Observe that τa-≤τ≤τa+ for all a>0, then we need to prove that, for all ω∈Bz and α>0, there exists γ(ω,α)>0 such that, for all 0<a<γ,
(5.7)0≤τa+(ω)-τ(ω)<α,0≤τ(ω)-τa-(ω)<α.
Let ω∈Bz and α>0, then Z(t,ω) is continuous and τ(ω)<∞. We first prove the continuity for τa+. Define
(5.8)γ+(α,ω):=sup⁡{t∈[τ(ω),τ(ω)+α),Z(t,ω)∉A¯}{d(Z(t,ω),∂A)}.
Since Z(t,ω) is continuous, then γ+(α,ω)>0. So, there exists t+∈[τ(ω),τ(ω)+α) such that Z(t+,ω)∉Aa+ for all 0<a≤(γ+/2). Let γ:=γ+/2, then, for all 0<a≤γ, we get that τa+(ω)∈[τ(ω),τ(ω)+α) and so
(5.9)0≤τa+(ω)-τ(ω)<α.
For τa-, we proceed in a similar way. Let
(5.10)β(ω):=inf⁡t≤τ(ω)-α{d(Z(t,ω),∂A)}.
Since τ(ω)=τ′(ω) and Z(t,ω) is continuous, we get that β(ω)>0. Define
(5.11)γ-(α,ω):=sup⁡t∈(τ(ω)-α,τ(ω)]{d(Z(t,ω),∂A)}.
Again, it follows from the continuity of Z(t,ω) that γ-(α,ω)>0. So, there exists t-∈(τ(ω)-α,τ(ω)] such that Z(t-,ω)∉Aa- for all 0<a≤(β∧γ-)/2. Let γ:=(β∧γ-)/2, then, for all 0<a≤γ, we get that τa-(ω)∈(τ(ω)-α,τ(ω)] and so
(5.12)0≤τ(ω)-τa-(ω)<α
and the proof is complete.Remark 5.2.
For the continuity ofτa-, we need the extra constant β. If we only consider the constant γ-, then we can guarantee that the process Z is “close” to ∂A in the interval (τ-α,τ] but we cannot guarantee that it is the first time that it does it and so we do not necessarily have that τa- and τ are “close enough.”Theorem 5.3.
Let{Z(t)}t≥0 be a stochastic process with Z(0)=z∈ℝd and A⊂ℝd an open, connected set with regular boundary. Let {zn} be a sequence such that
(5.13)zn⟶n→∞z0
with zn,z0∈A. Denote by Z0 and Zn the stochastic processes with initial conditions z0 and zn, respectively. Define
(5.14)τ:=inf⁡⁡{t>0∣Z0(t)∉A¯},τn:=inf⁡⁡{t>0∣Zn(t)∉A¯}.
Assume that τ<∞ and τ=τ′ a.s., where
(5.15)τ′:=inf⁡⁡{t>0∣Z0(t)∉A}.
Assume also that, for all T>0,
(5.16)‖Zn-Z‖T⟶n→∞P0.
Then,
(5.17)τn⟶n→∞Pτ.Proof.
Letϵ>0 and η>0. We need to prove that there exists N(ϵ,η)∈ℕ such that, for all n≥N(ϵ,η), we have
(5.18)P[|τn-τ|≤ϵ]≥1-η.
Since τ<∞ a.s., then there exists M=M(η) such that
(5.19)P[τ≤M]≥1-η3.
Thanks to Theorem 5.1, we get the existence of a0=a0(ϵ,η) such that, for all a≤a0(ϵ,η), it is satisfied that
(5.20)P[τa+≤τ+ϵ,τa-≥τ-ϵ]≥1-η3.
Let a1:=a0/2. By hypothesis, we have the existence of N=N(M,a1,ϵ,η)=N(ϵ,η) such that for all n≥N(ϵ,η)(5.21)P[‖Zn-Z‖M+ϵ≤a1]≥1-η3.
So, over the intersection of the three sets, we get
(5.22){τ≤M}∩{τa1+≤τ+ϵ,τa1-≥τ-ϵ}∩{‖Zn-Z‖M+ϵ≤a1}⊂{τa1-≤τn≤τa1+}∩{τa1+≤τ+ϵ,τa1-≥τ-ϵ}⊂{τ-ϵ≤τn≤τ+ϵ}.
The first inclusion follows from the fact that, over the set {τ≤M}∩{∥Zn-Z∥M+ϵ≤a1}, we have that the process Zn cannot leave the set A before the process Z leaves the set Aa1- and so we get the τa1-≤τn. On the intersection of the three sets, we have that τa1+≤M+ϵ. Considering a similar argument, we obtain the remaining part, that is, τn≤τa1+.
Finally, we get from inequalities (5.19), (5.20), (5.21) and Bonferroni's inequality that for all n≥N(ϵ,η),(5.23)P[τ-ϵ≤τn≤τ+ϵ]≥1-η
and the proof is complete.


## 5.2. Additional Results

Theorem 5.4.
Let{(tn,xn)}n∈ℕ⊂[0,∞)×ℝd be a sequence such that (tn,xn)→n→∞(t,x). Denote by Xn and X the solutions of the following equations:
(5.24)dXn(s)=b(tn-s,Xn(s))ds+σ(tn-s,Xn(s))dW(s),Xn(0)=xn,dX(s)=b(t-s,X(s))ds+σ(t-s,X(s))dW(s),X(0)=x.
Then, for all T>0,
(5.25)‖Xn-X‖T⟶n→∞P0.Proof.
This theorem is consequence of Theorem 1.5 in ChapterV of [26].The following theorem is Theorem 9 of Chapter 3 in [3]. Let 0≤T0<T1 and let A⊂ℝd be a bounded open set with C2 boundary. Since σ, b, c, and f are locally Lipschitz, then they are locally Hölder of any order β∈(0,1)Theorem 5.5.
AssumeH1 and H2. Consider the following differential equation:
(5.26)-ut(t,x)+L[u](t,x)+c(t,x)u(t,x)=-f(t,x)(t,x)∈[T0,T1]×A,u(T0,x)=g(T0,x)forx∈A,u(t,x)=g(t,x)for(t,x)∈(T0,T1]×∂A.
If g is continuous, then there exists a classical solution w∈C([T0,T1)×A¯)∩C1,2,β((T0,T1)×A) of (5.26).Remark 5.6.
Letw be the solution of (5.26) and define z as w(t,x)=ec0tz(t,x) in [T0,T1]×A. Then, z fulfils (5.26) with c′=c-c0 and f′(t,x)=e-c0tf(t,x). And so the hypotheses of Theorem 9 of Chapter 3 in [3] are satisfied.



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*Source: 469806-2011-06-22.xml*