A Probabilistic Approach to Dirichlet Problems of Semilinear Elliptic

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d ∂v − ˆbi(x)u dx − q(x)u(x)v(x) dx. Z ∂x Z Xi=1 D i D We refer readers to [14, 18] and [24] for details of the operator A. Probabilistic approaches to boundary value problems of second order differential operators have been adopted by many people. The earlier work went back as early as 1944 in [15]. See the books [1, 7] and references therein. If f = 0 (i.e., the linear case), and moreover ˆb = 0, the solution u to problem (1.1) can be solved by a Feynman–Kac formula

τD u(x)= Ex exp q(X(s)) ds ϕ(X(τD)) for x ∈ D, Z0 where X(t), t ≥ 0 is the diﬀusion process associated with the inﬁnitesimal generator

d d 1 ∂ ∂ ∂ (1.4) L = a (x) + b (x) , 1 2 ∂x ij ∂x i ∂x i,jX=1 i j Xi=1 i

τD is the first exit time of the diffusion process X(t),t ≥ 0 from the domain D. Very general results are obtained in the paper [6] for this case. When ˆb 6= 0, “div(ˆb·)” in (1.2) is just a formal writing because the divergence does not really exist for the merely measurable vector field ˆb. It should be interpreted in the distributional sense. It is exactly due to the nondifferentiability of ˆb, all the previous known probabilistic methods in solving the elliptic boundary value problems such as those in [1, 6, 15] and [13] could not be applied. We stress that the lower order term div(ˆb·) cannot be handled by Girsanov transform or Feynman–Kac transform either. In a recent work [5], we show that the term ˆb in fact can be tackled by the time-reversal of Girsanov 0 transform from the first exit time τD from D by the symmetric diffusion X associated with L = 1 d ∂ (a (x) ∂ ), the symmetric part of A. The 0 2 i,j=1 ∂xi ij ∂xj solution to equation (1.1P) (when f = 0 ) is given by

τD 0 0 −1 0 0 u(x)= Ex ϕ(X (τD)) exp h(a b)(X (s)), dM (s)i Z0

τD −1ˆ 0 0 + h(a b)(X (s)), dM (s)i ◦ rτD Z0 (1.5) 1 τD − (b − ˆb)a−1(b − ˆb)∗(X0(s)) ds 2 Z0 τD + q(X0(s)) ds , Z0 A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 3

0 0 where M (s) is the martingale part of the diffusion X , rt denotes the reverse operator, and h·, ·i stands for the inner product in Rd. Nonlinear elliptic PDEs [i.e., f 6= 0 in (1.1)] are generally very hard to solve. One can not expect explicit expressions for the solutions. However, in recent years backward stochastic differential equations (BSDEs) have been used effectively to solve certain nonlinear PDEs. The general approach is to represent the solution of the nonlinear equation (1.1) as the solution of certain BSDEs associated with the diffusion process generated by the linear operator A. But so far, only the cases where ˆb =0 and b being bounded were considered. The main difficulty for treating the general operator A in (1.2) with ˆb 6= 0, q 6= 0 is that there are no associated diffusion processes anymore. The mentioned methods used so far in the literature ceased to work. Our approach is to transform the problem (1.1) to a similar problem for which the operator A does not have the “bad” term ˆb. See below for detailed description. There exist many papers on BSDEs and their applications to nonlinear PDEs. We mention some related earlier results. The first result on probabilistic interpretation for solutions of semilinear parabolic PDE’s was obtained by Peng in [19] and subsequently in [21]. In [8], Darling and Pardoux obtained a viscosity solution to the Dirichlet problem for a class of semilinear elliptic PDEs (through BSDEs with random terminal time) for which the linear operator A is of the form

d d 1 ∂2 ∂ A = a (x) + b (x) , 2 ij ∂x ∂x i ∂x i,jX=1 j i Xi=1 i

2 1 where aij ∈ Cb (D) and b ∈ Cb (D). BSDEs associated with Dirichlet processes and weak solutions of semi-linear parabolic PDEs were considered by Lejay in [16] where the linear operator A is assumed to be

d d 1 ∂ ∂ ∂ A = a (x) + b (x) , 2 ∂x ij ∂x i ∂x i,jX=1 i j Xi=1 i for bounded coeﬃcients a and b. BSDEs associated with symmetric Markov processes and weak solutions of semi-linear parabolic PDEs were studied by Bally, Pardoux and Stoica in [2] where the linear operator A is assumed to be symmetric with respect to some measure m. BSDEs and solutions of semi-linear parabolic PDEs were also considered by Rozkosz in [23] for the linear operator A of the form

d 1 ∂ ∂ A = a (t,x) . 2 ∂x ij ∂x i,jX=1 i j 4 T. ZHANG

Now we describe the contents of this paper in more details. Our strategy is to transform the problem (1.1) by a kind of h-transform to a problem of a similar kind, but with an operator A that does not have the “bad” term ˆb. The first step will be to solve (1.1) assuming ˆb = 0. In Section 2, d we introduce the Feller diffusion process (Ω, F, Ft, X(t), Px,x ∈ R ) whose infinitesimal generator is given by d d 1 ∂ ∂ ∂ (1.6) L = a (x) + b (x) . 1 2 ∂x ij ∂x i ∂x i,jX=1 i j Xi=1 i In general, X(t), t ≥ 0 is not a semimartingale. But it has a nice martingale part M(t), t ≥ 0. In this section, we prove a martingale representation theorem for the martingale part M(t), which is crucial for the study of BSDEs in subsequent sections. In Section 3, we solve a class of BSDEs associated with the martingale part M(t), t ≥ 0: T T (1.7) Y (t)= ξ + f(s,Y (s),Z(s)) ds − hZ(s), dM(s)i. Zt Zt The random coefficient f(t,y,z,ω) satisfies a certain monotonicity condition which is particularly fulfilled in the situation we are interested. The BSDEs with deterministic terminal time were solved first and then the BSDEs with random terminal time were studied. In Section 4, we consider the Dirichelt problem for the second order differential operator d d 1 ∂ ∂ ∂ (1.8) L = a (x) + b (x) + q(x), 2 2 ∂x ij ∂x i ∂x i,jX=1 i j Xi=1 i p β d where bi ∈ L for some p > d and q ∈ L for some β> 2 . We first solve the linear problem with a given function F L u = F, in D, (1.9) 2 u = ϕ, on ∂D, and then the nonlinear problem L u = −g(x, u(x), ∇u(x)), in D, (1.10) 2 u = ϕ, on ∂D, with the help of BSDEs. Finally, in Section 5, we study the Dirichlet problem Au(x)= −f(x, u(x)), ∀x ∈ D, (1.11) u(x)|∂D = ϕ, ∀x ∈ ∂D, where A is a general second order differential operator given in (1.2). We apply a transform we introduced in [5] to transform the above problem to a problem like (1.10) and then a reverse transformation will solve the final problem. A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 5

2. Preliminaries. Let A be an elliptic operator of the following general form: d d 1 ∂ ∂ ∂ A = a (x) + b (x) − “div(ˆb·)”+ q(x), 2 ∂x ij ∂x i ∂x i,jX=1 i j Xi=1 i

d×d where a = (aij(x)) : D → R (d> 2) is a measurable, symmetric matrix- valued function which satisﬁes the uniform elliptic condition

d 2 2 d (2.1) λ|ξ| ≤ aij(x)ξiξj ≤ Λ|ξ| , ∀ξ ∈ R and x ∈ D i,jX=1

d for some constant λ, Λ > 0, b = (b1, . . ., bd),ˆb = (ˆb1,...,ˆbd) : D → R and q : D → R are measurable functions which could be singular and such that |b|2 ∈ Lp(D), |ˆb|2 ∈ Lp(D) and q ∈ Lp(D), d d for some p> 2 . Here D is a bounded domain in R whose boundary is x x regular, that is, for every x ∈ ∂D, P (τD = 0) = 1, where τD is the ﬁrst exit time of a standard Brownian motion started at x from the domain D. Let f : Rd × R × Rd → R be a measurable nonlinear function. Consider the following nonlinear Dirichlet boundary value problem: Au(x)= −f(x, u(x), ∇u(x)), ∀x ∈ D, (2.2) u(x)|∂D = ϕ, ∀x ∈ ∂D. Let W 1,2(D) denote the usual Sobolev space of order one: W 1,2(D)= {u ∈ L2(D) : ∇u ∈ L2(D; Rd)}.

Definition 2.1. We say that u ∈ W 1,2(D) is a continuous, weak solution of (2.2) if: 1,2 (i) for any φ ∈ W0 (D), d d 1 ∂u ∂φ ∂u aij(x) dx − bi(x) φ dx 2 Z ∂x ∂x Z ∂x i,jX=1 D i j Xi=1 D i

d ∂φ − ˆbi(x)u dx − q(x)u(x)φ dx = f(x, u, ∇u)φ dx, Z ∂x Z Z Xi=1 D i D D (ii) u ∈ C(D¯), (iii) limy→x u(y)= ϕ(x), ∀x ∈ ∂D.

Next we introduce two diﬀusion processes which will be used later. 6 T. ZHANG

d Let (Ω, F, Ft, X(t), Px,x ∈ R ) be the Feller diffusion process whose infinitesimal generator is given by d d 1 ∂ ∂ ∂ (2.3) L = a (x) + b (x) , 1 2 ∂x ij ∂x i ∂x i,jX=1 i j Xi=1 i where Ft is the completed, minimal admissible filtration generated by X(s), s ≥ 0. The associated nonsymmetric, semi-Dirichlet form with L1 is defined by

Q1(u, v) = (−L1u, v)L2(Rd) (2.4) d d 1 ∂u ∂v ∂u = aij(x) dx − bi(x) v(x) dx. 2 Z d ∂x ∂x Z d ∂x i,jX=1 R i j Xi=1 R i The process X(t), t ≥ 0 is not a semimartingale in general. However, it is known (see, e.g., [6, 10, 12] and [17]) that the following Fukushima’s decomposition holds:

(2.5) X(t)= x + M(t)+ N(t) Px-a.s., where M(t) is a continuous square integrable martingale with sharp bracket being given by t i j (2.6) hhM , M iit = ai,j(X(s)) ds, Z0 and N(t) is a continuous process of zero quadratic variation. Later we also write Xx(t), Mx(t) to emphasize the dependence on the initial value x. Let M denote the space of square integrable martingales w.r.t. the filtration Ft, t ≥ 0. The following result is a martingale representation theorem whose proof is a modification of that of Theorem A.3.20 in [12]. It will play an important role in our study of the backward stochastic differential equations associated with the martingale part M.

Theorem 2.1. For any L ∈ M, there exist predictable processes Hi(t), i = 1,...,d such that

d t i (2.7) Lt = Hi(s) dM (s). Z Xi=1 0

Proof. It is sufficient to prove (2.7) for 0 ≤ t ≤ T , where T is an arbitrary, but fixed constant T . Recall that M is a Hilbert space w.r.t. the inner product (K1, K2)M = E[hhK1, K2iiT ], where hhK1, K2ii denotes the sharp bracket of K1 and K2. Let M1 denote the subspace of square integrable martingales of the form (2.7). Let Rα,α> 0 be the resolvent operators of the d diffusion process X(t), t ≥ 0. Fix any g ∈ Cb(R ), we know that Rαg ∈ D(L1) A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 7 and L1Rαg = αRαg − g. Moreover, it follows from [12] and [17] that t Rαg(X(t)) − Rαg(X(0)) = h∇Rαg(X(s)), dM(s)i Z0 t + (αRαg − g)(X(s)) ds. Z0 Hence, t −αs Jt := e h∇Rαg(X(s)), dM(s)i Z0 t −αt −αs = e Rαg(X(t)) − Rαg(X(0)) + e g(X(s)) ds Z0 is a bounded martingale that belongs to M1. The theorem will be proved ⊥ ⊥ if we can show that M1 = {0}. Take K ∈ M1 . Since M1 is stable under stopping, by Lemma 2 in Chapter IV in [22], we deduce hhK,Lii = 0 for all L ∈ M1. In particular, hhK, Jii = 0. From here, we can follow the same proof of Theorem A.3.20 in [12] to conclude K = 0.

0 0 0 0 d We will denote by (Ω, F , Ft , X (t), Px ,x ∈ R ) the diﬀusion process generated by d 1 ∂ ∂ (2.8) L = a (x) . 0 2 ∂x ij ∂x i,jX=1 i j The corresponding Fukushima’s decomposition is written as X0(t)= x + M 0(t)+ N 0(t), t ≥ 0. For v ∈ W 1,2(Rd), the Fukushima’s decomposition for the Dirichlet process v(X0(t)) reads as (2.9) v(X0(t)) = v(X0(0)) + M v(t)+ N v(t), v t 0 0 v where M (t)= 0 ∇v(X (s)) · dM (s), N (t) is a continuous process of zero energy (the zeroR energy part). See [3, 4, 12] for details of symmetric Markov processes.

3. Backward SDEs with singular coefficients. Let (Ω, F, Ft) be the probability space carrying the diffusion process X(t) described in Section 2. Re- call M(t), t ≥ 0 is the martingale part of X. In this section, we will study backward stochastic differential equations (BSDEs) with singular coefficients associated with the martingale part M(t).

3.1. BSDEs with deterministic terminal times. Let f(s,y,z,ω) : [0, T ] × R×Rd ×Ω → R be a given progressively measurable function. For simplicity, we omit the random parameter ω. Assume that f is continuous in y and satisﬁes: 8 T. ZHANG

2 (A.1) (y1 − y2)(f(s,y1, z) − f(s,y2, z)) ≤−d1(s)|y1 − y2| , (A.2) |f(s,y,z1) − f(s,y,z2)|≤ d2|z1 − z2|, (A.3) |f(s,y,z)| ≤ |f(s, 0, z)| + K(s)(1 + |y|), where d1(·), K(s) are a progressively measurable stochastic process and d2 2 is a constant. Let ξ ∈ L (Ω, FT , P ). Let λ be the constant deﬁned in (2.1).

T T − R0 2d1(s) ds 2 Theorem 3.1. Assume E[e |ξ| ] < ∞, E[ 0 K(s) ds] < ∞ and T R − s 2d (u) du 2 E e R0 1 |f(s, 0, 0)| ds < ∞. Z0

Then, there exists a unique (Ft-adapted) solution (Y,Z) to the following BSDE: T T (3.1) Y (t)= ξ + f(s,Y (s),Z(s)) ds − hZ(s), dM(s)i, Zt Zt where Z(s) = (Z1(s),...,Zd(s)).

Proof. We ﬁrst prove the uniqueness. Set d(s)= −2d1(s). Suppose (Y 1(t),Z1(t)) and (Y 2(t),Z2(t)) are two solutions to equation (3.1). Then d(|Y 1(t) − Y 2(t)|2) = −2(Y 1(t) − Y 2(t))(f(t,Y 1(t),Z1(t)) − f(t,Y 2(t),Z2(t))) dt (3.2) + 2(Y 1(t) − Y 2(t))hZ1(t) − Z2(t), dM(t)i + ha(X(t))(Z1(t) − Z2(t)),Z1(t) − Z2(t)i dt. By the chain rule, using the assumptions (A.1), (A.2) and Young’s inequality, we get

t 1 2 2 R d(s) ds |Y (t) − Y (t)| e 0

T s R d(u) du 1 2 1 2 + e 0 ha(X(s))(Z (s) − Z (s)),Z (s) − Z (s)i ds Zt

T s R d(u) du 1 2 2 = − e 0 |Y (s) − Y (s)| d(s) ds Zt

T s R d(u) du 1 2 + 2 e 0 (Y (s) − Y (s)) Zt × (f(s,Y 1(s),Z1(s)) − f(s,Y 2(s),Z2(s))) ds

T s R d(u) du 1 2 1 2 − 2 e 0 (Y (s) − Y (s))hZ (s) − Z (s), dM(s)i Zt A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 9

T s d(u) du 1 2 2 (3.3) ≤− eR0 |Y (s) − Y (s)| d(s) ds Zt

T s R d(u) du 1 2 2 − 2 e 0 |Y (s) − Y (s)| d1(s) ds Zt

T s R d(u) du 1 2 1 2 + 2 e 0 d2|Y (s) − Y (s)||Z (s) − Z (s)| ds Zt

T s R d(u) du 1 2 1 2 − 2 e 0 (Y (s) − Y (s))hZ (s) − Z (s), dM(s)i Zt

T s R d(u) du 1 2 2 ≤ Cλ e 0 |Y (s) − Y (s)| ds Zt

T s 1 R d(u) du 1 2 1 2 + e 0 ha(X(s))(Z (s) − Z (s)), (Z (s) − Z (s))i ds 2 Zt

T s R d(u) du 1 2 1 2 − 2 e 0 (Y (s) − Y (s))hZ (s) − Z (s), dM(s)i. Zt Take expectation in above inequality to get

t T s 1 2 2 R d(s) ds R d(u) du 1 2 2 E[|Y (t) − Y (t)| e 0 ] ≤ Cλ E[e 0 |Y (s) − Y (s)| ] ds. Zt By Gronwall’s inequality, we conclude Y 1(t)= Y 2(t) and hence Z1(t)= Z2(t) by (3.3). Next, we prove the existence. Take an even, nonnegative function φ ∈ ∞ C0 (R) with R φ(x) dx = 1. Deﬁne R fn(t,y,z)= f(t,x,z)φn(y − x) dx, ZR where φn(x)= nφ(nx). Since f is continuous in y, it follows that fn(t,y,z) → f(t,y,z) as n →∞. Furthermore, it is easy to see that for every n ≥ 1,

(3.4) |fn(t,y1, z) − fn(t,y2, z)|≤ Cn|y1 − y2|, y1,y2 ∈ R, for some constant Cn. Consider the following BSDE:

T T (3.5) Yn(t)= ξ + fn(s,Yn(s),Zn(s)) ds − hZn(s), dM(s)i. Zt Zt In view of (3.4) and the assumptions (A.2), (A.3), it is known (e.g., [20]) that the above equation admits a unique solution (Yn,Zn). Our aim now is to show that there exists a convergent subsequence (Ynk ,Znk ). To this end, 10 T. ZHANG we need some estimates. Applying Itˆo’s formula, in view of assumptions (A.1)–(A.3) it follows that

t T s 2 R d(s) ds R d(u) du |Yn(t)| e 0 + e 0 ha(X(s))Zn(s),Zn(s)i ds Zt

T T s 2 R d(s) ds R d(u) du 2 = |ξ| e 0 − e 0 Yn (s)d(s) ds Zt

T s R d(u) du + 2 e 0 Yn(s)fn(s,Yn(s),Zn(s)) ds Zt

T s R d(u) du − 2 e 0 Yn(s)hZn(s), dM(s)i Zt

T T s 2 R d(s) ds R d(u) du 2 ≤ |ξ| e 0 − e 0 Yn (s)d(s) ds Zt

T s T s R d(u) du 2 R d(u) du − 2 e 0 d1(s)Yn (s) ds + 2C e 0 |Yn(s)||Zn(s)| ds Zt Zt

T s R d(u) du + 2 e 0 |Yn(s)|f(s, 0, 0) ds Zt

T s R d(u) du (3.6) − 2 e 0 Yn(s)hZn(s), dM(s)i Zt

T T s 2 R d(s) ds R d(u) du 2 ≤ |ξ| e 0 + Cλ e 0 Yn (s) ds Zt

T s 1 R d(u) du + e 0 ha(X(s))Zn(s),Zn(s)i ds 2 Zt

T s T s R d(u) du 2 R d(u) du 2 + e 0 Yn (s) ds + e 0 |f(s, 0, 0)| ds Zt Zt

T s R d(u) du − 2 e 0 Yn(s)hZn(s), dM(s)i. Zt Take expectation in (3.6) to obtain

t T s 2 R d(s) ds 1 R d(u) du E[|Yn(t)| e 0 ]+ E e 0 ha(X(s))Zn(s),Zn(s)i ds 2 Zt

T T s 2 R d(s) ds R d(u) du 2 (3.7) ≤ E[|ξ| e 0 ]+ Cλ E[e 0 Yn (s)] ds Zt

T s R d(u) du 2 + E e 0 |f(s, 0, 0)| ds . Zt A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 11

Gronwall’s inequality yields t 2 R d(s) ds sup sup E[|Yn(t)| e 0 ] n 0≤t≤T (3.8) T T s 2 R d(s) ds R d(u) du 2 ≤ C E[|ξ| e 0 ]+ E e 0 |f(s, 0, 0)| ds Z0 and also

T s R d(u) du (3.9) sup E e 0 ha(X(s))Zn(s),Zn(s)i ds < ∞. n Z0 Moreover, (3.6)–(3.9) further imply that there exists some constant C such that t 2 R d(s) ds E sup Yn (t)e 0 h0≤t≤T i

t s R d(u) du ≤ C + CE sup e 0 Yn(s)hZn(s), dM(s)i 0≤t≤T Z0 1/2 T s 2 R d(u) du 2 ≤ C + CE e 0 Yn (s)ha(X(s))Zn(s),Zn(s)i ds Z0

s (1/2) R d(u) du ≤ C + CE sup (e 0 |Yn(s)|) 0≤s≤T 1/2 T s R d(u) du (3.10) × e 0 ha(X(s))Zn(s),Zn(s)i ds Z0

s 1 R d(u) du 2 ≤ C + E sup (e 0 Yn (s)) 2 h0≤s≤T i

T s R d(u) du + C1E e 0 ha(X(s))Zn(s),Zn(s)i ds . Z0 In view of (3.9), this yields t 2 R d(s) ds (3.11) supE sup Yn (t)e 0 < ∞. n h0≤t≤T i t (1/2) R0 d(s) ds By (3.9) and (3.11), we can extract a subsequence nk such that Ynk (t)e converges to some Yˆ (t) in L2(Ω,L∞[0, T ]) equipped with the weak star t (1/2) R0 d(s) ds ˆ 2 topology and Znk (t)e converges weakly to some Z(t) in L (ΩT ; R), where ΩT = [0, T ] × Ω. Observe that t (1/2) R0 d(s) ds Ynk (t)e

T T s (1/2) R0 d(s) ds (1/2) R0 d(u) du = e ξ + e fnk (s,Ynk (s),Znk (s)) ds Zt 12 T. ZHANG

(3.12) T 1 s (1/2) R0 d(u) du − e Ynk (s)d(s) ds 2 Zt

T s (1/2) R0 d(u) du − e hZnk (s), dM(s)i. Zt Letting k →∞ in (3.12), using the monotonicity of f, following the same arguments as that in the proof of Proposition 2.3 in Darling and Pardoux in [8], we can show that the limit (Yˆ , Zˆ) satisﬁes T (1/2) R d(s) ds Yˆ (t)= e 0 ξ

T s s s (1/2) R d(u) du −(1/2) R d(u) du −(1/2) R d(u) du (3.13) + e 0 f(s, e 0 Yˆ (s), e 0 Zˆ(s)) ds Zt 1 T T − Yˆ (s)d(s) ds − hZˆ(s), dM(s)i. 2 Zt Zt Set t t −(1/2) R d(u) du −(1/2) R d(u) du Y (t)= e 0 Yˆ (t),Z(t)= e 0 Zˆ(t). An application of Itˆo’s formula yields that T T Y (t)= ξ + f(s,Y (s),Z(s)) ds − hZ(s), dM(s)i, Zt Zt namely, (Y,Z) is a solution to the backward equation (3.1). The proof is complete.

3.2. BSDEs with random terminal times. Let f(t,y,z) satisfy (A.1)– 2 (A.3) in Section 3.1. In this subsection, set d(s)= −2d1(s)+ δd2. The following result provides existence and uniqueness for BSDEs with random terminal time. Let τ be a stopping time. Suppose ξ is Fτ -measurable.

τ τ R0 d(s) ds 2 Theorem 3.2. Assume E[e |ξ| ] < ∞, E[ 0 K(s) ds] < ∞ and

τ s R R d(u) du 2 (3.14) E e 0 |f(s, 0, 0)| ds < ∞, Z0 1 for some δ> λ , where λ is the constant appeared in (2.1). Then, there exists a unique solution (Y,Z) to the BSDE τ τ (3.15) Y (t)= ξ + f(s,Y (s),Z(s)) ds − hZ(s), dM(s)i. Zτ∧t Zτ∧t Furthermore, the solution (Y,Z) satisﬁes

τ s τ s R d(u) du 2 R d(u) du 2 (3.16) E e 0 Y (s) ds < ∞, E e 0 |Z(s)| ds < ∞, Z0 Z0 A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 13 and s R d(u) du 2 (3.17) E sup {e 0 Y (s)} < ∞. h0≤s≤τ i

Proof. After the preparation of Theorem 3.1, the proof of this theorem is similar to that of Theorem 3.4 in [8], where d1(s), d2 were both assumed to be constants. We only give a sketch highlighting the diﬀerences. For every n ≥ 1, from Theorem 3.1 we know that the following BSDE has a unique solution (Yn,Zn) on 0 ≤ t ≤ n: τ∧n τ∧n (3.18) Yn(t)= E[ξ|Fn]+ f(s,Yn(s),Zn(s)) ds − hZn(s), dM(s)i. Zτ∧t Zτ∧t

Extend the deﬁnition of (Yn,Zn) to all t ≥ 0 by setting

Yn(t)= E[ξ|Fn],Zn(t)=0 for t ≥ n.

Then the extended (Yn,Zn) satisﬁes a bsde similar to (3.18) with f replaced by χ{s≤n∧τ}f(s,y,z). Let n ≥ m. By Itˆo’s formula, we have

t∧τ 2 R d(s) ds |Yn(t ∧ τ) − Ym(t ∧ τ)| e 0

n∧τ s R d(u) du + e 0 ha(X(s))(Zn(s) − Zm(s)χ{s≤m∧τ}), Zt∧τ

Zn(s) − Zm(s)χ{s≤m∧τ}i ds n∧τ R d(s) ds 2 (3.19) = e 0 (E[ξ|Fn] − E[ξ|Fm])

n∧τ s∧τ R d(u) du 2 − e 0 |Yn(s ∧ τ) − Ym(s ∧ τ)| d(s) ds Zt∧τ

n∧τ s∧τ R d(u) du + 2 e 0 (Yn(s ∧ τ) − Ym(s ∧ τ)) Zt∧τ

× (f(s,Yn(s ∧ τ),Zn(s ∧ τ))

− f(s,Ym(s ∧ τ),Zm(s ∧ τ))) ds

n∧τ s∧τ R d(u) du + 2 e 0 (Yn(s ∧ τ) − Ym(s ∧ τ)) Zm∧τ

× f(s,Ym(s ∧ τ),Zm(s ∧ τ)) ds

n∧τ s∧τ R d(u) du − 2 e 0 (Yn(s ∧ τ) − Ym(s ∧ τ))hZn(s ∧ τ), dM(s)i Zt∧τ

m∧τ s∧τ R d(u) du + 2 e 0 (Yn(s ∧ τ) − Ym(s ∧ τ))hZm(s ∧ τ), dM(s)i. Zt∧τ 14 T. ZHANG

1 Choose δ1, δ2 such that λ < δ1 < δ and 0 < δ2 < δ − δ1. In view of the (A.1) and (A.2), we have

n∧τ s∧τ R d(u) du 2 e 0 (Yn(s ∧ τ) − Ym(s ∧ τ)) Zt∧τ

× (f(s,Yn(s ∧ τ),Zn(s ∧ τ)) − f(s,Ym(s ∧ τ),Zm(s ∧ τ))) ds

n∧τ s∧τ R d(u) du 2 ≤−2 e 0 (Yn(s ∧ τ) − Ym(s ∧ τ)) d1(s) ds Zt∧τ (3.20) n∧τ s∧τ 2 R d(u) du 2 + δ1d2 e 0 (Yn(s ∧ τ) − Ym(s ∧ τ)) ds Zt∧τ

n∧τ s 1 R d(u) du + e 0 ha(X(s))(Zn(s) − Zm(s)χ{s≤m∧τ}), λδ1 Zt∧τ

Zn(s) − Zm(s)χ{s≤m∧τ}i ds. On the other hand, by (A.3), it follows that

n∧τ s∧τ R d(u) du 2 e 0 (Yn(s ∧ τ) − Ym(s ∧ τ))f(s,Ym(s ∧ τ),Zm(s ∧ τ)) ds Zm∧τ

n∧τ s∧τ 2 R d(u) du 2 (3.21) ≤ δ2d2 e 0 (Yn(s ∧ τ) − Ym(s ∧ τ)) ds Zt∧τ n∧τ 1 s∧τ 2 R0 d(u) du + 2 e (|f(s, 0, 0)| + K + K|E[ξ|Fm]|) ds. δ2d2 Zm∧τ Take expectation and utilize (3.19)–(3.21) to obtain

t∧τ 2 R d(s) ds E[|Yn(t ∧ τ) − Ym(t ∧ τ)| e 0 ]

n∧τ s 1 R d(u) du + 1 − E e 0 ha(X(s))(Zn(s) − Zm(s)χ{s≤m∧τ}), λδ1 Zt∧τ

Z (s) − Z (s)χ i ds n m {s≤m∧τ} (3.22) n∧τ s 2 R d(u) du 2 + (δ − δ1 − δ2)d2E e 0 (Yn(s ∧ τ) − Ym(s ∧ τ)) ds Zt∧τ n∧τ R d(s) ds 2 ≤ E[e 0 (E[ξ|Fn] − E[ξ|Fm]) ] n∧τ 1 s∧τ 2 R0 d(u) du + 2 E e (|f(s, 0, 0)| + K + K|E[ξ|Fm]|) ds . δ2d2 Zm∧τ Since the right-hand side tends to zero as n,m →∞, we deduce that t∧τ t∧τ (1/2) R d(s) ds (1/2) R d(s) ds {(e 0 Yn(t), e 0 Zn(t))} A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 15 converges to some (Yˆ , Zˆ) in M 2(0,τ; R × Rd). Furthermore, for every t ≥ 0, t∧τ (1/2) R d(s) ds 2 e 0 Yn(t) converges in L . We may as well assume t∧τ (1/2) R d(s) ds (3.23) Yˆ (t) = lim e 0 Yn(t) n→∞ for all t. Observe that for any n ≥ t ≥ 0, t∧τ (1/2) R d(s) ds e 0 Yn(t)

n∧τ n∧τ s∧τ (1/2) R d(s) ds (1/2) R d(u) du = e 0 E[ξ|Fn]+ e 0 f(s,Yn(s),Zn(s)) ds Zτ∧t (3.24) n∧τ s∧τ 1 (1/2) R d(u) du − e 0 Yn(s)d(s) ds 2 Zτ∧t

n∧τ s∧τ (1/2) R d(u) du − e 0 hZn(s), dM(s)i. Zτ∧t Letting n →∞ yields that

τ τ s∧τ s∧τ (1/2) R d(s) ds (1/2) R d(u) du −(1/2) R d(u) du Yˆ (t)= e 0 ξ + e 0 f(s, e 0 Yˆ (s), Zτ∧t s∧τ −(1/2) R d(u) du (3.25) e 0 Zˆ(s)) ds 1 τ τ − Yˆ (s)d(s) ds − hZˆ(s), dM(s)i. 2 Zτ∧t Zτ∧t Put t∧τ t∧τ −(1/2) R d(s) ds −(1/2) R d(s) ds Y (t)= e 0 Yˆ (t),Z(t)= e 0 Zˆ(t). An application of Itˆo’s formula and (3.25) yield that

τ τ (3.26) Y (t)= ξ + f(s,Y (s),Z(s)) ds − hZ(s), dM(s)i. Zτ∧t Zτ∧t Hence, (Y,Z) is a solution to the bsde (3.15) proving the existence. To obtain the estimates (3.16) and (3.17), we proceed to get an uniform estimate for Yn(s) and then pass to the limit. Let δ1, δ2 be chosen as before. Similar to the proof of (3.8), by Itˆo’s formula, we have

t∧τ n∧τ s 2 R d(s) ds R d(u) du |Yn(t ∧ τ)| e 0 + e 0 ha(X(s))Zn(s),Zn(s)i ds Zt∧τ

n∧τ n∧τ s 2 R d(s) ds R d(u) du 2 ≤ |E[ξ|Fn]| e 0 − e 0 |Yn(s)| d(s) ds Zt∧τ

n∧τ s R d(u) du 2 − 2 e 0 d1(s)Yn (s) ds Zt∧τ 16 T. ZHANG

n∧τ s R d(u) du + 2 e 0 d2|Yn(s)||Zn(s)| ds Zt∧τ

n∧τ s R d(u) du + 2 e 0 |Yn(s)||f(s, 0, 0)| ds Zt∧τ

n∧τ s R d(u) du (3.27) − 2 e 0 Yn(s)hZn(s), dM(s)i Zt∧τ

n∧τ n∧τ s 2 R d(s) ds R d(u) du 2 2 ≤ |E[ξ|Fn]| e 0 − e 0 δd2Yn (s) ds Zt∧τ

n∧τ s R d(u) du 2 2 + e 0 δ1d2Yn (s) ds Zt∧τ

n∧τ s 1 R d(u) du + e 0 ha(X(s))Zn(s),Zn(s)i ds δ1λ Zt∧τ n∧τ n∧τ s 1 s 2 R0 d(u) du 2 R0 d(u) du 2 + δ2d2e Yn (s) ds + 2 e |f(s, 0, 0)| ds Zt∧τ δ2d2 Zt∧τ

n∧τ s R d(u) du − 2 e 0 Yn(s)hZn(s), dM(s)i. Zt∧τ

Recalling the choices of d(s), δ1 and δ2, using Burkholder’s inequality, we obtain from (3.27) that

t∧τ 2 R d(s) ds E sup |Yn(t ∧ τ)| e 0 h0≤t≤n i τ n∧τ s 1 2 R0 d(s) ds R0 d(u) du 2 (3.28) ≤ E[|ξ| e ]+ E e 2 |f(s, 0, 0)| ds Z0 δ2d2 1/2 n∧τ s 2 R d(u) du 2 + 2CE e 0 Yn (s)ha(X(s))Zn(s),Zn(s)i ds Z0 τ n∧τ s 1 2 R0 d(s) ds R0 d(u) du 2 ≤ E[|ξ| e ]+ E e 2 |f(s, 0, 0)| ds Z0 δ2d2

t∧τ 1 2 R d(s) ds + E sup |Yn(t ∧ τ)| e 0 2 h0≤t≤n i

n∧τ s R d(u) du + C1E e 0 ha(X(s))Zn(s),Zn(s)i ds . Z0 In view of (3.27), as the proof of (3.9), we can show that

n∧τ s R d(u) du (3.29) sup E e 0 ha(X(s))Zn(s),Zn(s)i ds < ∞. n Z0 A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 17

(3.29) and (3.28) together with our assumptions on f and ξ imply t∧τ 2 R d(s) ds (3.30) sup E sup |Yn(t ∧ τ)| e 0 < ∞. n h0≤t≤n i Applying Fatou lemma, (3.17) follows.

3.3. A particular case. Let f(x,y,z) : Rd × R × Rd → R be a Borel measurable function. Assume that f is continuous in y and satisﬁes: 2 (B.1) (y1 − y2)(f(x,y1, z) − f(x,y2, z)) ≤−c1(x)|y1 − y2| , where c1(x) is a measurable function on Rd. (B.2) |f(x,y,z1) − f(x,y,z2)|≤ c2|z1 − z2|. (B.3) |f(x,y,z)| ≤ |f(x, 0, z)| + c3(x)(1 + |y|). Let D be a bounded regular domain. Deﬁne x (3.31) τD = inf{t ≥ 0 : Xx(t) ∈/ D}. d Given g ∈ Cb(R ). Consider for each x ∈ D the following BSDE: x τD x Yx(t)= g(Xx(τD)) + f(Xx(s),Yx(s),Zx(s)) ds x Zt∧τD (3.32) x τD − hZx(s), dMx(s)i, x Zt∧τD where Mx(s) is the martingale part of Xx(s). As a consequence of Theo- rem 3.2, we have the following theorem.

P d Theorem 3.3. Suppose c3 ∈ L (D) for p> 2 , x τD 2 Ex exp (−2c1(X(s)) + δc2) ds < ∞, Z0 1 for some δ> λ and x τD 2 Ex |f(X(s), 0, 0)| ds < ∞. Z0 The BSDE (3.32) admits a unique solution (Yx(t),Zx(t)). Furthermore,

(3.33) sup |Yx(0)| < ∞. x∈D¯

4. Semilinear PDEs. As in previous sections, (X(t), Px) will denote the diﬀusion process deﬁned in (2.5).

4.1. Linear case. Consider the second order diﬀerential operator d d 1 ∂ ∂ ∂ (4.1) L = a (x) + b (x) + q(x). 2 2 ∂x ij ∂x i ∂x i,jX=1 i j Xi=1 i 18 T. ZHANG

Let D be a bounded domain with regular boundary (w.r.t. the Laplace operator ∆) and F (x) a measurable function satisfying (4.2) |F (x)|≤ C + C|q(x)|. Take ϕ ∈ C(∂D) and consider the Dirichlet boundary value problem L u = F, in D, (4.3) 2 u = ϕ, on ∂D.

Theorem 4.1. Assume (4.2) and that there exists x0 ∈ D such that x0 τD

Ex0 exp q(X(s)) ds < ∞. Z0 Then there is a unique, continuous weak solution u to the Dirichlet boundary value problem (4.3) which is given by x τD t x R q(X(s)) ds (4.4) u(x)= Ex ϕ(X(τD)) + e 0 F (X(t) dt . Z0 Proof. Write x u1(x)= Ex[ϕ(X(τD))], and x τD t R q(X(s)) ds u2(x)= Ex e 0 F (X(t)) dt . Z0

We know from Theorem 4.3 in [6] that u1 is the unique, continuous weak solution to the problem L u = 0, in D, (4.5) 2 u = ϕ, on ∂D.

So it is suﬃcient to show that u2 is the unique, continuous weak solution to the following problem: L u = F, in D, (4.6) 2 u = 0, on ∂D.

By Lemma 5.7 in [6] and Proposition 3.16 in [7], we know that u2 belong to C0(D). Let Gβ, β ≥ 0 denote the resolvent operators of the generator L2 on D with Dirichlet boundary condition, that is, x τD t −βt R q(X(s)) ds Gβf(x)= Ex e e 0 f(X(t)) dt . Z0 By the Markov property, it is easy to see that

β(u2(x) − βGβu2(x)) = βGβF (x). A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 19

Since Gβ is strong continuous, it follows that

lim β(u2 − βGβu2)= F β→∞ 2 1,2 in L (D). This shows that u2 ∈ D(L2) ⊂ W (D) and L2u2 = F . The proof is complete.

4.2. Semilinear case. Let g(x,y,z) : Rd × R × Rd → R be a Borel measurable function that satisﬁes: 2 (C.1) (y1 − y2)(g(x,y1, z) − g(x,y2, z)) ≤−k1(x)|y1 − y2| , (C.2) |g(x,y,z1) − g(x,y,z2)|≤ k2|z1 − z2|, (C.3) |g(x,y,z)|≤ C + C|q(x)|, where k1(x) is a measurable function and k2,C are constants. Consider the semilinear Dirichlet boundary value problem L u = −g(x, u(x), ∇u(x)), in D, (4.7) 2 u = ϕ, on ∂D, where ϕ ∈ C(∂D).

Theorem 4.2. Assume x τD 2 Ex exp (q(X(s)) − 2k1(X(s)) + δk2) ds < ∞, Z0 1 for some δ> λ . The Dirichlet boundary value problem (4.7) has a unique continuous weak solution.

Proof. Set f(x,y,z)= q(x)y +g(x,y,z). According to Theorem 3.3, for every x ∈ D the following BSDE: x τD x Yx(t)= φ(Xx(τD)) + f(Xx(s),Yx(s),Zx(s)) ds x Zt∧τD (4.8) x τD − hZx(s), dMx(s)i, x Zt∧τD admits a unique solution (Yx(t),Zx(t)),t ≥ 0. Put u0(x)= Yx(0) and v0(x)= Zx(0). By the strong Markov property of X and the uniqueness of the BSDE (4.8), it is easy to see that x (4.9) Yx(t)= u0(Xx(t)),Zx(t)= v0(Xx(t)), 0 ≤ t ≤ τD. Now consider the following problem: L u = −f(x, u (x), v (x)), in D, (4.10) 1 0 0 u = ϕ, on ∂D, 20 T. ZHANG where L1 is deﬁned as in Section 2. By Theorem 4.1, problem (4.10) has a unique continuous weak solution u(x). As u ∈ W 1,2(D), it follows from the x decomposition of the Dirichlet process u(X(t ∧ τD)) (see [12, 17]) that x τD x x x x u(X(t ∧ τD)) = ϕ(Xx(τD)) + f(Xx(s), u0(X(s ∧ τD)), v0(X(s ∧ τD))) ds x Zt∧τD x τD x − h∇u(X(s ∧ τD)), dMx(s)i x Zt∧τD (4.11) x τD x = ϕ(Xx(τD)) + f(Xx(s),Yx(s),Zx(s))) ds x Zt∧τD x τD x − h∇u(X(s ∧ τD)), dMx(s)i. x Zt∧τD Take conditional expectation both in (4.11) and (4.8) to discover x x Yx(t ∧ τD)= u(X(t ∧ τD)) x τD x x = E ϕ(Xx(τD)) + f(Xx(s),Yx(s),Zx(s)) ds Ft∧τD . Zt∧τ x D

In particular, let t = 0 to obtain u(x)= u0(x). On the other hand, comparing (4.8) with (4.11) and by the uniqueness of decomposition of semimartingales, we deduce that x x τD τD x h∇u(X(s ∧ τD)), dMx(s)i = hZx(s), dMx(s)i x x Zt∧τD Zt∧τD for all t. By Itˆo’s isometry, we have ∞ 2 E h(∇u(X(s)) − Z (s))χ x , dM (s)i x {s<τD} x Z0 ∞ 2 = E h(∇u(X(s)) − v (X (s)))χ x , dM (s)i 0 x {s<τD} x Z0 (4.12) ∞ = E ha(Xx(s))(∇u(X(s)) − v0(Xx(s))), Z0

∇u(X(s)) − v0(Xx(s))iχ x ds = 0. {s<τD} By Fubini theorem and the uniform ellipticity of the matrix a(x), we deduce that D 2 2 P (|∇u − v | )= E[|∇u(X(s)) − v (X (s))| χ x ] = 0 s 0 0 x {s<τD} A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 21

D a.e. in s with respect to the Lebesgue measure, where Ps h(x)= Ex[h(X(t)), x D t<τD]. The strong continuity of the semigroup Ps ,s ≥ 0 implies that 2 D 2 (4.13) |∇u − v0| (x) = lim Ps (|∇u − v0| ) = 0 s→0 a.e. Returning to problem (4.10), we see that u actually is a weak solution to the nonlinear problem: L u = −f(x, u(x), ∇u(x)), in D, (4.14) 0 u = φ, on ∂D. Supposeu ¯ is another solution to the problem (4.14). By the decomposition of the Dirichlet processu ¯(Xx(s)), we ﬁnd that (¯u(Xx(s)), ∇u¯(Xx(s))) is also a solution to the BSDE (4.8). The uniqueness of the BSDE implies thatu ¯(Xx(s)) = Yx(s). In particular,u ¯(x)= u0(x)= Yx(0). This proves the uniqueness.

5. Semilinear elliptic PDEs with singular coeﬃcients. In this section, we study the semilinear second order elliptic PDEs of the following form: Au(x)= −f(x, u(x)), ∀x ∈ D, (5.1) u(x)|∂D = ϕ, ∀x ∈ ∂D, where the operator A is given by d d 1 ∂ ∂ ∂ A = a (x) + b (x) − “div(ˆb·)”+ q(x) 2 ∂x ij ∂x i ∂x i,jX=1 i j Xi=1 i as in Section 2 and ϕ ∈ C(∂D). Consider the following conditions: 2 (D.1) (y1 − y2)(f(x,y1) − f(x,y2)) ≤−J1(x)|y1 − y2| , (D.2) |f(x,y,z)|≤ C, where J1(x) is a measurable function, C is a constant. The following theorem is the main result of this section.

Theorem 5.1. Suppose that (D.1), (D.2) hold and

τD 0 −1 0 0 Ex exp h(a b)(X (s)), dM (s)i Z0

τD −1ˆ 0 0 + h(a b)(X (s)), dM (s)i ◦ rτD Z0 1 τD (5.2) − (b − ˆb)a−1(b − ˆb)∗(X0(s)) ds 2 Z0 τD τD 0 0 + q(X (s)) ds − 2 J1(X (s)) ds Z0 Z0 < ∞ 22 T. ZHANG

0 for some x ∈ D, where X is the diﬀusion generated by L0 as in Section 2 0 and τD is the ﬁrst exit time of X from D. Then there exists a unique, continuous weak solution to equation (5.1).

Proof. Set t t −1 0 0 −1 0 0 Zt = exp h(a b)(X (s)), dM (s)i + h(a ˆb)(X (s)), dM (s)i ◦ rt Z0 Z0 1 t t (5.3) − (b − ˆb)a−1(b − ˆb)∗(X0(s)) ds + q(X0(s)) ds 2 Z0 Z0 t 0 − 2 J1(X (s)) ds . Z0 Put t Mˆ (t)= h(a−1ˆb)(X0(s)), dM 0(s)i for t ≥ 0. Z0

Let R> 0 so that D ⊂ BR := B(0,R). By Lemma 3.2 in [5] (see also [9]), 1,p 1,2 there exits a bounded function v ∈ W0 (BR) ⊂ W0 (BR) such that v (Mˆ (t)) ◦ rt = −Mˆ (t)+ N (t), where N v is the zero energy part of the Fukushima decomposition for the Dirichlet process v(X0(t)). Furthermore, v satisﬁes the following equation in the distributional sense:

(5.4) div(a∇v)= −2 div(ˆb) in BR. Note that by Sobolev embedding theorem, v ∈ C(Rd) if we extend v =0 on Dc. This implies that Mˆ and N v are continuous additive functionals of X0 in the strict sense (see [9, 12]), and so is t → (Mˆ (t)) ◦ rt. Thus, t −1 0 0 h(a ˆb)(X (s)), dM (s)i ◦ rt Z0 t = − h(a−1ˆb)(X0(s)), dM 0(s)i + N v(t) Z0 t = − h(a−1ˆb)(X0(s)), dM 0(s)i + v(X0(t)) − v(X0(0)) − M v(t) Z0 t = − h(a−1ˆb)(X0(s)), dM 0(s)i + v(X0(t)) − v(X0(0)) Z0 t − h∇v(X0(s)), dM 0(s)i. Z0 A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 23

Hence, 0 ev(X (t)) Zt = ev(X0(0)) t t −1 0 0 0 × exp ha (b − ˆb − a∇v)(X (s)), dM (s)i− 2 J1(X (s)) ds Z0 Z0 (5.5) t 1 + q − (b − ˆb − a∇v)a−1(b − ˆb − a∇v)∗ (X0(s)) ds Z0 2 t 1 + (∇v)a(∇v)∗ − hb − ˆb, ∇vi (X0(s)) ds . Z0 2 0 v(x) Note that Zt is well defined under Px for every x ∈ D. Set h(x)= e . Introduce d d 1 ∂ ∂ ∂ Aˆ = a (x) + [b (x) − ˆb (x) − (a∇v) (x)] 2 ∂x ij ∂x i i i ∂x i,jX=1 i j Xi=1 i 1 − hb − ˆb, ∇vi(x)+ (∇v)a(∇v)∗(x)+ q(x). 2 d Let (Ω, F, Ft, Xˆ (t), Pˆx,x ∈ R ) be the diffusion process whose infinitesimal generator is given by d d 1 ∂ ∂ ∂ Lˆ = a (x) + [b (x) − ˆb (x) − (a∇v) (x)] . 2 ∂x ij ∂x i i i ∂x i,jX=1 i j Xi=1 i ˆ 0 It is known from [17] that Px is absolutely continuous with respect to Px and dPˆ x = Zˆ , dP 0 t x Ft where t −1 0 0 Zˆt = exp ha (b − ˆb − a∇v)(X (s)), dM (s)i Z0 (5.6) t 1 − (b − ˆb − a∇v)a−1(b − ˆb − a∇v)∗ (X0(s)) ds . Z0 2 Put fˆ(x,y)= h(x)f(x, h−1(x)y). Then 2 (y1 − y2)(f(x,y1) − f(x,y2)) ≤−J1(x)|y1 − y2| . 24 T. ZHANG

Consider the following nonlinear elliptic partial diﬀerential equation: ˆ ˆ (5.7) Auˆ(x)= −f(x, uˆ(x)), ∀x ∈ D, uˆ(x)|∂D = h(x)v(x), ∀x ∈ ∂D. In view of (5.5), condition (5.2) implies that

τD τD 0 0 Eˆx exp −2 J1(X (s)) ds + q(X (s)) ds Z0 Z0 (5.8) τD 1 + (∇v)a(∇v)∗ − hb − ˆb, ∇vi (X0(s)) ds < ∞, Z0 2 where Eˆx indicates that the expectation is taken under Pˆx. From Theo- rem 4.2, it follows that equation (5.7) admits a unique weak solutionu ˆ. Set u(x)= h−1(x)ˆu(x). We will verify that u is a weak solution to equation (5.1). 1,2 Indeed, for ψ ∈ W0 (D), sinceu ˆ(x)= h(x)u(x) is a weak solution to equation (5.7), it follows that

d 1 ∂[h(x)u(x)] ∂[h−1(x)ψ] aij(x) dx 2 Z ∂x ∂x i,jX=1 D i j

d ∂[h(x)u(x)] −1 − [bi(x) − ˆbi(x) − (a∇v)i(x)] h (x)ψ dx Z ∂x Xi=1 D i

+ hb − ˆb, ∇v(x)iu(x)ψ(x) dx ZD 1 − (∇v)a(∇v)∗(x)u(x)ψ dx − q(x)u(x)ψ(x) dx 2 ZD ZD

= f(x, u(x))ψ(x) dx. ZD

Denote the terms on the left of the above equality, respectively, by T1, T2, T3, T4, T5. Clearly, d d 1 ∂u(x) ∂ψ 1 ∂u(x) ∂v T1 = aij(x) dx − aij(x) ψ dx 2 Z ∂x ∂x 2 Z ∂x ∂x i,jX=1 D i j i,jX=1 D i j

d 1 ∂v ∂ψ (5.9) + aij(x) u(x) dx 2 Z ∂x ∂x i,jX=1 D i j

d 1 ∂v ∂v − aij(x) ψu(x) dx. 2 Z ∂x ∂x i,jX=1 D i j A PROBABILISTIC APPROACH TO DIRICHLET PROBLEMS OF PDES 25

Using chain rules, rearranging terms, it turns out that d d ∂u(x) ∂ψ T2 + T3 = − bi(x) ψ dx − ˆbi(x) u(x) dx Z ∂x Z ∂x Xi=1 D i Xi=1 D i d ∂[ψu(x)] (5.10) + [ˆbi(x) + (a∇v)i(x)] dx Z ∂x Xi=1 D i d d ∂ψ ∂v − (a∇v)i(x) u(x) dx + (a∇v)i(x) u(x)ψ dx. Z ∂x Z ∂x Xi=1 D i Xi=1 D i In view of (5.4), d ∂[ψu(x)] [ˆbi(x) + (a∇v)i(x)] dx Z ∂x Xi=1 D i (5.11) d 1 ∂[ψu(x)] = (a∇v)i(x) dx. 2 Z ∂x Xi=1 D i Thus, d d ∂u(x) ∂ψ T2 + T3 = − bi(x) ψ dx − ˆbi(x) u(x) dx Z ∂x Z ∂x Xi=1 D i Xi=1 D i d 1 ∂[ψu(x)] (5.12) + (a∇v)i(x) dx 2 Z ∂x Xi=1 D i d d ∂ψ ∂v − (a∇v)i(x) u(x) dx + (a∇v)i(x) u(x)ψ dx. Z ∂x Z ∂x Xi=1 D i Xi=1 D i After cancelations, it is now easy to see that d 1 ∂u(x) ∂ψ T1 + T2 + T3 + T4 + T5 = aij(x) dx 2 Z ∂x ∂x i,jX=1 D i j

d ∂u(x) − bi(x) ψ dx Z ∂x Xi=1 D i (5.13) d ∂ψ − ˆb u(x) dx − q(x)u(x)ψ(x) dx Z ∂x Z Xi D i D

= f(x, u(x))ψ(x) dx. ZD 26 T. ZHANG

Since ψ is arbitrary, we conclude that u is a weak solution of equation (5.1). Suppose u is a continuous weak solution to equation (5.1). Putu ˆ(x)= h(x)u(x). Reversing the above process, we see thatu ˆ is a weak solution to equation (5.7). The uniqueness of the solution of equation (5.1) follows from that of equation (5.7).

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