Mixed Boundary Value Problems of Semilinear Elliptic PDEs and BSDEs with Singular Coefficients
aa r X i v : . [ m a t h . P R ] D ec Mixed Boundary Value Problems of SemilinearElliptic PDEs and BSDEs with Singular Coefficients
Xue Yang Tusheng Zhang ∗ School of Mathematics, University of Manchester,Manchester, UK, M13 9PL
Abstract
In this paper, we prove that there exists a unique weak solution to the mixed boundaryvalue problem for a general class of semilinear second order elliptic partial differentialequations with singular coefficients. Our approach is probabilistic. The theory of Dirichletforms and backward stochastic differential equations with singular coefficients and infinitehorizon plays a crucial role.
Keywords:
Dirichlet forms; Quadratic forms; Fukushima’s decomposition; Mixed bound-ary value problem; Backward stochastic differential equations; Reflecting diffusion processes. ∗ Corresponding author (E-mail: [email protected]). Introduction
In this paper, our aim is to use probabilistic methods to solve the mixed boundary valueproblem for semilinear second order elliptic partial differential equations (called PDEs forshort) of the following form: ( Lu ( x ) = − F ( x, u ( x ) , ∇ u ( x )) , on D ∂u∂γ ( x ) − b B · n ( x ) u ( x ) = Φ( x ) on ∂D (1.1)The elliptic operator L is given by : L = 12 ∇ · ( A ∇ ) + B · ∇ − ∇ · ( ˆ B · ) + Q (1.2)= 12 d X i,j =1 ∂∂x i (cid:18) a ij ( x ) ∂∂x j (cid:19) + d X i =1 B i ( x ) ∂∂x i − div ( ˆ B · ) + Q ( x )on a d-dimensional smooth bounded Euclidean domain D . A ( x ) = ( a ij ) ≤ i,j ≤ d : R d → R d ⊗ R d is a smooth, symmetric matrix-valued function which isuniformly elliptic. That is, there is a constant λ > λ I d × d ≤ A ( · ) ≤ λI d × d . (1.3)Here B = ( B , ..., B d ) and ˆ B = ( ˆ B , ..., ˆ B d ) : R d → R d are Borel measurable functions, whichcould be singular, and Q is a real-valued Borel measurable function defined on R d such that,for some p > d , I D ( | B | + | ˆ B | + | Q | ) ∈ L p ( D ) .L is rigorously determined by the following quadratic form: Q ( u, v ) := ( − Lu, v ) L ( D ) = 12 X i,j Z D a ij ( x ) ∂u∂x i ∂v∂x j dx − X i Z D B i ( x ) ∂u∂x i v ( x ) dx − X i Z D ˆ B i ( x ) ∂v∂x i u ( x ) dx − Z D Q ( x ) u ( x ) v ( x ) dx. Details about the operator L can be found in [9], [16] and [20].The function F ( · , · , · ) in (1 .
1) is a nonlinear function defined on R d × R × R d and Φ( x ) is abounded measurable function defined on the boundary ∂D and γ = An , where n denotes theinward normal vector field defined on the boundary ∂D .To solve the problem (1.1), it turns out that we need to establish the existence and unique-ness of solutions of backward stochastic differential equations (BSDEs) with singular coeffi-cients and infinite horizon, which is of independent interest.Probabilistic approaches to boundary value problem of second order differential operatorshave been adopted by many authors and the earliest work went back as early as 1944 in [12].2here has been a lot of study on the Dirichlet boundary problem (see [1], [8], [3],[6], [11] and[22]). However, there are not many articles on the probabilistic approaches to the Neumannboundary problem.When A = I , B = 0 and ˆ B = 0, the following Neumann boundary problem (cid:26) △ u ( x ) + qu ( x ) = 0 , on D ∂u∂n ( x ) = φ ( x ) on ∂D was solved in [1] and [11], which also gives the solution the following representation: u ( x ) = E x [ Z ∞ e R t q ( B u ) du φ ( B t ) dL t ] , where ( B t ) t> is the reflecting Brownian motion on the domain D associated with the infinites-imal generator G = 12 △ , and L t , t > L t = R t I ∂D ( B s ) dL s .But when ˆ B = 0, the term ∇· ( ˆ B · ) is just a formal way of writing because the divergence doesnot exist as ˆ B is only a measurable vector field. It should be interpreted in the distributionalsense. For this reason, the term ∇ · ( ˆ B · ) can not be handled by Girsanov transform or Feyman-Kac transform.The study of the boundary value problems for the general operator L in the PDE literature(see e.g. [9], [20]) was always carried out under the extra condition: − div ( ˆ B ) + Q ( x ) ≤ F = 0, i.e. the linear case, problem (1 .
1) was studied in [4]( see also [3] for theDirichlet boundary problem). The term ∇ · ( ˆ B · ) is tackled using the time-reversal of Girsanovtransform of the symmetric reflecting diffusion (Ω , P x , X t , t >
0) associated with the operator L = 12 ∇ · ( A ∇ ) . The semigroup S t associated with the operator L has the following representation (see [5]): S t f ( x ) = E x [ f ( X t ) exp( Z t ( A − B ) ∗ ( X s ) dM s + ( Z t ( A − ˆ B ) ∗ ( X s ) dM s ) ◦ γ t − Z t ( B − ˆ B ) A − ( B − ˆ B ) ∗ ( X s ) ds + Z t Q ( X s ) ds )] , where M is the martingale part of the diffusion X and γ t is the reverse operator.3he main purpose of this paper is to study the nonlinear equation (1 . F = 0), whichcan not be handled by the methods used for the linear case. Our approach is first to solve abackward stochastic differential equation (BSDE) with singular coefficients and infinite horizonto produce a candidate for the solution of the boundary value problem and then to show thatthe candidate is indeed a solution. The results we obtained for BSDEs with infinite horizonare of independent interest.We would like to mention that the first results on BSDEs and probabilistic interpretationof solutions of semilinear parabolic PDEs via BSDEs were obtained by Peng and pardoux in[19], [17] and [18]. There the operator L is smooth and the solution is a viscosity solution. Westress that the solutions we considered for PDEs in this paper are Soblev (also called weak)solutions, not viscosity solutions.In [22], the corresponding Dirichlet problem for the semilinear elliptic PDEs: (cid:26) Lu ( x ) = − F ( x, u ( x ) , ∇ u ( x )) , on Du ( x ) = Φ( x ) on ∂D (1.4)was solved. The strategy in [3], [22] is to transform the general operator L by a kind of h-transform to an operator of the form: L = ∇ ( A ∇ ) + b · ∇ + q which does not have the ”bad”term such as ∇ ( ˆ B · ). This idea is used in current paper too.The BSDEs we studied are inspired by the ones in [10] where the author gave a probabilisticinterpretation of the solution to the following Neumann problem: (cid:26) ( △ − ν ) u ( x ) = 0 , on D ∂u∂n = φ, on ∂D The content of the paper as follows. In Section 2, we study the following BSDEs withinfinite horizon: dY ( t ) = − F ( X ( t ) , Y ( t ) , Z ( t )) dt + e R t q ( X ( u )) dt Φ( X ( s )) dL t + h Z ( t ) , dM ( t ) i , lim t →∞ e R t d ( X ( u )) du Y t = 0 in L (Ω) , (1.5)where ( X ( t )) t> is the reflecting diffusion associated with an infinitesimal generator of theform: A = ∇ ( A ∇ ) + b · ∇ , M ( t ) is the martingale part of X ( t ), L t is the boundary localtime of X and d ( · ) is an appropriate measurable function. The existence and uniqueness of an L -solution ( Y, Z ) is obtained.In Section 3, we solve the linear PDEs of the form: (cid:26) ∇ ( A ∇ u )( x ) + b · ∇ u ( x ) + qu ( x ) = F ( x ) , on D ∂u∂γ ( x ) = φ ( x ) on ∂D . (1.6)under the condition: E x [ Z ∞ e R t q ( X ( u )) du dL t ] < ∞ x ∈ ¯ D . Useful estimates for local time and Girsanov density are proved which willalso be used in subsequent sections.In Section 4, we obtain the solution of the semilinear PDE: (cid:26) ∇ ( A ∇ u )( x ) + b · ∇ u ( x ) + qu ( x ) = G ( x, u ( x ) , ∇ u ( x )) , on D ∂u∂γ ( x ) = φ ( x ) on ∂D . (1.7)To this end, we first use the solution ( Y x ( t ) , Z x ( t )) of the BSDE (1 .
5) to produce a candidate u ( x ) = E x [ Y x (0)] and then find a solution u of an equation like (1.6) with a given F ( x ) := G ( x, u ( x ) , v ( x )). Finally we identify u with u . In Section 5, we consider the general problem: ( Lu ( x ) = − F ( x, u ( x )) , on D ∂u∂γ ( x ) − b B · n ( x ) u ( x ) = Φ( x ) on ∂D . (1.8)We apply the transformation introduced in [3] to transform the problem (1 .
8) to a problemlike (1 . .
8) under thecondition that the L p norm of ˆ B is sufficiently small.To remove some of the restrictions imposed on ˆ B in Section 5, in Section 6, we study the L -solutions of the BSDEs (1 .
5) under appropriate conditions. Our approach is inspired bythe one in [2]. The study of L -solutions and L -solutions of the BSDEs (1.5) are carried outin Section 2 and Section 6 separately because the methods used for these two cases are quitedifferent. Consider the operator L = 12 d X i,j =1 ∂∂x i (cid:18) a ij ( x ) ∂∂x j (cid:19) + d X i =1 b i ( x ) ∂∂x i on the domian D equipped with the Neumann boundary condition: ∂∂γ := h An, ∇·i = 0 , on ∂D.
By [13], there exists a unique reflecting diffusion process denoted by (Ω , F t , X x ( t ) , P x , θ t , x ∈ D )associated with the generator L .Here θ : Ω → Ω is the shift operator defined as follows: X x ( s )( θ t · ) = X x ( t + s ) , s, t ≥ . Let E x denote the expectation under the measure P x .Set ˜ b = { ˜ b , ..., ˜ b d } , where ˜ b i = P j ∂a ij ∂x j + b i .Then the process X x ( t ) has the following decomposition: X x ( t ) = X x (0) + M x ( t ) + Z t ˜ b ( X x ( s )) ds + Z t An ( X x ( s )) dL s , P x − a.s.. (2.1)5ere M x ( t ) is a F t square integrable continuous martingale additive functional. And L t is apositive increasing continuous additive functional satisfying L t = R t I { X x ( s ) ∈ ∂D } dL s .We write X x ( t ) as X ( t ) for short in the following discussion.In this section, we will study the backward stochastic differential equations with singularcoefficients and infinite horizon associated with the martingale part M x ( t ) and the local time L t . A unique L solution of such BSDEs is obtained.Let g ( ω, t, y, z ) : Ω × R + × R × R d → R be a progressively measurable function. Considerthe following conditions: (A.1) ( y − y )( g ( t, y , z ) − g ( t, y , z )) ≤ − a ( t ) | y − y | , (A.2) | g ( t, y, z ) − g ( t, y, z ) | ≤ a | z − z | , (A.3) | g ( t, y, z ) | ≤ | g ( t, , | + a ( t )(1 + | y | ).Here a ( t ) and a ( t ) are two progressively measurable processes and a is a constant.Set a ( t ) = − a ( t ) + δa , for some constant δ > λ , where λ is the constant appeared in (1 . Lemma 2.1
Assume the conditions (A.1)-(A.3) and E x [ Z ∞ e R t a ( u ) du | g ( t, , | dt ] < ∞ . Then there exists a unique solution ( Y x ( t ) , Z x ( t )) to the following backward stochastic differ-ential equation: Y x ( t ) = Y x ( T ) + Z Tt g ( s, Y x ( s ) , Z x ( s )) ds − Z Tt < Z x ( s ) , dM x ( s ) >, t < T ;lim t →∞ e R t a ( u ) du Y x ( t ) = 0 , in L (Ω) . (2.2) Moreover, E x [sup t e R t a ( u ) du | Y x ( t ) | ] < ∞ and E x [ Z ∞ e R s a ( u ) du | Z x ( s ) | ds ] < ∞ . (2.3) Proof.
Existence:The proof of this lemma is similar to that of Theorem 3.2 in [22], but the terminal condi-tions here are different. By Theorem 3.1 in [22], the following BSDE has a unique solution( Y nx ( t ) , Z nx ( t )): Y nx ( t ) = Z nt g ( s, Y nx ( s ) , Z nx ( s )) ds − Z nt < Z nx ( s ) , dM x ( s ) >, t ≤ n ; (2.4)and moreover, Y nx ( t ) = 0 , Z nx ( t ) = 0 , t > n. t > n > m > t . It follows that e R t a ( u ) du | Y nx ( t ) − Y mx ( t ) | + Z ∞ t e R s a ( u ) du h A ( X ( s ))( Z nx ( s ) − Z mx ( s )) , ( Z nx ( s ) − Z mx ( s )) i ds = − Z nt a ( s ) e R s a ( u ) du | Y nx ( s ) − Y mx ( s ) | ds +2 Z nt e R s a ( u ) du ( Y nx ( s ) − Y mx ( s ))( g ( s, Y nx ( s ) , Z nx ( s )) − g ( s, Y mx ( s ) , Z mx ( s ))) ds +2 Z nm e R s a ( u ) du ( Y nx ( s ) − Y mx ( s )) g ( s, , ds − Z nt e R s a ( u ) du ( Y nx ( s ) − Y mx ( s )) < Z nx ( s ) − Z mx ( s ) , dM x ( t ) > Choose two positive numbers δ and δ such that δ > λ and δ + δ < δ . Then from2 Z nt e R s a ( u ) du ( Y nx ( s ) − Y mx ( s ))( g ( s, Y nx ( s ) , Z nx ( s )) − g ( s, Y mx ( s ) , Z mx ( s ))) ds ≤ − Z nt a ( s ) e R s a ( u ) du | Y nx ( s ) − Y mx ( s ) | ds +2 δ a Z nt e R s a ( u ) du | Y nx ( s ) − Y mx ( s ) | ds + 12 λδ Z nt e R s a ( u ) du h A ( X ( s ))( Z nx ( s ) − Z mx ( s )) , ( Z nx ( s ) − Z mx ( s )) i ds and 2 Z nm e R s a ( u ) du ( Y nx ( s ) − Y mx ( s )) g ( s, , ds ≤ δ a Z nm e R s a ( u ) du | Y nx ( s ) − Y mx ( s ) | ds + 12 δ a Z nm e R s a ( u ) du | g ( s, , | ds, it follows that E x [ e R t a ( u ) du | Y nx ( t ) − Y mx ( t ) | ] + 1 λ (1 − λδ ) E x [ Z ∞ t e R s a ( u ) du | Z nx ( s ) − Z mx ( s ) | ds ] ≤ δ a E x [ Z nm e R s a ( u ) du | g ( s, , | ds ] . This implies that E x [ Z ∞ e R s a ( u ) du | Z nx ( s ) − Z mx ( s ) | ds ] → , as m, n → ∞ . Hence there exists ˜ Z x such that˜ Z x = lim n →∞ e R · a ( u ) du Z nx in L ([0 , ∞ ) × Ω) .
7t the same time, we also obtain the following estimates:sup t e R t a ( u ) du | Y nx ( t ) − Y mx ( t ) | ≤ δ a Z nm e R s a ( u ) du | g ( s, , | ds +2 sup t | Z nt e R s a ( u ) du ( Y nx ( s ) − Y mx ( s )) < Z nx ( s ) − Z mx ( s ) , dM x ( t ) > | . Taking expectation on both sides of the above inequality, by BDG inequality, we obtain E x [sup t e R t a ( u ) du | Y nx ( t ) − Y mx ( t ) | ] ≤ δ a E x [ Z nm e R s a ( u ) du | g ( s, , | ds ]+ C E x [ { Z nt e R s a ( u ) du | Y nx ( s ) − Y mx ( s ) | | Z nx ( s ) − Z mx ( s ) | ds } ] ≤ δ a E x [ Z nm e R s a ( u ) du | g ( s, , | ds ] + 12 E x [sup t e R t a ( u ) du | Y nx ( t ) − Y mx ( t ) | ]+ C E x [ Z ∞ e R s a ( u ) du | Z nx ( s ) − Z mx ( s ) | ds ]Thus E x [sup t e R t a ( u ) du | Y nx ( t ) − Y mx ( t ) | ] ≤ δ a E x [ Z nm e R s a ( u ) du | g ( s, , | ds ] + 2 C E x [ Z ∞ e R s a ( u ) du | Z nx ( s ) − Z mx ( s ) | ds ] → , as m, n → ∞ . So, there exists { ˜ Y x ( t ) } such thatlim n →∞ E x [sup t | ˜ Y x ( t ) − e R t a ( u ) du Y nx ( t ) | ] = 0 . For any ε >
0, there exist a positive number N such that for any n ≥ N , E x [sup t | ˜ Y x ( t ) − e R t a ( u ) du Y nx ( t ) | ] < ε . For t > N , noticing Y Nx ( t ) = 0, it follows that E x [ | ˜ Y x ( t ) | ] ≤ E x [ | ˜ Y x ( t ) − e R t a ( u ) du | Y Nx ( t ) | ] + 2 E x [ e R t a ( u ) du | Y Nx ( t ) | ] ≤ E x [sup t | ˜ Y x ( t ) − e R t a ( u ) du | Y Nx ( t ) | ] + 2 E x [ e R t a ( u ) du | Y Nx ( t ) | ] < ε. t → E x [ | ˜ Y x ( t ) | ] = 0.By chain rule, it is easy to see from (2 .
4) that Y x ( t ) = e − R t a ( u ) du ˜ Y x ( t ) and Z x ( t ) = e − R t a ( u ) du ˜ Z x ( t )satisfy the equation (2 .
2) andlim t →∞ E x [ e R t a ( u ) du | Y x ( t ) | ] = lim t →∞ E x [ | ˜ Y x ( t ) | ] = 0 . From the above proof, we also see that (2 .
3) holds.Uniqueness:Suppose that ( Y x , Z x ) and ( Y x , Z x ) are two solutions of the equation (2 . Y x ( t ) = Y x ( t ) − Y x ( t ) and ¯ Z x ( t ) = Z x ( t ) − Z x ( t ) . Then d ( e R t a ( u ) du ¯ Y x ( t )) = − e R t a ( u ) du ( g ( t, Y x ( t ) , Z x ( t )) − g ( t, Y x ( t ) , Z x ( t ))) dt + a ( t ) e R t a ( u ) du ¯ Y x ( t ) dt + e R t a ( u ) du h ¯ Z x ( t ) , dM x ( t ) i . (2.5)By Ito’s formula, we get, for any t < T , e R t a ( u ) du | ¯ Y x ( t ) | + Z Tt e R t a ( u ) du h A ( X ( s )) ¯ Z x ( s ) , ¯ Z x ( s ) i ds = e R T a ( u ) du | ¯ Y x ( T ) | + 2 Z Tt e R s a ( u ) du ¯ Y x ( s )( g ( s, Y x ( s ) , Z x ( s )) − g ( s, Y x ( s ) , Z x ( s ))) ds − Z Tt a ( s ) e R s a ( u ) du | ¯ Y x ( s ) | ds − Z Tt a ( s ) e R s a ( u ) du ¯ Y x ( s ) h ¯ Z x ( s ) , dM x ( s ) i (2.6)By condition (A.1) and (A.2), we have2 Z Tt e R s a ( u ) du ¯ Y x ( s )( g ( s, Y x ( s ) , Z x ( s )) − g ( s, Y x ( s ) , Z x ( s ))) ds = 2 Z Tt e R s a ( u ) du ¯ Y x ( s )( g ( s, Y x ( s ) , Z x ( s )) − g ( s, Y x ( s ) , Z x ( s ))) ds + 2 Z Tt e R s a ( u ) du ¯ Y x ( s )( g ( s, Y x ( s ) , Z x ( s )) − g ( s, Y x ( s ) , Z x ( s ))) ds ≤ − Z Tt a ( s ) e R s a ( u ) du | ¯ Y x ( s ) | ds + a Z Tt e R s a ( u ) du ¯ Y x ( s ) | ¯ Z x ( s ) | ds ≤ − Z Tt a ( s ) e R s a ( u ) du | ¯ Y x ( s ) | ds + c ′ a Z Tt e R s a ( u ) du | ¯ Y x ( s ) | ds + a c ′ λ Z Tt e R s a ( u ) du | ¯ Z x ( s ) | ds. (2.7)9hoosing c ′ = 2 δa , we obtain | e R t a ( u ) du ¯ Y x ( t ) | + (1 − δλ ) Z Tt e R t a ( u ) du h A ( X ( s )) ¯ Z x ( s ) , ¯ Z x ( s ) i ds ≤ e R T a ( u ) du | ¯ Y x ( T ) | − Z Tt a ( s ) e R s a ( u ) du ¯ Y x ( s ) h ¯ Z x ( s ) , dM x ( s ) i (2.8)Taking expectation on both sides of the above inequality, we get that, for any t < T , E x [ e R t a ( u ) du | ¯ Y x ( t ) | ] ≤ E x [ e R T a ( u ) du | ¯ Y x ( T ) | ] . For both Y and Y satisfy the terminal condition in (2 . T →∞ E x [ e R T a ( u ) du | ¯ Y x ( T ) | ] = 0 , which leads to E x [ e R t a ( u ) du | ¯ Y x ( t ) | ] = 0.We conclude that Y x ( t ) = Y x ( t ) and Z x ( t ) = Z x ( t ). (cid:3) We now want to apply Lemma 2 . F ( x, y, z ) : R d × R × R d → R be a Borel measurable function. Consider the followingconditions: (D.1) ( y − y )( F ( x, y , z ) − F ( x, y , z )) ≤ − d ( x ) | y − y | , (D.2) | F ( x, y, z ) − F ( x, y, z ) | ≤ d | z − z | , (D.3) | F ( x, y, z ) | ≤ | F ( x, , z ) | + K ( x )(1 + | y | ).Set d ( x ) = − d ( x ) + δd for some constant δ > λ .The follows result follows from Lemma 2 . Lemma 2.2
Assume the conditions (D.1)-(D.3) and E x [ Z ∞ e R t d ( X ( u )) du | F ( X ( t ) , , | dt ] < ∞ . Then there exists a unique solution ( Y x ( t ) , Z x ( t )) to the following equation: Y x ( t ) = Y x ( T ) + Z Tt F ( X ( s ) , Y x ( s ) , Z x ( s )) ds − Z Tt < Z x ( s ) , dM x ( s ) >, t < T ;lim t →∞ e R t d ( X ( u )) du Y x ( t ) = 0 , in L (Ω) . (2.9)Consider the following condition instead of ( D. (D.3) ′ | F ( X ( t ) , y, z ) | ≤ K ( t ), for any y ∈ R and z ∈ R d .Let Φ be a bounded measurable function defined on ∂D , and function ˜ q ∈ L p ( D ), for p > d .The following theorem is the main result in this section.10 heorem 2.1 Assume the conditions (D.1), (D.2) and (D.3)’, E x [ Z ∞ e R s ˜ q ( X ( u )) du dL s ] < ∞ for some x ∈ D and for x ∈ D , E x [ Z ∞ e R t d ( X ( u )) du { e R t ˜ q ( X ( u )) du + | K ( t ) | } dt ] < ∞ . (2.10) Then there exists a unique solution ( Y x , Z x ) to the following BSDE: Y x ( t ) = Y x ( T ) + Z Tt F ( X ( s ) , Y x ( s ) , Z x ( s )) ds − Z Tt e R s ˜ q ( X ( u )) dt Φ( X ( s )) dL s − Z Tt h Z x ( s ) , dM x ( s ) i , f or t < T, (2.11) and lim t →∞ e R t d ( X ( u )) du Y t = 0 in L (Ω) . (2.12) Proof.
Uniqueness:Suppose that ( Y x , Z x ) and ( Y x , Z x ) are two solutions of the equation (2 .
11) satisfying (2 . Y x ( t ) = Y x ( t ) − Y x ( t ) and ¯ Z x ( t ) = Z x ( t ) − Z x ( t ) . Then d ( e R t d ( X ( u )) du ¯ Y x ( t )) = − e R t d ( X ( u )) du ( F ( X ( t ) , Y x ( t ) , Z x ( t )) − F ( X ( t ) , Y x ( t ) , Z x ( t ))) dt + d ( X ( t )) e R t d ( X ( u )) du ¯ Y x ( t ) dt + e R t d ( X ( u )) du h ¯ Z x ( t ) , dM x ( t ) i . (2.13)By Ito’s formula, we get, for any t < T , e R t d ( X ( u )) du | ¯ Y x ( t ) | + Z Tt e R t d ( X ( u )) du h A ( X ( s )) ¯ Z x ( s ) , ¯ Z x ( s ) i ds = e R T d ( X ( u )) du | ¯ Y x ( T ) | + 2 Z Tt e R s d ( X ( u )) du ¯ Y x ( s )( F ( X ( s ) , Y x ( s ) , Z x ( s )) − F ( X ( s ) , Y x ( s ) , Z x ( s ))) ds − Z Tt d ( X ( s )) e R s d ( X ( u )) du | ¯ Y x ( s ) | ds − Z Tt d ( X ( s )) e R s d ( X ( u )) du ¯ Y x ( s ) h ¯ Z x ( s ) , dM x ( s ) i (2.14)By (D.1) and (D.2), we have2 Z Tt e R s d ( X ( u )) du ¯ Y x ( s )( F ( X ( s ) , Y x ( s ) , Z x ( s )) − F ( X ( s ) , Y x ( s ) , Z x ( s ))) ds
11 2 Z Tt e R s d ( X ( u )) du ¯ Y x ( s )( F ( X ( s ) , Y x ( s ) , Z x ( s )) − F ( X ( s ) , Y x ( s ) , Z x ( s ))) ds + 2 Z Tt e R s d ( X ( u )) du ¯ Y x ( s )( F ( X ( s ) , Y x ( s ) , Z x ( s )) − F ( X ( s ) , Y x ( s ) , Z x ( s ))) ds ≤ − Z Tt d ( X ( s )) e R s d ( X ( u )) du | ¯ Y x ( s ) | ds + d Z ∞ t e R s d ( X ( u )) du ¯ Y x ( s ) | ¯ Z x ( s ) | ds ≤ − Z Tt d ( X ( s )) e R s d ( X ( u )) du | ¯ Y x ( s ) | ds + cd Z Tt e R s d ( X ( u )) du | ¯ Y x ( s ) | ds + d cλ Z Tt e R s d ( X ( u )) du | ¯ Z x ( s ) | ds. (2.15)Choosing c = 2 δd , we obtain from (2 . | e − R t d ( X ( u )) du ¯ Y x ( t ) | + (1 − δλ ) Z Tt e − R t d ( X ( u )) du h A ( X ( s )) ¯ Z x ( s ) , ¯ Z x ( s ) i ds ≤ e R T d ( X ( u )) du | ¯ Y x ( T ) | − Z Tt d ( X ( s )) e − R s d ( X ( u )) du ¯ Y x ( s ) h ¯ Z x ( s ) , dM x ( s ) i (2.16)Taking expectation on both sides of the above inequality and letting T tend to infinity, weobtain that E x [ e R t d ( X ( u )) du | ¯ Y x ( t ) | ] = 0We conclude that Y x ( t ) = Y x ( t ) and hence from (2 . Z x ( t ) = Z x ( t ).Existence:First of all, the assumption (2 .
10) implies (see [ ? ])sup x E x [ Z ∞ e R s ˜ q ( X ( u )) du dL s ] < ∞ . ◦ : There exists ( p x ( t ) , q x ( t )) such that dp x ( t ) = e R t ˜ q ( X ( u )) du Φ( X ( t )) dL t + < q x ( t ) , dM x ( t ) >, (2.17)and e R t d ( X ( u )) du p x ( t ) → t → ∞ in L (Ω).In fact, let p x ( t ) := − E x [ Z ∞ t e R s ˜ q ( X ( u )) du Φ( X ( s )) dL s |F t ]= Z t e R s ˜ q ( X ( u )) du Φ( X ( s )) L s − E x [ Z ∞ e R s ˜ q ( X ( u )) du Φ( X ( s )) dL s |F t ] . (2.18)By the martingale representation theorem in [22], there exists a process q x ( t ), such that − E x [ Z ∞ e R s ˜ q ( X ( u )) du Φ( X ( s )) dL s |F t ] = − E x [ Z ∞ e R s ˜ q ( X ( u )) du Φ( X ( s )) dL s ]+ Z t < q x ( s ) , dM x ( s ) > . (2.19)12hen ( p x , q x ) satisfies the equation (2 . p x ( t ) := − E x [ Z ∞ t e R s ˜ q ( X ( u )) du Φ( X ( s )) dL s |F t ]= − e R t ˜ q ( X ( u )) du E x [ Z ∞ t e R st ˜ q ( X ( u )) du Φ( X ( s )) dL s |F t ]= − e R t ˜ q ( X ( u )) du E x [ Z ∞ e R s + tt ˜ q ( X ( u )) du Φ( X ( s + t )) dL s + t |F t ]= − e R t ˜ q ( X ( u )) du E x [ Z ∞ e R s ˜ q ( X ( u + t )) du Φ( X ( s + t )) dL s + t |F t ]= − e R t ˜ q ( X ( u )) du E X ( t ) [ Z ∞ e R l ˜ q ( X ( u )) du Φ( X ( l )) dL l ] (2.20)The last equality follows from the fact that L t + s = L t + L s ◦ θ t . Therefore,sup x | p x ( t ) | ≤ e R t ˜ q ( X ( u )) du sup x ∈ D | Φ( x ) | · sup x ∈ ¯ D E x [ Z ∞ e R t ˜ q ( X ( u )) du dL t ] . Set M = sup x ∈ D | Φ( x ) | · sup x ∈ ¯ D E x [ Z ∞ e R t ˜ q ( X ( u )) du dL t ].In view of (2 . t →∞ e R t ( d +˜ q )( X ( u )) du = 0 in L (Ω).Hence, e R t d ( X ( u )) du p x ( t ) ≤ M e R t ( d +˜ q )( X ( u )) du → as t → ∞ , in L (Ω) . (2.21)2 ◦ : Set g ( t, y, z ) = F ( X ( t ) , p x ( t ) + y, q x + z ). Then( y − y )( g ( t, y , z ) − g ( t, y , z ))= ( y − y )( F ( X ( t ) , p x ( t ) + y , q x + z ) − F ( X ( t ) , p x ( t ) + y , q x + z )) ≤ − d ( X ( t )) | y − y | . (2.22)and | g ( t, y, z ) − g ( t, y, z ) | = | F ( X ( t ) , p x ( t ) + y, q x + z ) − F ( X ( t ) , p x ( t ) + y, q x + z ) |≤ d | z − z | . (2.23)Moreover, E x [ Z ∞ e R t d ( X ( u )) du | g ( X ( t ) , , | dt ] ≤ E x [ Z ∞ e R t d ( X ( u )) du | F ( X ( t ) , p x ( t ) , q x ( t )) | dt ] ≤ E x [ Z ∞ e R t d ( X ( u )) du | K ( t ) | dt ] < ∞ . (2.24)13 satisfies all the conditions of the Lemma 2.2. Hence, there exist processes ( k x , l x ) such that dk x ( t ) = − g ( t, k x ( t ) , l x ( t )) dt + < l x ( t ) , dM x ( t ) >, and e R t d ( X ( u )) du k x ( t ) → , as t → ∞ .Putting Y x ( t ) = p x ( t ) + k x ( t ) and Z x ( t ) = q x ( t ) + l x ( t ), we find that ( Y x ( t ) , Z x ( t )) satisfies thefollowing equation dY x ( t ) = e R t ˜ q ( X ( u )) du φ ( X ( t )) dL t − F ( t, Y x ( t ) , Z x ( t )) dt + < Z x ( t ) , dM x > . and lim t →∞ e R t d ( X ( u )) du Y t = 0 . Corollary 2.1
Suppose all the assumptions in Theorem 2.1 hold. If, in addition, sup x E x [ Z ∞ e R t d ( X ( u )) du | K ( t ) | dt ] < ∞ , then it follows that sup x ∈ D | Y x (0) | < ∞ . Proof.
As shown in the proof of Theorem 2.1, Y x ( t ) has the decomposition: Y x ( t ) = p x ( t ) + k x ( t ).Setting t = 0 in (2 . | p x (0) | ≤ E X ( t ) [ | Z ∞ e R l ˜ q ( X ( u )) du Φ( X ( l )) dL l | ] ≤ k Φ k ∞ sup x E x [ Z ∞ e R l ˜ q ( X ( u )) du dL l ] < ∞ . (2.25)By Ito’s formula, we obtain de R t d ( X ( u )) du | k x ( t ) | = − e R t d ( X ( u )) du k x ( t ) g ( t, k x ( t ) , l x ( t )) dt + 2 e R t d ( X ( u )) du k x ( t ) d ( X ( t )) dt + 2 e R t d ( X ( u )) du k x ( t ) < l x ( t ) , dM x ( t ) > + e R t d ( X ( u )) du h A ( X ( t )) l x ( t ) , l x ( t ) i dt Choosing two positive numbers δ and δ such that δ > λ and δ + δ < δ , similar calculationsas in the proof of Theorem 2.1 yield that, for any t < T , E x [ e R t d ( X ( u )) du | k x ( t ) | ] + 1 λ (1 − λδ ) E x [ Z Tt e R s d ( X ( u )) du | l x ( s ) | ds ] ≤ E x [ e R Tt d ( X ( u )) du | k x ( T ) | ] + 12 δ d E x [ Z Tt e R s d ( X ( u )) du | g ( s, , | ds ] . t = 0, we have | k x (0) | = E x [ | k x (0) | ] ≤ E x [ e R T d ( X ( u )) du | k x ( T ) | ]+ 12 δ d E x [ Z T e R s d ( X ( u )) du | g ( s, , | ds ] . Let T → ∞ to obtain thatsup x | k x (0) | ≤ (cid:18) δ d sup x E x [ Z ∞ e R s d ( X ( u )) du | g ( s, , | ds ] (cid:19) < ∞ , where the fact that e R T d ( X ( u )) du k x ( T ) → T → ∞ , has been used. Hence, we havesup x | Y x (0) | ≤ sup x | p x (0) | + sup x | k x (0) | < ∞ . Set L = 12 ∇ · ( A ∇ ) + b · ∇ + q where b = ( b , ..., b d ) is a R d -valued Borel measurable function, and q is a Borel measurablefunction on R d such that: I D ( | b | + | q | ) ∈ L p ( D ) , p > d . In this section, we solve the following linear boundary value problem: (cid:26) ∇ · ( A ∇ u )( x ) + b · ∇ u ( x ) + q ( x ) u ( x ) = F ( x ) , on D ∂u∂γ ( x ) = φ on ∂D , (3.1)where F and φ are bounded measurable functions on D .It is well known that operator L defined on a bounded domain D with Neumann bound-ary condition ∂u∂γ ( x ) = 0 is associated with the quadratic form: E ( f, g ) : = − Z D L f ( x ) g ( x ) dx = 12 Z D h A ∇ f, ∇ g i dx − Z D b · ∇ f ( x ) g ( x ) dx − Z D q ( x ) f ( x ) g ( x ) dx Definition 3.1
A bounded continuous function u ( x ) defined on D is a weak solution of theproblem (3 . if u ∈ W , ( D ) , and for any g ∈ C ∞ ( D ) , E ( u, g ) = Z ∂D φ ( x ) g ( x ) σ ( dx ) − Z D F ( x ) g ( x ) dx, where σ denotes the d − dimensional Lebesgue measure on ∂D . L = 12 ∇ · ( A ∇ u ) (3.2)on domain D with boundary condition ∂u∂γ = 0 on ∂D . L is associated with a reflecting diffusion process ( X , P x ). By [13], X has the followingdecomposition: dX t = σ ( X t ) dW t + 12 ∇ A ( X t ) dt + γ ( X t ) dL t ,L t = Z t I { X s ∈ ∂D } dL s , (3.3)where the matrix σ ( x ) is the positive definite symmetric square root of the matrix A ( x ) and { W t } t> is a d-dimensional standard Brownian motion.It is well known that operator L is associated with the regular Dirichlet form: E ( u, v ) = 12 Z D a ij ∂u∂x i ∂v∂x j dx and the domain of E is W , ( D ) := { u ∈ L ( D ) : ∂u∂x i ∈ L ( D ) } .The following lemma can be proved similarly as the Corollary 3.8 in [11] using the heat kernelestimates in [21]. Lemma 3.1
There exists a constant
K > , such that sup x ∈ ¯ D E x [ L t ] ≤ K √ t and inf x ∈ ¯ D E x [ L t ] > . Moreover, we have sup x ∈ ¯ D E x [( L t ) n ] ≤ K n t n , for some constant K n > . Set M t = R t σ ( X s ) dW s and Z t = e R t − R t bA − b ∗ ( X s ) ds + R t q ( X s ) ds , (3.4)where b ∗ is the transpose of the row vector b .The proof of the following two lemmas are inspired by that of the Lemma 2.1 and Theorem2.2 in [11]. Lemma 3.2
For t > , there are two strictly positive functions M ( t ) and M ( t ) such that,for any x ∈ D , M ( t ) ≤ E x [ R t Z s dL s ] ≤ M ( t ) . Furthermore, M ( t ) → as t → . roof. ◦ : Put ˜ M ( t ) = e R t − R t bA − b ∗ ( X s ) ds , (3.5) e q ( t ) = e R t q ( X s ) ds ,M q ( t ) = sup x ∈ D E x [ Z t | q ( X s ) | ds ] . Then we havesup x ∈ D E x [ Z t Z s dL s ] = sup x ∈ D E x [ Z t ˜ M ( s ) e q ( s ) dL s ] ≤ sup x ∈ D E x [ max ≤ s ≤ t | ˜ M ( s ) | ] · sup x ∈ D E x [ e | q | ( t )( L t ) ] ≤ sup x ∈ D E x [ | ˜ M ( t ) | ] | {z } ( I ) · sup x ∈ D E x [ e | q | ( t )] | {z } ( II ) · sup x ∈ D E x [( L t ) ] | {z } ( III ) (3.6)By Khash’Minskii’s lemma and Theorem 2.1 in [15], ( I ) and ( II ) are bounded if t belongs toa bounded interval. Because of E x [( L t ) n ] ≤ K n t n , we see that M ( t ) := K ( I )( II ) √ t is therequired upper bound.2 ◦ : Since E x [ L t ] ≤ E x [ Z t ˜ M − ( s ) e − q ( s ) dL s ] · E x [ Z t ˜ M ( s ) e q ( s ) dL s ] , (3.7)we obtain E x [ Z t ˜ M ( s ) e q ( s ) dL s ] ≥ E x [ L t ] E x [ R t ˜ M − ( s ) e − q ( s ) dL s ] . (3.8)Here ˜ M − ( t ) = e − R t + R t bA − b ∗ ( X s ) ds = e − R t − R t bA − b ∗ ( X s ) ds · e R t bA − b ∗ ( X s ) ds := N ( t ) · e R t bA − b ∗ ( X s ) ds (3.9)By the proof of the first part, replacing ˜ M t , q by N t and bA − b ∗ − q respectively, it is seen thatthere exists K ( t ) > x ∈ D E x [ R t ˜ M − ( s ) e − q ( s ) dL s ] ≤ K ( t ).As inf x ∈ D E x [ L t ] >
0, we complete the proof of the lemma by setting M ( t ) = inf x ∈ D E x [ L t ] K ( t ) . (cid:3) Set G ( x ) := E x [ R ∞ Z s dL s ]. Lemma 3.3
If there is a point x ∈ D , such that G ( x ) < ∞ , then there are two positiveconstants K and β such that sup x ∈ D E x [ Z t ] ≤ Ke − βt . roof. By Girsanov Theorem and Feymann-Kac formula, L = ∇ · ( A ∇ ) + b · ∇ + q is associatedwith the semigroup { T t } t> , where T t f ( x ) = E x [ Z t f ( X t )] for f ∈ L ( D ).By the upper and lower bound estimates of the heat kernel p ( t, x, y ) associated with T t in[21], the following inequality holds, c − Z D f ( x ) dx ≤ E x [ Z f ( X )] ≤ c Z D f ( x ) dx, (3.10)where c is a positive constant. Since G ( x ) = ∞ X n =0 E x [ Z n E X n [ Z Z s L ( ds )]] ≥ M (1) ∞ X n =0 E x [ Z n ]and G ( x ) < ∞ , there is a positive integer number N such that12 c ≥ E x [ Z N ] = E x [ Z E X [ Z N − ]] ≥ c − Z D E x [ Z N − ] m ( dx ) . This implies Z D E x [ Z N − ] m ( dx ) ≤ c . Thus sup x ∈ D E x [ Z N ] = sup x ∈ D E x [ Z E X [ Z N − ]] ≤ c Z D E x [ Z N − ] m ( dx ) ≤ . (3.11)For any t >
0, there exists a positive number n such that tN ∈ [ n − , n ). Then by (3 . E x [ Z t ] ≤ n − E x [ Z t − N ( n − ] ≤ sup x ∈ D, ≤ t ≤ N E x [ Z t ] ! n − ≤ x ∈ D, ≤ t ≤ N E x [ Z t ] e − ln N t . (cid:3) (3.12) Theorem 3.1
If there exists x ∈ D such that G ( x ) < ∞ , then there exists a unique boundedcontinuous weak solution of the problem (3 . : Proof.
Existence :Due to Theorem 3.2 in [4], there exists a unique, bounded, continuous weak solution u of thefollowing problem: (cid:26) L u ( x ) = 0 , on D ∂u ∂γ ( x ) = φ on ∂D . (3.13)18hus by the linearity of the problem (3 . (cid:26) L u ( x ) = F ( x ) , on D ∂u ∂γ ( x ) = 0 on ∂D (3.14)The semigroup associated with operator L is { T t , t > } . By Lemma 3 .
3, we havesup x ∈ D | T t F ( x ) | = sup x ∈ D | E x [ Z t F ( X t )] | ≤ Ke − βt k F k ∞ . Then u ( x ) := Z ∞ T t F ( x ) dt is well defined and has the following bound:sup x ∈ D | u ( x ) | ≤ Kβ k F k ∞ . The function u ( x ) is also continuous on D.In fact, fixing any x ∈ D and ǫ >
0, we can firstly choose a constant t >
0, such thatsup z ∈ D | R t T s F ( z ) ds | < ǫ . And because T t u ( x ) is continuous, there exists a constant δ > y with | y − x | < δ , | T t u ( x ) − T t u ( y ) | ≤ ǫ .We find that T t u ( x ) = E x [ Z t u ( X t )] = E x [ Z t Z ∞ E X t [ Z s F ( X s )] ds ]= Z t E x [ Z t + s u ( X t + s )] ds = Z ∞ t T s F ( x ) ds = u ( x ) − Z t T s F ( x ) ds. (3.15)For any y satisfying | y − x | < δ , it follows that | u ( x ) − u ( y ) | ≤ | T t u ( x ) − T t u ( y ) | + | Z t T s F ( x ) ds | + | Z t T s F ( y ) ds | ≤ ǫ. (3.16)This implies that the function u is continuous on domain D .Denote the resolvents associated with operator L by { G β , β > } . Note that G β u ( x ) = Z ∞ e − βt T t u ( x ) dt = Z ∞ e − βt u ( x ) dt − Z ∞ e − βt Z t T s F ( x ) dsdt
19 1 β u ( x ) − Z ∞ Z t e − βt T s F ( x ) dsdt = 1 β u ( x ) − Z ∞ T s F ( x )( Z ∞ s e − βt dt ) ds = 1 β u ( x ) − β G β F ( x ) . (3.17)We have β ( u ( x ) − βG β u ( x )) = βG β F ( x ) . Therefore,lim β →∞ Z D β ( u ( x ) − βG β u ( x )) u ( x ) dx = lim β →∞ Z D βG β F ( x ) u ( x ) dx = Z D F ( x ) u ( x ) dx < ∞ . This implies that u ∈ D ( E ) (see [16]) and u is a weak solution of equation (3 . u = u + u is a bounded continuous weak solution of equation (3 . U niqueness :Let v and v be two bounded continuous weak solutions of the equation (3 . v − v is the solution of equation (3 .
13) with φ = 0. Then by the uniqueness of the equation (3 . v = v . (cid:3) Recall that L = 12 d X i,j =1 ∂∂x i (cid:18) a ij ( x ) ∂∂x j (cid:19) + d X i =1 b i ( x ) ∂∂x i and L = L + q are two operators both defined on the domain D and equipped with theNeumann boundary condition ∂∂γ = 0 on ∂D .(Ω , F t , X ( t ) , P x , x ∈ D ) is the reflecting diffusion process associated with the operator L withthe decomposition introduced in (2 . (cid:26) L u ( x ) = − G ( x, u ( x ) , ∇ u ( x )) , on D ∂u∂γ ( x ) = φ ( x ) on ∂D (4.1)Let E ( · , · ) be the quadratic form associated with the operator L : E ( u, v ) = 12 Z D < A ∇ u, ∇ v > dx − Z D < b, ∇ u > vdx − Z D quvdx. efinition 4.1 A bounded continuous function u ( x ) defined on D is called a weak solution ofthe equation (4 . if u ∈ W , ( D ) , and for any g ∈ C ∞ ( D ) , E ( u, g ) = Z ∂D φ ( x ) g ( x ) σ ( dx ) + Z D G ( x, u ( x ) , ∇ u ( x )) g ( x ) dx. Recall that L t is the boundary local time of X ( t ) defined in (2 .
1) and L t is the boundary localtime of X t in (3 . Lemma 4.1
Suppose that the function f satisfies E x [ R T e R t f ( X ( u )) du dL t ] < ∞ . Then it holdsthat E x [ Z T e R t f ( X ( u )) du dL t ] = E x [ Z T ˜ M t e R t f ( X u ) du dL t ] , where ˜ M t was defined in (3 . . The following lemma is deduced from Theorem 3.2 in [4].
Lemma 4.2
Suppose that the function ˜ q ∈ L p ( D ) and p > d . If there exists some point x ∈ D , such that E x [ Z ∞ e R t ˜ q ( X ( u )) du dL t ] < ∞ , (4.2) then it holds that sup x E x [ Z ∞ e R t ˜ q ( X ( u )) du dL t ] < ∞ . Let G ( x, y, z ) : R d × R × R d → R be a bounded Borel measurable function. Introduce thefollowing conditions: (H.1) ( y − y )( G ( x, y , z ) − G ( x, y , z )) ≤ − h ( x ) | y − y | , (H.2) | G ( x, y, z ) − G ( x, y, z ) | ≤ h | z − z | .Set h ( t ) = − h ( X ( t )) + δh + q ( X ( t )) and ˜ h ( t ) = − h ( X ( t )) + δh for some constant δ > λ . Theorem 4.1
Suppose that the conditions (H.1) and (H.2) are satisfied. Assume E x [ Z ∞ e R t ( q ( X ( u ))+˜ h ( u )) du dt ] < ∞ , f or some x ∈ D, (4.3) and there exists some point x ∈ D , such that E x [ Z ∞ e R t q ( X ( u )) du dL t ] < ∞ . (4.4) Then the semilinear Neumann boundary value problem (4 . has a unique continuous weaksolution. roof. Set ˜ G ( X ( t ) , y, z ) := e R t q ( X ( u )) dt G ( x, e − R t q ( X ( u )) dt y, e − R t q ( X ( u )) dt z ) . Then ( y − y )( ˜ G ( X ( t ) , y , z ) − ˜ G ( X ( t ) , y , z )) ≤ − h ( x ) | y − y | (4.5)and | ˜ G ( X ( t ) , y, z ) − ˜ G ( X ( t ) , y, z ) | ≤ h | z − z | . (4.6)Note that ˜ G ( X ( t ) , y, z ) ≤ e R t q ( X ( u )) dt k G k ∞ . By Theorem 2.1 there exists a unique process ( ˆ Y x , ˆ Z x ) satisfying d ˆ Y x ( t ) = − ˜ G ( X ( t ) , ˆ Y x ( t ) , ˆ Z x ( t )) dt + e R t q ( X ( u )) du φ ( X ( t )) dL ( t ) + h ˆ Z x ( t ) , dM x ( t ) i e R t ˜ h ( u ) du ˆ Y x ( t ) → as t → ∞ . Furthermore, Corollary 2 . x ˆ Y x (0) < ∞ .From Ito’s formula, it follows that d ( e − R t q ( X ( u )) dt ˆ Y x ( t ))= − q ( X ( t )) e − R t q ( X ( u )) dt ˆ Y x ( t ) dt − e − R t q ( X ( u )) dt ˜ G ( X ( t ) , ˆ Y x ( t ) , ˆ Z x ( t )) dt + φ ( X ( t )) dL t + < e − R t q ( X ( u )) dt ˆ Z x ( t ) , dM x ( t ) > . Setting Y x ( t ) := e − R t q ( X ( u )) dt ˆ Y x ( t ) and Z x ( t ) := e − R t q ( X ( u )) dt ˆ Z x ( t ), we obtain dY x ( t ) = − ( q ( X ( t )) Y x ( t ) + G ( X ( t ) , Y x ( t ) , Z x ( t ))) dt + φ ( X ( t )) dL t + < Z x ( t ) , dM x ( t ) > . Moreover, e R t h ( u ) dt Y x ( t ) = e R t h ( u ) dt e − R t q ( X ( u )) dt ˆ Y x ( t ) = e R t ˜ h ( X ( u )) dt ˆ Y x ( t ) → as t → ∞ . (4.7)So by Ito’s formula, we have that, for any t < T , e R t h ( u ) du Y x ( t )= e R T h ( u ) du Y x ( T ) + Z Tt e R s h ( u ) du ( G ( X x ( s ) , Y x ( s ) , Z x ( t )) + q ( X x ( s )) Y x ( s )) ds − Z Tt e R s h ( u ) du φ ( X ( s )) dL s − Z Tt h ( s ) e R s h ( u ) du Y x ( s ) ds − Z Tt e R s h ( u ) du h Z x ( t ) , dM x ( t ) i . (4.8)22ut u ( x ) = Y x (0) and v ( x ) = Z x (0).Since Y x (0) = ˆ Y x (0), we know that u is a bounded function on domain D . By the Markovproperty of X and the uniqueness of ( Y x , Z x ) , it is easy to see that Y x ( t ) = u ( X ( t )) , Z x ( t ) = v ( X ( t )) . So that sup x ∈ D,t> | Y x ( t ) | ≤ k u k ∞ < ∞ . Now consider the following problem: (cid:26) L u ( x ) = − G ( x, u ( x ) , v ( x )) , on D ∂u∂γ ( x ) = φ ( x ) on ∂D (4.9)By Theorem 3.1, problem (4 .
9) has a unique continuous weak solution u ( x ). Next we will showthat u = u .Since u belongs to the domain of the Dirichlet form associated with the process X ( t ), it followsfrom the Fukushima’s decomposition that: du ( X ( t ))= − [ G ( X ( t ) , u ( X ( t )) , v ( X ( t ))) + q ( X ( t )) u ( X ( t ))] dt + φ ( X ( t )) dL ( t ) + h∇ u ( X ( t )) , dM x ( t ) i = − [ G ( X ( t ) , Y x ( t ) , Z x ( t )) + q ( X ( t )) u ( X ( t ))] + φ ( X ( t )) dL ( t ) + h∇ u ( X ( t )) , dM x ( t ) i From the condition (4 .
3) and the boundedness of u ( x ), it follows thatlim t →∞ E x [ e R t h ( u ) du u ( X ( t ))] ≤ k u k ∞ lim t →∞ E x [ e R t (˜ h + q )( u ) du ] = 0 . By Ito’s formula, it follows that, for any t < T , e R t h ( u ) du u ( X ( t ))= e R T h ( u ) du u ( X ( T )) + Z Tt e R s h ( u ) du [ G ( X ( s ) , Y x ( s ) , Z x ( s )) + q ( X ( s )) u ( X ( s ))] ds − Z Tt e R s h ( u ) du φ ( X ( s )) dL ( s ) − Z Tt h ( s ) e R s h ( u ) du u ( X ( s )) ds − Z Tt e R s h ( u ) du h∇ u ( X ( t )) , dM x ( t ) i . (4.10)Set v x ( t ) = u ( X ( t )) − Y x ( t ) and R x ( t ) = ∇ u ( X ( t )) − Z x ( t ) . Subtracting the equations (4 .
8) from (4 . t < T , e R t h ( u ) du v ( X ( t ))= e R T h ( u ) du v ( X ( T )) + Z ∞ t ( q ( X ( u )) − h ( u )) e R s h ( u ) du v ( X ( s )) ds − Z ∞ t e R s h ( u ) du < R x ( t ) , , dM x ( t ) > e R T h ( u ) du v ( X ( T )) − Z Tt ˜ h ( s ) e R s h ( u ) du v ( X ( s )) ds − Z Tt e R s h ( u ) du < R x ( t ) , dM x ( t ) > . Set g ( t ) = e R t h ( u ) du v ( t ). Taking conditional expectation on both sides of (4 . g ( t ) = E x [ g ( T ) − Z Tt ˜ h ( s ) g ( s ) ds |F t ]= E x [ g ( T )(1 − Z Tt ˜ h ( s ) ds ) + Z Tt Z Ts ˜ h ( s )˜ h ( s ) g ( s ) ds ds |F t ]= E x [ g ( T )(1 − Z Tt ˜ h ( s ) ds + 12 ( Z Tt ˜ h ( s ) ds ) )+( − Z Tt Z Ts Z Ts ˜ h ( s )˜ h ( s )˜ h ( s ) g ( s ) ds ds ds |F t ] . Keeping iterating, we obtain g ( t ) = E x [ g ( T )( n X k =0 ( − R Tt ˜ h ( s ) ds ) n n ! )+( − n +1 Z Tt Z Ts Z Ts ... Z Ts n − ˜ h ( s )˜ h ( s ) ... ˜ h ( s n ) g ( s n ) ds n ...ds ds |F t ]Since E x [ | g ( T ) | e R Tt | ˜ h | ( s ) ds ] < ∞ , letting n → ∞ , by dominated convergence theorem, it followsthat g ( t ) = E x [ g ( T ) e − R Tt ˜ h ( s ) ds |F t ] . Then v ( t ) = E x [ v ( T ) e R Tt ( h ( s ) − ˜ h ( s )) ds |F t ] ≤ ( k u k ∞ + k u k ∞ ) E x [ e R Tt q ( X ( s )) ds |F t ] . (4.11)Hence, it follows that0 ≤ e R t q ( X ( s )) ds | v ( t ) | ≤ ( k u k ∞ + k u k ∞ ) lim T →∞ E x [ e R T q ( X ( s )) ds |F t ] . (4.12)Since the condition (4 .
4) implies lim T →∞ E x [ e R T q ( X ( s )) ds ] = 0 , we deduce that E x [ e R t q ( X ( s )) ds | v ( t ) | ] = 0 and hence v ( t ) = 0, P x − a.s. .Therefore, for any t >
0, we have u ( X ( t )) = Y x ( t ) and ∇ u ( X ( t )) = Z x ( t ) by the uniquenessof the Doob-Meyer decomposition of semimartingales. In particular, u ( x ) = E x [ u ( X x (0))] =24 x [ Y x (0)] = u ( x ). This shows that u ( x ) is a weak solution of the equation (4 . u is another solution of the problem (4 . Y x ( t ) := ˜ u ( X ( t )) and ˜ Z x ( t ) := ∇ ˜ u ( X ( t )) satisfy the following equation d ˜ Y x ( t ) = − G ( X ( t ) , ˜ Y x ( t ) , ˜ Z x ( t )) dt − φ ( X ( t )) dL t + < ˜ Z x ( t ) , dM x ( t ) > . (4.13)Set ¯ Y x ( t ) = e R t q ( X ( u )) du ˜ Y x ( t ) and ¯ Z x ( t ) = e R t q ( X ( u )) du ˜ Z x ( t ).By chain rule, it follows that d ¯ Y x ( t ) = − ˜ G ( X ( t ) , ¯ Y x ( t ) , ¯ Z x ( t )) dt + e R t q ( X ( u )) du φ ( X ( t )) dL ( t ) + h ¯ Z x ( t ) , dM x ( t ) i Moreover, because ˜ u is bounded, we havelim t →∞ e R t ˜ h ( u ) du ¯ Y x ( t ) = lim t →∞ e R t h ( u ) du ˜ u ( X ( t )) = 0 . Therefore, from the uniqueness of the solution of the BSDE in Theorem 2.1, we have˜ Y x ( t ) = Y x ( t ) ˜ Z x ( t ) = Z x ( t ) . In particular, ˜ u ( x ) = E x [ ˜ Y x ( t )] = E x [ Y x ( t )] = u ( x ) . Recall the operator L = 12 ∇ · ( A ∇ ) + B · ∇ − ∇ · ( ˆ B · ) + Q on the domain D equipped with the mixed boundary condition on ∂D :12 ∂u∂γ − h ˆ B, n i u ( x ) = 0 . The quadratic form associated with L is given by: Q ( u, v ) := ( − Lu, v ) = 12 X i,j Z D a ij ( x ) ∂u∂x i ∂v∂x j dx − X i Z D B i ( x ) ∂u∂x i v ( x ) dx − X i Z D ˆ B i ( x ) ∂v∂x i u ( x ) dx − Z D Q ( x ) u ( x ) v ( x ) dx, where ( · . · ) stands for the inner product in L ( D ).The domain of the quadratic form is D ( Q ) = W , ( D ) := { u : u ∈ L ( D ) , ∂u∂x i ∈ L ( D ) , i = 1 , ..., d } . { S t , t ≥ } denote the semigroup generated by L .In this section, our aim is to solve the following equation: ( Lf ( x ) = − F ( x, f ( x )) , on D ∂f∂γ ( x ) − < b B, n > ( x ) f ( x ) = Φ( x ) on ∂D (5.1) Definition 5.1
A bounded continuous function f ( x ) defined on D is called a weak solution ofthe equation (5 . if f ∈ W , , and for any g ∈ C ∞ ( ¯ D ) , Q ( u, g ) = Z ∂D Φ( x ) g ( x ) σ ( dx ) + Z D F ( x, u ( x )) g ( x ) dx. Here the function F : R d × R → R is a bounded measurable function and satisfies the followingcondition: (E.1) ( y − y )( F ( x, y ) − F ( x, y )) ≤ − r ( x ) | y − y | .Recall the following regular Dirichlet form ( E ( u, v ) = P i,j R D a ij ( x ) ∂u∂x i ∂v∂x j dx,D ( E ) = W , ( D ) (5.2)associated with the operator L = ∇ ( A ∇ ) equipped with the Neumann boundary condition ∂∂γ = 0 on ∂D .The associated reflecting diffusion process is denoted by { Ω , F t , X t , θ t , γ t , P x } . Here θ t and γ t are the shift and reverse operators defined by X s ( θ t ( ω )) = X t + s ( ω ) , s, t ≥ X s ( γ t ( ω )) = X t − s ( ω ) , s ≤ t. The process ( X t ) t ≥ has the decomposition in (3 . X t is M t = R t σ ( X s ) dW s .The following probabilistic representation of semigroup S t was proved in [5] S t f ( x ) = E x [ f ( X t ) exp( Z t ( A − B ) ∗ ( X s ) dM s + ( Z t ( A − ˆ B ) ∗ ( X s ) dM s ) ◦ γ t − Z t ( B − ˆ B ) A − ( B − ˆ B ) ∗ ( X s ) ds + Z t Q ( X s ) ds )] (5.3) E x denotes the expectation under P x .Set ˆ Z t = exp( Z t ( A − B ) ∗ ( X s ) dM s + ( Z t ( A − ˆ B ) ∗ ( X s ) dM s ) ◦ γ t − Z t ( B − ˆ B ) A − ( B − ˆ B ) ∗ ( X s ) ds + Z t Q ( X s ) ds ) . (5.4)26y [3] and [21], there exists a bounded, continuous functions v ∈ W ,p ( D ) satisfying that( Z t ( A − ˆ B ) ∗ ( X s ) dM s ) ◦ γ t = − Z t ∇ v ( X s ) dM s + v ( X t ) − v ( X ) − Z t ( A − ˆ B ) ∗ ( X s ) dM s (5.5)Moreover, v satisfies the following equations: for g ∈ C ( ¯ D ), Z D < A ∇ v, ∇ g > ( x ) dx = Z D < ˆ B, ∇ g > ( x ) dx. (5.6)Thus the representation of S t becomes: S t f ( x ) = e − v ( x ) E x [ f ( X t ) e v ( X t ) exp( Z t ( A − ( B − ˆ B − A ∇ v )) ∗ dM s − Z t ( B − ˆ B − A ∇ v ) ∗ A − ( B − ˆ B − A ∇ v )( X s ) ds + Z t ( Q + 12 ( ∇ v ) A ( ∇ v ) ∗ − h B − ˆ B, ∇ v i )( X s ) ds )]= e − v ( x ) ˜ S t [ f e v ]( x ) . (5.7)Here, setting b := B − ˆ B − ( A ∇ v ) and q := Q + ( ∇ v ) A ( ∇ v ) ∗ − h B − ˆ B, ∇ v i , we see that ˜ S t is the semigroup generated by the following operator: L = 12 ∇ · ( A ∇ ) + ( B − ˆ B − ( A ∇ v )) · ∇ + ( Q + 12 ( ∇ v ) A ( ∇ v ) ∗ − h B − ˆ B, ∇ v i )= 12 ∇ · ( A ∇ ) + b · ∇ + q equipped with the boundary condition ∂∂γ = 0.In this section, we will stick to this particular choice of b and q .Recall that ˜ M ( t ) = e R t A − b ( X s ) dM s − R t bA − b ∗ ( X s ) ds and set Z t = ˜ M ( t ) e R t q ( X s ) ds .Then from (5 . Z ( t ) = Z t e v ( X t ) − v ( X ) . Recall the operator L = ∇ · ( A ∇ ) + b · ∇ with Neumann boundary condition, which isassociated with the reflecting diffusion ( X ( t ) , P x ). It is known from [14] that dP x | F t = ˜ M t dP x | F t , and X ( t ) = x + Z t σ ( X ( s )) dW s + Z t ( 12 ∇ A + b )( X ( s )) ds + Z t γ ( X ( s )) dL s , P x − a.s. { W t } is a d-dimensional Brownian motion and L t is the local time satisfying that L t = R t I ∂D ( X ( s )) dL s . Lemma 5.1
Assume that there exists x ∈ D , such that E x [ Z ∞ | ˆ Z t | e R t (2 Q − r )( X u ) du dL t ] < ∞ . (5.8) Then there exists a positive number ε > , if k ˆ B k L p ≤ ε , the following inequality holds: sup x ∈ D E x [ Z ∞ e R t ( − r + q )( X ( u )) du dt ] < ∞ . (5.9) Proof. E x [ e R t ( − r + q )( X ( u )) du ] = E x [ ˜ M ( t ) e R t ( − r + q )( X u ) du ]= E x [ Z ( t ) e R t ( − r + q )( X ( u )) du ] ≤ C E x [ ˆ Z ( t ) e − R t ( r ( X ( u )) du e R t ( Q + )( X u ) du ] ≤ C E x [ ˆ Z ( t ) e R t ( Q − r )( X u ) du ] · E x [ e R t ( X u ) du ] By Lemma 3.3 and condition (5 . c , β > x ∈ D E x [ ˆ Z ( t ) e R t ( Q − r )( X u ) du ] < c e − βt . Moreover, for p > d , by the Theorem 2.1 in [15], there exist two positive constants c and c such that E x [ e R t ( X u ) du ] ≤ c e c t , where c = c k < A ∇ v − B − ˆ B ) , ∇ v > k L p/ .Since |∇ v | L p ≤ C | ˆ B | L p ( D ) (see [21]), there exists ε >
0, such that | ˆ B | L p ( D ) ≤ ε implies c < β .Thus (5 .
9) holds. (cid:3)
Theorem 5.1
Assume (5 . and for some point x ∈ DE x [ Z ∞ ˆ Z s dL t ] < ∞ (5.10) Then there exists ε > such that if k ˆ B k L p ≤ ε , the problem (5 . has a unique, bounded,continuous weak solution u ( x ) . Proof.
Existence: Set ˜ F ( x, y ) = e v ( x ) F ( x, e − v ( x ) y ) and φ ( x ) = e v ( x ) Φ( x ).From the boundedness of v , ˜ F is also bounded.And ˜ F satisfies ( y − y )( ˜ F ( x, y ) − ˜ F ( x, y )) ≤ − r ( x ) | y − y | . c >
0, such that ∞ > E x [ Z ∞ ˆ Z s dL s ] = E x [ Z ∞ Z s e v ( X s ) − v ( X ) dL s ] ≥ cE x [ Z ∞ Z s dL s ] = cE x [ Z ∞ ˜ M s e R s q ( X u ) du dL s ] (5.11)By Lemma 4 .
1, we know that, at x ∈ D , E x [ Z ∞ e R s q ( X u ) du dL s ] < ∞ . (5.12)Furthermore, by Lemma 4 .
2, it follows thatsup x E x [ Z ∞ e R t q ( X ( u )) du dL t ] < ∞ . (5.13)By Lemma 5 .
1, the following condition is satisfied : E x [ Z ∞ e R t ( q − r )( X ( u )) du dt ] < ∞ , (5.14)So ˜ F satisfies all of the conditions in Theorem 4 . G by ˜ F . Thus the followingproblem ( L u ( x ) = − ˜ F ( x, u ( x )) , on D ∂u∂γ ( x ) = φ on ∂D (5.15)has a unique bounded continuous weak solution u ( x ).Set f ( x ) = e − v ( x ) u ( x ). Then we claim the function f ( x ) is the weak solution of the equation(5 . v is continuous and bounded, f ( x ) is also continuous. From the fact thatfunction u is the weak solution of the problem (5 . ψ ∈ C ∞ ( D ), E ( u, e − v ψ ) = 12 Z D < A ∇ u, ∇ ( e − v ψ ) > − < b, ∇ u > e − v ψ − e − v quψdx = Z ∂D e − v φψdσ + Z D ˜ F ( x, u ( x )) e − v ψdx. (5.16)As in the proof of Theorem 5.1 in [22], we can show that the left side of the equation (5 . Q ( f, ψ ) = 12 Z D [ < A ∇ f, ∇ ψ > − < B, ∇ u > ψ − < ˆ B, ∇ ψ > f − Qf ψ ] dx. At the same time, by the definition of the function φ and ˜ F , the right side of the equation(5 .
16) equals to Z ∂D Φ ψdσ + Z D F ( x, f ( x )) ψdx. ψ ∈ C ∞ ( D ), Q ( f, ψ ) = Z ∂D Φ ψdσ + Z D F ( x, f ( x )) ψdx. which proves that function f is a weak solution of the problem (5 . f is another solution of the problem (5 . u := e v f can be shown to be the solutionof the equation (5 . .
15) proved in the Theorem4 .
1, we find ¯ u = u . Therefore, f = ¯ f . (cid:3) L solutions of the BSDE and Semilinear PDEs Recall the operator L = 12 d X i,j =1 ∂∂x i (cid:18) a ij ( x ) ∂∂x j (cid:19) + d X i =1 b i ( x ) ∂∂x i on the domian D equipped with the Neumann boundary condition ∂∂γ = 0, on ∂D .And (Ω , F t , X ( t ) , P x , x ∈ D ) is the reflecting diffusion process associated with the generator L .Then the process X ( t ) has the following decomposition: X ( t ) = X (0) + M ( t ) + Z t ˜ b ( X ( s )) ds + Z t An ( X ( s )) dL s , P x − a.s.. Here ˜ b = { ˜ b , ..., ˜ b d } with ˜ b i = P j ∂a ij ∂x j + b i . M ( t ) is the F t square integrable continuousmartingale additive functional.In this section, we will consider the L solutions of the BSDEs in Section 2 and use thisresult to solve the nonlinear elliptic partial differential equation with the mixed boundary con-dition.Let f : Ω × R + × R → R be progressively measurable. Consider the following conditions: (I.1) ( y − y ′ )( f ( t, y ) − f ( t, y ′ )) ≤ d ( t ) | y − y ′ | , where d ( t ) is a progressively measurable process; (I.2) E [ R ∞ e R s d ( u ) du | f ( s, | ds ] < ∞ ; (I.3) P x − a.s. , for any t > y → f ( t, y ) is continuous; (I.4) ∀ r > , T > ψ r ( t ) := sup | y |≤ r | f ( t, y ) − f ( t, | ∈ L ([0 , T ] × Ω , dt × dP x ) . The following lemma is deduced from Corollary 2.3 in [2].
Lemma 6.1
Suppose a pair of progressively measurable processes ( Y, Z ) with values in R × R d such that t → Z t belongs to L ([0 , T ]) and t → f ( t, Y t ) belongs to L ([0 , T ]) , P x − a.s. .If Y t = ξ + Z Tt f ( r, Y r ) dr − Z Tt < Z r , dM r >, (6.1)30 hen the following inequality holds, for ≤ t < u ≤ T , | Y t | ≤ | Y u | + Z ut ˆ Y s f ( s, Y s ) ds − Z ut ˆ Y s h Z r , dM r i . where ˆ y = y | y | I { y =0 } . The following lemma can be proved by modifying the proof of Proposition 6.4 in [2].
Lemma 6.2
Assume that conditions (I.1)-(I.4) with d ( t ) ≡ . Then there exists a uniquesolution ( Y, Z ) of the BSDE Y t = Z Tt f ( r, Y r ) dr − Z Tt h Z r , dM r i , f or t ≤ T. (6.2) Moreover, for each β ∈ (0 , , E [sup t ≤ T | Y t | β ] + E [( R T | Z r | dr ) β ] < ∞ . Suppose β ∈ (0 , S β denotes the set of real-valued, adapted and continuous process { Y t } t ≥ such that k Y k β := E [sup t> | Y t | β ] < ∞ . It is known that k · k β deduces a complete metric on S β . M β denotes the set of R d -valued predictable processes { Z t } such that k Z k M β := E [( Z ∞ | Z t | dt ) β ] < ∞ .M β is also a complete metric space with the distance deduced by k · k M β . Lemma 6.3
Under the same assumption as the Lemma . , there exists a unique solution ( Y, Z ) of the BSDE Y t = Y T + Z Tt f ( r, Y r ) dr − Z Tt h Z r , dM r i , any t ≤ T ;lim t →∞ Y t = 0 , P − a.s.. (6.3) Proof.
Existence:By the Lemma 6.2 above, there exists ( Y n , Z n ) such that, for 0 ≤ t ≤ n , Y nt = Z nt f ( r, Y nr ) dr − Z nt h Z nr , dM r i , and Y nt = Z nt = 0, for t ≥ n .Fix t > t < n < n + i , then Y n + it − Y nt = Z n + it ( f ( r, Y n + ir ) − f ( r, Y nr )) dr − Z n + it h ( Z n + ir − Z nr ) , dM r i + Z n + in f ( r, dr F n ( r, y ) = f ( r, y + Y nr ) − f ( r, Y nr ) + f ( r, I { r>n } , y nt = Y n + it − Y nt and z nt = Z n + it − Z nt .Then ( y nt , z nt ) is the solution of the following BSDE: y nt = Z n + it F ( r, y nr ) dr − Z n + it h z nr , dM r i . (6.4)So that by the condition (I.1) with d ( t ) ≡
0, it follows from Lemma 6 . | y nt | ≤ Z n + it h ˆ y nr , F n ( r, y nr ) i dr − Z n + it h ˆ y nr , z nr dM r i≤ Z n + it I { y nr =0 } | y nr | h y nr , f ( r, y nr + Y nr ) − f ( r, Y nr ) i dr + Z n + in | f ( s, | ds − Z n + it h ˆ y nr , z nr dM r i≤ Z n + in | f ( s, | ds − Z n + it h ˆ y nr , z nr dM r i . (6.5)Taking conditional expectation on both side of the inequality, we got | y nt | ≤ E [ Z n + in | f ( s, | ds |F t ] := M nt , where M nt is a martingale. Then by Doob’s inequality and condition (I.2), it follows that, for β ∈ (0 , E [sup t | y nt | β ] ≤ E [sup t ( M nt ) β ] ≤ − β E [ Z n + in | f ( s, | ds ] β → , as n → ∞ . (6.6)Therefore, { Y n } is a Cauchy sequence under the norm k · k β ∞ . So that there is a process Y such that E [sup t | Y t − Y nt | β ] → Y t →
0, as t → ∞ , P x − a.s. .Moreover, by the equation (6 . | y nt | + Z n + it h A ( X ( r )) z nr , z nr i dr = 2 Z n + it h y nr , F n ( r, y nr ) i dr − Z n + it h y nr , z nr dM r i≤ Z n + in h y nr , f n ( r, i dr + 2 | Z n + it h y nr , z nr dM r i|≤ sup r | y nr | + ( Z n + in | f ( r, | dr ) + 2 | Z ∞ t h y nr , z nr dM r i| , and thus that( Z n + it | z nr | dr ) β ≤ c [sup r | y nr | β + ( Z n + in | f ( r, | dr ) β + | Z n + it < y nr , z nr dM r > | β ] . E [( Z n + it | z nr | dr ) β ] ≤ c ( E [sup r | y nr | β ] + E [( Z n + it | f ( r, | dr ) β ]) + c E [( Z n + it | y nr | | z nr | dr ) β ] ≤ c ( E [sup r | y nr | β ] + E [( Z n + in | f ( r, | dr ) β ]) + c E [(sup r | y nr | β Z n + it | z nr | dr ) β ] ≤ ( c + c E [sup r | y nr | β ] + E [( Z n + in | f ( r, | dr ) β ]) + 12 E [( Z n + it | z nr | dr ) β ] . Therefore, we know that there is a constant
C >
0, such that E [( Z ∞ | z ns | ds ) β ] ≤ CE [sup t | y nt | β + ( Z n + in | f ( s, | ds ) β ] ≤ CE [sup t | y nt | β ] + CE [ Z n + in | f ( s, | ds ] β → as n → ∞ . So that { Z nt } is a Cauchy sequence in M β . Let Z denote the limit of { Z n } .At last, by the condition (I.3), we find that Z T f ( t, Y nt ) dt → Z T f ( t, Y t ) dt, P x − a.s.. (6.7)Therefore, (Y,Z) is the solution satisfies the BSDE (6 . Y, Z ) and ( Y ′ , Z ′ ) are two solutions to (6 . ∀ t > , | Y t − Y ′ t | = 0 , P − a.s.. (cid:3) (I.5) The process d ( t ) is a progressively measurable process satisfying d ( · ) ∈ L [[0 , T ] × Ω , dt ⊗ P ] , f or any T > . Theorem 6.1
Assume the conditions (I.1)-(I.4). Then there exists a unique process ( Y, Z ) such that, Y t = Y T + Z Tt f ( r, Y r ) dr − Z Tt < Z r , dM r >, f or any t < T ;lim t →∞ e R t d ( u ) du Y t = 0 , P − a.s. (6.8)33 roof. Existence:Set ˆ f ( t, y ) = e R t d ( u ) du f ( t, e − R t d ( u ) du y ) − d ( t ) y . Then(1) ( y − y ′ )( ˆ f ( t, y ) − ˆ f ( t, y ′ )) ≤ f ( t,
0) = e R t d ( u ) du f ( t, E [ R ∞ | ˆ f ( s, | ds ] = E [ R ∞ e R t d ( u ) du | f ( t, | ds ] < ∞ . (3) sup | y |≤ r | ˆ f ( t, y ) − ˆ f ( t, | ≤ ψ r ( t )+ | d ( t ) | r , where the process ψ r ( t )+ | d ( t ) | r ∈ L ([0 , T ] × Ω , dt ⊗ P ),for T > f satisfies all the conditions of the Lemma 6 .
3. So there exists a pair of processes( ˆ
Y , ˆ Z ) satisfying the equation:ˆ Y t = ˆ Y T + Z Tt ˆ f ( r, ˆ Y r ) dr − Z Tt h ˆ Z r , dM r i , and obviously lim t →∞ ˆ Y t = 0.By the chain rule and the definition of the function ˆ f , it follows that de − R t d ( u ) du ˆ Y t = − f ( t, e − R t d ( u ) du ˆ Y t ) dt + h e − R t d ( u ) du ˆ Z t , dM t i . Set Y t = e − R t d ( u ) du ˆ Y t and Z t = e − R t d ( u ) du ˆ Z t . Then the process ( Y, Z ) is the solution to theequation (6 . .
8) follows from the uniqueness of the solution to equation(6 . (cid:3) Let G ( x, y ) : R d × R → R be a bounded Borel measurable function. Consider the followingconditions: (H.1) ′ ( y − y )( G ( x, y , z ) − G ( x, y , z )) ≤ − h ( x ) | y − y | , where h ∈ L p ( D ) for p > d . (H.2) ′ y → G ( x, y ) is continuous. Theorem 6.2
Assume the Conditions ( H. ′ and ( H. ′ and that there is some point x ∈ D ,such that E x [ Z ∞ e R s q ( X ( u )) du dL s ] < ∞ . (6.9) Then the semilinear Neumann boundary value problem (cid:26) L u ( x ) = − G ( x, u ( x )) , on D ∂u∂γ ( x ) = φ ( x ) on ∂D (6.10) has a unique continuous weak solution. roof. Step 1Set ˜ G ( X ( t ) , y ) = e R t q ( X ( u )) dt G ( x, e − R t q ( X ( u )) dt y ) . Then there exists a unique solution ( ˆ Y x , ˆ Z x )to the following BSDE:for any T > < t < T ,ˆ Y x ( t ) = ˆ Y x ( T ) + Z Tt ˜ G ( X ( t ) , ˆ Y x ( s )) ds − Z Tt e R s q ( X ( u )) dt φ ( X ( s )) dL s − Z Tt h ˆ Z x ( s ) , dM x ( s ) i and lim t →∞ e − R t h ( X ( u )) du ˆ Y t = 0 P x − a.s. The uniqueness follows from the uniqueness proved in Theorem 6.1. Only the existence ofsolution ( ˆ Y x , ˆ Z x ) needs to be proved:(a) Similarly as the proof of Theorem 2.1, we can show that there exists ( p x ( t ) , q x ( t )) such that dp x ( t ) = e R t q ( X ( u )) du φ ( X ( t )) dL t + < q x ( t ) , dM x ( t ) >,e − R t h ( X ( u )) du p x ( t ) → , as t → ∞ , P x − a.s.. (6.11)(b) Set g ( x, y ) = ˜ G ( x, y + p x ( t )). Then it follows that( y − y ′ )( g ( x, y ) − g ( x, y ′ )) ≤ − h ( x ) | y − y ′ | . The condition (6 .
9) and Lemma 3.3 imply, for x ∈ D , E x [ Z ∞ e R s ( − h + q )( X ( u )) du ds ] < ∞ . Furthermore, as the function G is bounded, we see that condition ( I.
2) is satisfied: E x [ Z ∞ e − R s h ( X ( u )) du | g ( X ( s ) , | ds ]= E x [ Z ∞ e − R s h ( X ( u )) du | ˜ G ( X ( s ) , p x ( s )) | ds ]= E x [ Z ∞ e R s ( − h + q )( X ( u )) du | G ( X ( s ) , e − R s q ( X ( u )) du p x ( s )) | ds ] ≤ k G k ∞ E x [ Z ∞ e R t ( − h + q )( X ( u )) du dt ] < ∞ . (6.12)Obviously condition (I.3) is satisfied, i.e., y → g ( x, y ) is continuous.Moreover, the condition (I.4) is also satisfied. In fact, for any r > ψ r ( t ) = sup r | ˜ G ( X ( t ) , y ) − ˜ G ( X ( t ) , | ≤ k G k ∞ e R t q ( X t ) dt , T >
0, by the fact that q ∈ L p ( D ) with p > d and Theorem 2.1 in [15], E x [ R T e R t q ( X u ) du dt ] < ∞ .Therefore, the function g ( x, y ) satisfies all of the conditions of Theorem 6 . y x ( t ) , z x ( t )) such that for any T > < t < T , y x ( t ) = y x ( T ) + Z Tt g ( X ( s ) , y x ( s )) ds − Z Tt h z x ( s ) , dM x ( s ) i (6.13)and lim t →∞ e − R t h ( X ( u )) du y x ( t ) = 0 P x − a.s. (6.14)Put ˆ Y x ( t ) = p x ( t ) + y x ( t ) and ˆ Z x ( t ) = q x ( t ) + z x ( t ). It follows that ( ˆ Y x ( t ) , ˆ Z x ( t )) satisfies thefollowing equation d ˆ Y x ( t ) = e R t q ( X ( u )) du φ ( X ( t )) dL t − ˜ G ( t, ˆ Y x ( t )) dt + < ˆ Z x ( t ) , dM x >, lim t →∞ e − R t h ( X ( u )) du ˆ Y t = 0 P x − a.s.. Step 2.Put Y x ( t ) := e − R t q ( X ( u )) dt ˆ Y x ( t ) and Z x ( t ) := e − R t q ( X ( u )) dt ˆ Z x ( t ), we have dY x ( t ) = − F ( X ( t ) , Y x ( t )) + φ ( X ( t )) dL t + < Z x ( t ) , dM x ( t ) >, where F ( x, y ) = q ( x ) y + G ( x, y ). Moreover, e R t ( − h + q )( X ( u )) du Y x ( t ) = e R t ( − h + q )( X ( u ))( u ) dt e − R t q ( X ( u )) dt ˆ Y x ( t )= e − R t h ( X ( u )) dt ˆ Y x ( t ) → as t → ∞ . Put u ( x ) = Y x (0) and v ( x ) = Z x (0).Now as in the proof of Theorem, 4.1, we can solve the following equation (cid:26) L u ( x ) = − G ( x, u ( x )) , on D ∂u∂γ ( x ) = φ ( x ) on ∂D (6.15)and prove that the solution u coincides with u ( x ). This completes the proof of the theorem. (cid:3) Suppose that F : R d × R → R is a bounded measurable function and r ∈ L p ( D ). Con-sider the following conditions : (E.1) ( y − y )( G ( x, y , z ) − G ( x, y , z )) ≤ − r ( x ) | y − y | ; (E.3) y → F ( x, y ) is continuous;Now, after establishing Theorem 6.2, following the same proof as that of Theorem 5.1, wefinally have the following main result. 36 heorem 6.3 Suppose that the function F satisfies the condition (E.1) and (E.2), and thereexists x ∈ D such that E x [ Z ∞ ˆ Z s dL t ] < ∞ . (6.16) Then the following problem ( Lu ( x ) = − F ( x, u ( x )) , on D ∂u∂γ ( x ) − < b B, n > ( x ) u ( x ) = Φ( x ) on ∂D (6.17) has a unique, bounded, continuous weak solution. eferences [1] R. F. Bass and P. Hsu, Some potential theory for reflecting Brownian motion in H¨older andLipschitz domains , Ann. Probab. (1991), 486-508.[2] Ph. Briand. B. Delyon. Y. Hu. E. Pardoux. L. Stoica, L p solutions of backward stochastic differ-ential equations , Stochastic Process. Appl. 108 (2003), 109-129.[3] Z. Q. Chen and T. S. Zhang, Time-reversal and elliptic boundary value problems , Ann. Probab. (2009), 1008-1043.[4] Z. Q. Chen and T. S. Zhang, A probabilistic approach to mixed boundary value problems for ellipticoperators with singular coefficients , Preprint.[5] Z. Q. Chen, P. J. Fitzsimmons, K. Kuwae and T. S. Zhang,
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