A probabilistic approach to mixed boundary value problems for elliptic operators with singular coefficients Zhen-Qing Chen∗ and
Tusheng Zhang
Abstract In this paper, we establish existence and uniqueness of solutions of a class of mixed boundary value problems for elliptic operators with singular coefficients. Our approach is probabilistic. The theory of Dirichlet forms plays an important role.
AMS Subject Classification: Primary 60H15 Secondary 93E20, 35R60.
1
Introduction
The pioneering work by Kakutani [17] in 1944 on representing the solution to the classical Dirichlet boundary value problem { ∆u = 0 in D, u=ϕ
on ∂D,
using Brownian motion started a new era in the very fruitful interplay between probability theory and analysis. Here D is a bounded connected open subset of Rd . Since then, in place of Laplacian ∆, more general second order elliptic differential operators have been studied in connection with probabilistic approach. In another direction, other boundary conditions (Nuemann and mixed boundary) have been investigated using probabilistic methods. For example, in [16], Hsu studied solutions to the classical Neumann boundary value problems on a bounded domain D in Rd with C 3 boundary: { ∆u = 0 in D, ∂u ∂n
=ϕ
on ∂D,
using reflecting Brownian motion on D, where n is the unit inward normal vector field of D on ∂D. We refer the reader to [6, 7, 9, 12, 16, 19] and [21] for a brief history on the related subject. Let d ≥ 1 be an integer and D a bounded Lipschitz domain in Rd . Let A = (aij ) : D → Rd × Rd be a Borel measurable, symmetric matrix-valued function that is uniformly elliptic and bounded, that is, there is are constants λ2 ≥ λ1 > 0 such that λ1
d ∑ i=1
∗
ξi2
≤
d ∑
aij (x)ξi ξj ≤ λ2
i,j=1
d ∑
ξi2
for every x, ξ ∈ Rd .
i=1
Research partially supported by NSF Grants DMS-0906743 and DMR-1035196.
1
(1.1)
Let n(x) = (n1 (x), ..., nd (x)) be the unit inward normal vector on ∂D. If A(x) is defined on the → boundary ∂D, it determines a conormal vector field − γ = (γ1 , · · · , γd ) on ∂D by 12 An, that is, ∑ b = (bb1 , · · · , bbd ) γi (x) = 21 dj=1 aij (x)nj (x) for i = 1, · · · , d. Suppose that b = (b1 , · · · , bd ) and b are measurable Rd -valued functions on D and q is a measurable real-valued function on D so that b ∈ Lp2 (D; dx) |b|
|b| ∈ Lp1 (D; dx),
and q ∈ Lp3 (D; dx)
(1.2)
b and q to Rd for some p1 > d, p2 > d, p3 > d2 . We always extend the definition of the functions b, b by setting their values to be zero off D. In this paper, we study the mixed boundary value problem { Lu = 0 on D (1.3) ∂u b ∂γ − ⟨b, n⟩u = ϕ on ∂D in the bounded Lipschitz domain D for the second order elliptic operator L of the following form: L = =
1 b + q, ∇ · (A∇) + b · ∇ − div(b·) 2 ( ) ∑ n n 1 ∑ ∂ ∂ ∂ b + q(x) aij (x) + bi (x) − “div(b·)” 2 ∂xi ∂xj ∂xi i,j=1
(1.4)
i=1
b in (1.4) is just a formal writing because the divergence really does not exist Note that “div(b·)” b It should be interpreted in the distributional sense. It is for the merely measurable vector field b. b the lower order term div(b·) b can not be handled by exactly due to the non-differentiability of b, Girsanov transform or Feynman-Kac transform. In [6], we studied the Dirichlet boundary value b by a time-reversal of problem for such operator L, where tackled the lower order term div(b·) Girsanov transform from the first exit time τD from D by the symmetric diffusion X associated with 21 ∇(A∇). We show in this paper that a similar strategy works for the mixed boundary value problem (1.3) but we need to run the reflecting symmetric diffusion X on D associated with the differential operator 12 ∇(A∇) instead [3]. Now we give a precise definition of solutions to the mixed boundary value problem (1.3). Let := {u ∈ L2 (D; dx) : ∇u ∈ L2 (D; dx)} be the Sobolev space on D of order (1, 2) equipped with norm (∫ )1/2 ( ) 2 2 ∥u∥1,2 := |∇u(x)| + u(x) dx .
W 1,2 (D)
D
Define for u, v ∈
W 1,2 (D),
a bilinear form
Q(u, v) :=
d ∫ d ∫ ∑ 1 ∑ ∂u ∂v ∂u dx − v(x)dx aij (x) bi (x) 2 ∂xi ∂xj ∂xi D D i,j=1
−
i=1
d ∫ ∑ i=1
bbi (x) ∂v u(x)dx − ∂xi D
∫ q(x)u(x)v(x)dx.
(1.5)
D
Under the conditions of (1.1) and (1.2), the quadratic form Q on W 1,2 (D) is well defined and has the following properties: there are constants K ≥ 1 and α > 0 so that for u, v ∈ W 1,2 (D), K −1 ∥u∥21,2 ≤ Qα (u, u) ≤ K∥u∥21,2
and |Q(u, v)| ≤ K∥u∥1,2 ∥v∥1,2 , 2
(1.6)
∫ where Qα (u, v) := Q(u, v) + α D u(x)v(x)dx. We then define the differential operator L as follows: u ∈ Dom(L) if and only if u ∈ W 1,2 (D) and there is a function g ∈ L2 (Rd ; x) so that ∫ Q(u, v) = − g(x)v(x) dx for every v ∈ W 1,2 (D). (1.7) Rd
Clearly such g is unique and we define Lu = g. We call L the infinitesimal generator of the quadratic form (Q, W 1,2 (D)) on L2 (D; dx). If aij ∈ W 1,2 (D), then it admits a quasi-continuous version on D and so the conormal vector field γ(x) := 12 A(x)n(x) is quasi-everywhere defined in ∂D. An integration by part (see [3]) implies that for a function u ∈ Dom(L), Lu has the expression (1.4) and satisfies the following mixed boundary condition in the distributional sense: ∂u b n⟩u(x) = 0 (x) − ⟨b, ∂γ
on
∂D
(1.8)
Here for two vectors α and β in Rd , we use α · β or ⟨α, β⟩ to denote their inner product. We remark that the condition aij ∈ W 1,2 (D) is not needed in many results of this paper. It is only needed when we want to interpret the boundary condition of L in an explicit form (1.8) Put C 2 (D) = {f |D ; f ∈ C 2 (Rd )}. Definition 1.1 We say that a measurable function u is a weak solution of the boundary value problem (1.3) if u ∈ W 1,2 (D) and ∫ Q(u, f ) = ϕ(x)f (x)σ(dx) for every f ∈ C 2 (D). (1.9) ∂D
Here Q is the quadratic form defined in (1.5) and σ(dx) is the Lebesgue surface measure on the boundary ∂D. To see that the above formulation of weak solution is sensible, observe that if all functions involved are smooth, then using Green-Gauss formula, we have every f ∈ C 2 (D), ) ∫ ∫ ( ∂u b Q(u, f ) = (x) − ⟨b, n⟩u(x) f (x)σ(dx). −Lu(x)f (x)dx + D ∂D ∂γ It is not difficulty to see that the differential operator L enjoys the maximum principle if and b + q ≤ 0 on D in the following distributional sense: only if −div(b) n ∫ ∑ i=1
bbi (x) ∂ϕ dx + ∂xi D
∫ q(x)ϕ(x)dx ≤ 0,
(1.10)
D
for all nonnegative function ϕ in Cc2 (D). We point out that we do not impose such a Markovian condition in this paper. The aim of the present paper is twofold. The first is to give a probabilistic representation for the weak solution of the mixed boundary value problem (1.3). This is highly non-trivial because there is no longer a Markov process associated with the operator L due to the appearance of the lower b nor can that lower order term be handled via Girsanov transform or Feynmanorder term div(b·), Kac transform. Motivated by [6, 5, 19], our idea is to use the symmetric reflected diffusion process 3
X associated with the divergence form operator 12 ∇(A∇), the symmetric part of L, and treat L as its lower order perturbation via a combination of Girsanov and Feynman-Kac transforms and a time-reversal of Girsanov transform. Based on the new probabilistic representation, our second aim is to establish the existence and uniqueness of the weak solution to mixed boundary problem (1.3) without the Markovian assumption (1.10). To this end, we introduce a kind of h-transformation which transforms the solution of the mixed boundary problem (1.3) to the solution of a mixed b The boundary value problem for operators which do not involve the adjoint vector field like b. time reversal and the theory of Dirichlet forms play an essential role throughout this paper. Let X be the symmetric reflecting diffusion on D with infinitesimal generator 12 ∇(A∇). It is known (cf. [3] and [15]) that X is a conservative Feller process on Rd that has H¨older continuous transition density function which admits a two-sided Aronson’s Gaussian type estimate. In general, X is a not a semimartingale but it admits the following Fukushima’s decomposition (cf. [11]): Xt = X0 + Mt + Nt ,
t ≥ 0,
(1.11)
where M = (M , · · · , M n ) is a martingale additive functional of X with quadratic co-variation ∫t ⟨M i , M j ⟩t = 0 aij (Xs )ds and N = (N 1 , · · · , N n ) is a continuous additive functional (CAF in abbreviation) of X locally of zero quadratic variations. Note that, since X has a continuous density function and each aij is bounded on D, by [10, Theorem 2], M and N can be refined to be CAFs of X in the strict sense without exceptional set and (1.11) holds under Px for every x ∈ D. Without loss of generality, we work on the canonical continuous path space C([0, ∞), D) of X and, for t > 0, denote by rt the reverse operator of X from time t. Define (∫ t ∫ t −1 ∗ b ∗ (Xs )dMs ) ◦ γt (A−1 b) Zt = exp (A b) (Xs )dMs + 0 0 ) ∫ ∫ t 1 t −1 ∗ b b (Xs )ds + − (1.12) (b − b)A (b − b) q(Xs )ds . 2 0 0 All the vectors in this paper are row vectors and we use b∗ to denote the transpose of a vector b. Let {Tt , t ≥ 0} be the semigroup associated with the quadratic form (Q, W 1,2 (D)). It is established in [5] that for every f ∈ L2 (D; dx), Tt f (x) = Ex [Zt f (Xt )] for a.e. x ∈ D. Let Lt be the positive continuous additive functional (PCAF in abbreviation) of X having the surface measure σ(dx) of ∂D as its Revuz measure. The PCAF L is called the boundary local time of the reflecting diffusion X. We introduce the following condition. Condition (A): there is some p > d so that for every F = (F1 , . . . , Fd ) ∈ Lp (D; dx) and every v ∈ W 1,2 (D) that satisfies ∫ ∑ ∫ ∑ d d ∂v(x) ∂ϕ(x) ∂ϕ(x) aij (x) dx = aij (x)Fi (x) dx for every ϕ ∈ W 1,2 (D), (1.13) ∂xi ∂xj ∂xj D D i,j=1
we have v ∈
i,j=1
W 1,p (D)
:= {f ∈
Lp (D; dx)
: ∇f ∈ Lp (D; dx)}.
Note that (1.13) says that v is a solution in the distributional sense to the following elliptic equation with Neumann boundary condition: { div(A∇v) = div(AF ) on D (1.14) ∂v on ∂D. ∂γ = ⟨F, γ⟩ 4
Recall λk , k = 1, 2, are the constants in (1.1). By [14, Theorem 1 and Remark 1], Condition (A) is satisfied when D when D is a bounded C 1 domain and λ2 /λ2 is close to 1. Moreover, by [25], Condition (A) is satisfied when each aij is smooth and D is a bounded Lipschitz domain in Rd . The following is the main result of this paper. [∫ ∞ ] Theorem 1.2 Suppose that Condition (A) holds. Assume that Ex0 0 Zs dLs < ∞ for some point x0 ∈ D and let ϕ be a bounded measurable function on ∂D. Then [∫ ∞ ] u(x) := Ex Zs ϕ(Xs )dLs 0
is bounded continuous and is the unique bounded weak solution of the mixed boundary value problem (1.3). The remaining of the paper is organized as follows. In Section 2, we present some basic properties of symmetric reflecting diffusion X, including the definition of time-reversal operator rt A crucial h-transform is given there that transform the differential operator L into Le (see (2.13) below) that b u). In Section 3, we first show that, under does not involve the adjoint vector field like div(b [∫ ∞ ] Condition (A), the gauge function g(x) := Ex 0 Zs dLs is either bounded or identically infinite on D. We then proceed to give the proof for the main result, Theorem 1.2. In this paper, we use “:=” as a way of definition. A statement is said to hold quasi-everywhere (q.e. in abbreviation) on some set A ⊂ Rd if there is an exceptional set N of zero capacity so that the statement holds on A \ N . For the general theory of Dirichlet forms and Markov processes and their terminology, we refer readers to [11] and [20] .
2
Preliminaries
Let D be a bounded Lipschitz domain in Rd . Define d ∫ 1 ∑ ∂u ∂v E (u, v) = aij (x) dx 2 ∂x i ∂xj D 0
for u, v ∈ W 1,2 (D).
(2.1)
i,j=1
It is known that (E 0 , W 1,2 (D)) is a regular Dirichlet form on L2 (D; dx). So there is a symmetric continuous Hunt process X := {Ω, F, Xt , θt , γt , Px , x ∈ D} associated with it, which is called reflected diffusion on D (cf. [3]). By a similar argument as that in [2] or by Theorem 3.34 in [15], X has a jointly continuous transition density function p(t, x, y) on (0, ∞) × D × D; moreover there are constants c1 , c2 ≥ 1 so that for every t ∈ (0, 1] and x, y ∈ D, ( ) ( ) c2 |x − y|2 |x − y|2 −1 −d/2 −d/2 c1 t exp − ≤ p(t, x, y) ≤ c1 t exp − . (2.2) t c2 t Consequently, X can be refined to start from every point on D as a Feller process having strong Feller property. Without loss of generality, we can take Ω to be the canonical space C([0, ∞) → D) of continuous functions on D and Xt the coordinate projection process. Let {θt , t ≥ 0} and {rt , t ≥ 0} be the shift and reverse operators defined on Ω by Xs (θt (ω)) = Xt+s (ω) for s, t ≥ 0
and 5
Xs (rt (ω)) = Xt−s (ω) for s ≤ t.
(2.3)
We will use Ex to denote the expectation with respect to Px . For any u ∈ W 1,2 (D), it is well known that the following Fukushima’s decomposition holds: u(Xt ) − u(X0 ) = Mtu + Ntu ,
t ≥ 0,
Px -a.s.
(2.4)
for q.e. x ∈ D, where Mtu is an {Ft }t≥0 -square integrable continuous martingale additive functional and Ntu is a continuous additive functional of zero energy. Here {Ft }t≥0 is the minimum augmented σ-field generated by the reflected diffusion X on D. In particular, taking u(x) = xi yields that Xt = x + Mt + Nt ,
t ≥ 0,
Px -a.s.
(2.5)
for every x ∈ D, where M = (M 1 , ..., M d ) is a square integrable continuous martingale additive functional of X with ∫ t
⟨M i , M j ⟩t =
t ≥ 0,
aij (Xs )ds,
(2.6)
0
and each component of N is a continuous additive functional of X of zero energy. Let {Tt , t ≥ 0} be the semigroup associated with the quadratic form (Q, W 1,2 (D)). It admits the following probabilistic representation, which is established in [5]. Proposition 2.1 Let Zt be defined as in (1.12). Then for every non-negative f on D, Tt f (x) = Ex [Zt f (Xt )] a.e. on D. For a martingale additive functional K = {Kt , t ≥ 0}, recall that the functional Γ(K)t of local zero energy defined in [22] and [8]: 1 Γ(K)t = − (Kt + Kt ◦ rt ). 2 Lemma 2.2 Suppose that Condition (A) holds. Let F = (F1 , . . . , Fd ) be a function on D so ∫ ct = t F (Xs )dMs for t ≥ 0. Then there exists a function that |F | ∈ L2 (D; dx) and define M 0 v ∈ W 1,2 (D)) such that c)t = Γ(M v )t = Ntv . Γ(M (2.7) and (1.13) is satisfied. Moreover, if |F | ∈ Lp (D; dx) for some p > d, then the above v ∈ W 1,p (D) and therefore it has a bounded and continuous version on D. c is a martingale additive functional of X of finite energy. Thus by Corollary Proof. Note that M 3.2 in [8], there is some v ∈ W 1,2 (D) and a martingale additive functional M 1 of X satisfying Γ(M 1 ) = 0, e(M 1 , M ϕ ) := µ⟨M 1 ,M ϕ ⟩ (D) = 0 for all ϕ ∈ W 1,2 (D), and that ct = Mtv + Mt1 , M Hence, we have
t ≥ 0.
c)t = Γ(M v )t = Ntv . Γ(M
On the other hand, by the representation of martingale additive functionals ([11]), ∫ t ∫ t v 1 Mt = ∇v(Xs )dMs , Mt = G(Xs )dMs 0
0
6
(2.8)
for some measurable vector field G = (G1 (x), ..., Gd (x)) : D → Rd . Since µ⟨M 1 ,M ϕ ⟩ (D) = 0 for all ϕ ∈ W 1,2 (D)), it follows that ∫
d ∑
aij (x)
D i,j=1
ct = On the other hand, since M
∫t 0
∂ϕ Gj (x)dx = 0 ∂xi F (Xs )dMs =
∫t
0 (∇v
for every ϕ ∈ W 1,2 (D).
(2.9)
+ G)(Xs ))dMs for every t ≥ 0, we have
F (x) = ∇v(x) + G(x)
a.e.,
from which we conclude that v satisfies (1.13). Now assume that F ∈ Lp (D; dx) for some p > d. Condition (A) implies that v ∈ W 1,p (D). Now by the Sobolev embedding theorem, v has a bounded and continuous version on D. 2 The function v in Lemma 2.2 allows us to drop the time-reversal operator rt in the expression of Z in (1.12) and rewrite it as a combination of h-transform, Girsanov transform and Feynmanb ∈ Lp (D; dx), by Lemma 2.2 there is some v ∈ W 1,2 (D) with a bounded Kac transform Since b continuous version so that ∫ t ∫ t −1 b ∗ v b ∗ (Xs )dMs (2.10) (A b) (Xs )dMs ◦ rt = Nt − (A−1 b) 0 0 ∫ t ( ) b ∗ (Xs )dMs . = v(Xt ) − v(X0 ) − ∇v + A−1 b 0
Hence the multiplicative function Z in (1.12) can be expressed as Zt =
ev(Xt ) e Zt , ev(X0 )
(2.11)
where et Z
) ∫ t ∫ t 1 −1 ∗ b − ∇v (Xs )dMs − b b (Xs )ds + A (b − b) q(Xs )ds = exp (b − b)A (b − b) 2 0 0 0 (∫ t ∫ ( −1 )∗ ) ( ) 1 t( b b − A∇v · A−1 b − b b − A∇v (Xs )ds A (b − b) − ∇v (Xs )dMs − = exp b−b 2 0 0 ) ∫ t( ) 1 b + ∇v · A∇v − ⟨b − b, ∇v⟩ + q (Xs )ds . (2.12) 2 0 (∫
t(
Let Tet be defined by
−1
)∗
[ ] et f (Xt ) (= e−v(x) Tt (ev f )(x)). Tet f (x) = Ex Z
Observe that Zet is a combination of Girsanov transform followed by a Feynman-Kac transform, and so {Tet ; t ≥ 0} is the semigroup associated with differential operator ( ) b − A∇v) · ∇u + 1 ∇v · A∇v − ⟨b − b, b ∇v⟩ + q u e := 1 ∇ · (A∇u) + (b − b Lu 2 2
7
(2.13)
3
Mixed boundary value problem
In this section, we consider the following boundary value problem (1.3). We assume throughout this section that Condition (A) holds. Recall that Zt is the multiplicative functional of X defined by (1.12) and L = {Ls , s ≥ 0} be the boundary local time of the reflected diffusion process {Xt , t ≥ 0} on D, i.e. the continuous additive functional associated with the smooth measure σ(dx). Theorem 3.1 Assume that Condition (A) holds. The function ] [∫ ∞ Zs dLs < ∞ x 7→ Ex
(3.1)
0
is either bounded or identically infinite on D. Proof. In view of (2.11), the function Ex
[∫ ∞ 0
[∫ ] ] ∞ e Zs dLs is comparable to u(x) := Ex 0 Z dL s s
and so it suffices to show that either u is bounded or identically infinite on D. In view of (2.2), one can deduce by a similar argument as that for [7, Theorem 3.1] together with Khasminskii’s lemma that [ 2] [ 2] e < ∞ and e = 1. lim sup Ex Z sup Ex Z t t t→0
x∈D
x∈D
Moreover, es ≥ exp inf Z
0≤s≤t
(
∫ t ∫ t( ) b − ∇v ∗ (Xs )dMs − 1 b −1 (b − b) b ∗ (Xs )ds − sup A−1 (b − b) (b − b)A 2 s≤t 0 0 ) ∫ t − |q(Xs )|ds . 0
In addition, Ex [Lt ] =
∫ t∫ p(t, x, y)σ(dy)ds. 0
∂D
It follows from (2.2) that there is a constant c > 0 so that √ Ex [Lt ] ≤ c t for every x ∈ D and t ≤ t0 . With these ingredients, one can show that either u is bounded or identically infinite on D in a similar way as that for [16, Theorem 2.3] thus its proof is omitted here. 2 ∫∞ Proof of Theorem 1.2. By Theorem 3.1, the gauge function Ex [ 0 Zs dLs ] is bounded on D, so is u. For any t > 0, by the Markov property of X, we have [∫ t ] u(x) = Ex Zs ϕ(Xs )dLs + Tt u(x). 0
b − A∇v| ∈ Lp (D; dx) for Since Tt is the semigroup for the differential operator Le in (2.13), |b − b b ∇v⟩ + q ∈ Lq (D; dx) for some q > d/2, one can derive in a some p > d and 12 ∇v · A∇v − ⟨b − b, similar way as that in [7] that Tet f is a bounded continuous function on D for every f ∈ L2 (D; dx) 8
and t > 0. Consequently, Tt u(x) = e−v(x) Tet (ev u)(x) is a bounded continuous function on D for every t > 0. On the other hand, by the proof of [16, Lemma 2.1], [∫ t ] lim Ex Zs ϕ(Xs )dLs = 0 uniformly on D. t→0
0
It follows that u(x) is continuous on D. Under Condition (A), by Lemma 2.2, there is some e be defined by (1.12) and (2.12). Then in v ∈ W 1,2 (D) ∩ C(D) so that (2.10) holds. Let Z and Z −v(x) view of (2.11), u = e u0 (x) with [∫ ∞ ] v(X ) s u0 (x) := Ex Zes e ϕ(Xs )dLs . 0
Let et g(Xt )] Tet g(x) := Ex [Z [∫ ∫ ∞ −βt eβ g(x) := R e Tet g(x)dt = Ex 0
∞
−βt
e
]
et g(Xt )dt . Z
(3.2)
0
e Dom(Q)) e associated with From the expression (2.12), it is easy to see that the quadratic form (Q, the semigroup {Tet , t ≥ 0} is given by e g) Q(f, ∫ ∫ 1 b − A∇v, ∇f ⟩ g dx = ⟨A∇f, ∇g⟩dx − ⟨b − b 2 D D ) ∫ ( 1 b ⟨A∇, ∇v⟩ − ⟨b − b, ∇v⟩ + q f g dx, − D 2
(3.3)
e = W 1,2 (D). To prove that u ∈ W 1,2 (D), it is sufficient to show u0 ∈ W 1,2 (D). For and Dom(Q) notational simplicity, let f (x) = ev(x) ϕ(x). By the Markov property of X and Fubini’s theorem, ] ] [∫ ∞ [∫ ∞ −βt e e e Rβ u0 (x) = Ex e Zt Ex Zs f (Xs )dLs ◦ θt Ft dt ] ] [ ∫0 ∞ [ ∫ ∞0 −βt = Ex e Ex Zes f (Xs )dLs Ft dt t [ ∫0 ∞ ] ∫ s −βt e = Ex Zs f (Xs )dLs e dt 0 0 [∫ ∞ ] 1 1 es f (Xs )dLs . u0 (x) − Ex e−βs Z (3.4) = β β 0 By Lemma 4.1 in [6], we know that [∫ ∞ ] −βs e eβ (f σ)(x), Ex e Zs f (Xs )dLs = U 0
eβ (f σ) is the β-potential of the smooth measure f σ associated with the quadratic form Q, e where U namely, it satisfies that for h ∈ W 1,2 (D), ∫ e e Qβ (Uβ (f σ), h) = h(x)f (x)σ(dx), (3.5) ∂D
9
e β (·, ·) = Q(·, e ·) + β(·, ·). Rearrange (3.4) to obtain where Q eβ u0 (x) = U eβ (f σ)(x). u0 (x) − β R b(1) the adjoint resolvent operator of R eβ on L2 (D, dx). Then, For β > 0, denote by R β eβ u0 , u0 ) = β(U (1) (f σ), u0 ) = β Q e β (U eβ (f σ), R b β u0 ) β(u0 − β R β ∫ bβ u0 )(x)f (x)σ(dx). (R = β ∂D
eβ is generated by a semigroup that has the same form as the The adjoint resolvent operator of R semigroup in (1.12). Hence we have bβ u0 (x)| ≤ C∥u0 ∥∞ sup β|R β
This yields that
eβ u0 , u0 )} < ∞. sup{β(u0 − β R β
e = W 1,2 (D) (see [18]). Moreover, for h ∈ W 1,2 (D) ∩ C(D), Therefore, we conclude that u0 ∈ D(Q) eβ h(x) = h(x) and supβ β|R eβ h(x)| ≤ C∥h∥∞ , we have as limβ→∞ β R e 0 , h) = Q(u
eβ u0 , h) lim β(u0 − β R ∫ b(1) h)(x)f (x)σ(dx) = lim β (R β β→∞ ∂D ∫ = h(x)f (x)σ(dx). β→∞
(3.6)
∂D
Next we show that u satisfies (1.9). For g ∈ C 2 (D), we have Q(u, g) d ∫ d ∫ ∑ 1 ∑ ∂(u0 e−v ) ∂g ∂(u0 e−v ) = aij (x) dx − bi (x) gdx 2 ∂xi ∂xj ∂xi D D i,j=1
∫
bbi (x) ∂g u0 e−v dx − q(x)u0 (x)e−v gdx ∂x i D i=1 D ∫ d d ∫ 1 ∑ ∂u0 ∂(e−v g) 1 ∑ ∂u0 ∂e−v aij (x) dx − aij (x) gdx 2 ∂xi ∂xj 2 ∂xi ∂xj D D
− =
i=1
d ∫ ∑
i,j=1
1 + 2 −
i,j=1
d ∑ i,j=1
∫
∑ ∂e−v ∂g u0 dx − aij (x) ∂xi ∂xj D
d ∫ ∑
∫i=1 −
d
i=1
∑ ∂e−v u0 gdx − bi (x) ∂xi D d
i=1
q(x)u0 (x)e−v gdx
∫
∫ bi (x) D
∂u0 −v e gdx ∂xi
bbi (x) ∂g u0 e−v dx ∂xi D (3.7)
D
10
Applying (3.6) to h = ge−v we obtain that d ∫ ∂u0 ∂(ge−v ) 1 ∑ aij (x) dx 2 ∂xi ∂xj D i,j=1 ∫ b = ⟨b(x) − b(x) − (A∇v)(x), ∇u0 (x)⟩g(x)e−v dx D∫ ∫ b ∇v⟩(x)u0 (x)ge−v dx + q(x)u0 (x)ge−v dx − ⟨b − b, D D ∫ ∫ 1 ∗ + (∇v)A(∇v) (x)u0 ge−v dx + ge−v ev ϕ(x)σ(dx). 2 D ∂D
Substituting this expression in (3.7) it follows after cancelations that ∫ ∫ b u0 g⟩e−v dx − ⟨A∇v), ∇u0 ⟩ge−v dx Q(u, g) = − ⟨b, D ∫D ∫ −v b ∇v⟩gu0 e dx + 1 + ⟨b, ⟨∇v, A∇v⟩ u0 ge−v dx 2 D D∫ ∫ 1 1 ⟨A∇u, ∇e−v ⟩gdx + ⟨A∇g, ∇e−v ⟩u0 dx − 2 D 2 D ∫ + g(x)ϕ(x)σ(dx)
(3.8)
(3.9)
∂D
In the sequel, we write div(·) for the divergence in the sense of distributions. Now, ∫ b ∇(u0 g)⟩e−v dx − ⟨b, D ∫ ∫ −v b b n⟩u0 gσ(dx) = div(e b)u0 gdx − e−v ⟨b, D ∂D ∫ ∫ ∫ −v b −v b b n⟩u0 gσ(dx). = div(e b)u0 gdx − ⟨b, ∇v⟩gu0 e dx − e−v ⟨b, D
D
(3.10) (3.11)
∂D
b in place of F , we have Furthermore, since (1.13) holds for v with b ∫ ∫ 1 −v b div(b)e u0 gdx = − div(A∇v)e−v u0 gdx 2 D∫ D ∫ 1 1 = ⟨A∇v, ∇(e−v u0 g)⟩dx − ⟨A∇v, n⟩e−v u0 gdx 2 D 2 ∂D ∫ ∫ 1 1 −v ⟨A∇v, ∇v⟩e u0 gdx + ⟨A∇v, ∇u0 ⟩e−v g)dx = − 2 D 2 D ∫ ∫ 1 −v b n⟩u0 gσ(dx). + ⟨A∇v, ∇g⟩e u0 gdx + e−v ⟨b, 2 D ∂D Combining (3.7)-(3.12) together we arrive at ∫ ∫ 1 −v Q(u, g) = − ⟨A∇v, ∇u0 ⟩ge dx − ⟨A∇u0 , ∇e−v ⟩ g dx 2 D ∫ ∫D 1 −v ⟨A∇v, ∇u0 ⟩e gdx + gϕ(x)σ(dx) + 2 D ∂D 11
(3.12)
∫ gϕ(x)σ(dx),
=
(3.13)
∂D
which completes the proof of the existence. Now we show the uniqueness. Let u be any bounded, continuous weak solution to the boundary value problem (1.3). Define u0 (x) = ev(x) u(x). Then it is easy to verify that u0 satisfies (3.6). Under the assumption (3.1), using a similar proof to that of [16, Theorem 3.3] we can show that u0 is uniquely determined, so is u. This completes the proof of uniqueness. We finish the paper with a probabilistic characterization of weak solutions of mixed boundary value problem. Theorem 3.2 Assume that Condition (A) holds. Suppose u ∈ W 1,2 (D) is bounded and that [∫ ∞ ] Ex0 0 Zs dLs < ∞ for some x0 ∈ D. Then the following statements are equivalent. (i) u is a weak solution of the boundary value problem (1.3); ∫t (ii) u(Xt )Zt − u(X0 ) + 0 Zs ϕ(Xs )dLs is a Px -martingale for a.e. x ∈ D. Proof. We first prove (i)⇒(ii). By Theorem 1.2, the continuous version of the unique bounded [∫ ∞ ] weak solution of (1.3) is given by u(x) = Ex 0 Zs ϕ(Xs )dLs . Using this version of u, we get a martingale [∫ ∞ ] ∫ t Zs ϕ(Xs )dLs + Zt u(Xt ) Zs ϕ(Xs )dLs Ft = Kt := Ex 0
0
by the Markov property of X and that Z is a multiplicative functional of X. It follows that ∫t u(Xt )Zt − u(X0 ) + 0 Zs ϕ(Xs )dLs = Kt − u(X0 ) is a Px -martingale for every x ∈ D, so the asserted implication (i)⇒(ii) follows. Next we show that (ii)⇒(i). Suppose that u is a bounded function in W 1,2 (D) so that u(Xt )Zt − [∫ ∞ ] ∫t u(X0 ) + 0 Zs ϕ(Xs )dLs is a Px -martingale for a.e. x ∈ D. Let v(x) := Ex 0 Zs ϕ(Xs )dLs , which, by Theorem 1.2 is a continuous bounded weak solution for (1.3). From above, we know that ∫t 0 Zs ϕ(Xs )dLs + Zt v(Xt ) is a Px -martingale for every x ∈ D. It follows then (u(Xt ) − v(Xt ))Zt is a Px -martingale for a.e. x ∈ D. Consequently, u(x) − v(x) = Ex [(u(Xt ) − v(Xt ))Zt ] = Tt (u − v)(x). Since u − v ∈ W 1,2 (D), this implies that Q(u − v, f ) = 0 for every f ∈ C 2 (D). In other words, ∫ Q(u, f ) = Q(v, f ) = f (x)ϕ(x)σ(dx) for every f ∈ C 2 (D) ∂D
2
and so u is a weak solution of (1.3).
Acknowledgement. We are grateful to the referee’s helpful comments that improves the exposition of this paper, especially the comments that leads to the current proof of Theorem 3.2.
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Zhen-Qing Chen Department of Mathematics, University of Washington, Seattle, WA 98195, USA E-mail:
[email protected] Tusheng Zhang Department of Mathematics, University of Manchester, Oxford Road, Manchester M13 9PL, England, U.K. Email:
[email protected]
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