Some Results on Stochastic Differential Equations with Reflecting Boundary Conditions

Some results on stochastic differential equations with reflecting boundary conditions Pedro Mar´ın-Rubio1,2and Jos´e Real3 Abstract Some results related to stochastic differential equations with reflecting boundary conditions are obtained. Existence and uniqueness of strong solution is ensured under the relaxation on the drift coefficient (instead of the Lipschitz character, a monotonicity condition is supposed). KEY WORDS: Skorokhod problem; Reflected Stochastic Differential Equations; Monotonicity condition; Strong solution 1 Introduction In this paper we extend some results of Tanaka(6) and Lions and Sznitman(4) on existence and uniqueness of strong solutions for stochastic differential equations with reflecting boundary conditions (SDER) to the case in which the drift coefficient bsatisfies the monotonicity condition (x−x0, b(t, x)−b(t, x0)) ≤Lbx|x−x0|2, instead of the classical Lipschitz condition. As far as we know, for this type of drift coefficient there is not in the literature a general result of existence of strong solutions for SDER (an exception is the 1-dimensional case, cf. Zhang(7) and Matoussi(5)). In the case of stochastic differential equations without reflection the same kind of problem has been previously solved, for instance, in Jacod(3) and Gy¨ongy and Krylov(2). 1To whom correspondence should be addressed. 2Dpto. de Matem´aticas, Universidad de Huelva, Avda. Fuerzas Armadas s/n, Apdo. de Correos 21071, Huelva (SPAIN). E-mail: p[email protected]u.es 3Dpto. de Ecuaciones Diferenciales y An´alisis Num´erico, Universidad de Sevilla, Apdo. de Correos 1160, 41080-Sevilla (SPAIN). E-mail: [email protected] 1 In section 2 we give the framework, definitions and claim the main result. Section 3 is devoted to prove a previous and similar result on the deterministic Skorokhod problem. Finally, the stochastic version is treated in section 4. 2 Statement of the problem and main result Let (Ω,F, P) be a complete probability space, {Ft}t≥0an increasing and right continuous family of sub-σ-algebras of Fsuch that F0contains all the P-null sets of F, and {Wt;t≥0}an m-dimensional standard {Ft}-Wiener process. Let Obe an open connected bounded subset of Rdgiven by O={φ > 0}, with φ∈C2(Rd), and such that ∂O={φ= 0},with |∇φ(x)|= 1 for all x∈∂O. Observe that in particular φ,∇φand D2φare bounded in ¯ O. Also, observe that n(x), the unit outward normal to ∂Oat x, coincides with −∇φ(x), and that we can assert that there exists a constant C0>0 such that 2(x0−x, ∇φ(x)) + C0|x0−x|2≥0,∀x∈∂O,∀x0∈¯ O.(2.1) We are also given a final time T > 0, and two random functions: b: Ω ×[0, T]ׯ O → Rd, σ: Ω ×[0, T]ׯ O → Rd×m, such that (i) band σare uniformly bounded; (ii) for all x∈¯ Othe processes b(·,·, x) and σ(·,·, x) are {Ft}-progressively measurable; (iii) for all t∈[0, T] and a.s. ω, the function b(ω, t, ·) is continuous on ¯ O; (iv) there exist two constants Lbx∈Rand Lσx≥0 such that for all t∈[0, T] and all x, x0∈¯ O, (x−x0, b(ω, t, x)−b(ω, t, x0)) ≤Lbx|x−x0|2, a.s., kσ(ω, t, x)−σ(ω, t, x0)k ≤ Lσx|x−x0|, a.s., where | · | and k · k denote the usual Euclidean and trace norm for vectors and matrices respectively. From now on, in general we will omit the explicit dependence of the processes on ω. 2 Remark 1. Observe that if bsatisfies the conditions (i)-(iii) above, then, reasoning as in the proof of Tietze’s Extension Theorem, and using the theorems 8.1.4 and 8.2.9 in Aubin and Frankowska(1), one can see that there exists an extension of b, ˜ b: Ω ×[0, T]×Rd→Rd, such that ˜ bis also uniformly bounded and satisfies (ii) and (iii) in Rdinstead of ¯ O. We seek strong solutions for the problem: Xt=x0+Zt 0 b(s, Xs)ds +Zt 0 σ(s, Xs)dWs−kt,(2.2) kt=−Zt 0 ∇φ(Xs)d|k|s,|k|t=Zt 0 1{Xs∈∂O} d|k|s, t ∈[0, T],(2.3) where x0∈¯ Ois given, and |k|tstands for the total variation of kon [0, t]. Definition 1. A strong solution to the above problem is a pair of {Ft}- adapted and continuous processes (X, k)defined on Ω×[0, T], the first one with values in ¯ O, the second one with values in Rdand paths of bounded variation in [0, T], satisfying the equations (2.2)-(2.3) a.s. for all t∈[0, T]. We now state our main result, which generalizes that given in Lions and Sznitman(4) when bis Lipschitz. Theorem 1. Under the assumptions (i)-(iv), for each x0∈¯ Ogiven there exists a unique pair (X, k), strong solution of (2.2)-(2.3). To prove this theorem, we will analyze a deterministic problem which generalizes the Skorokhod problem studied in Lions and Sznitman(4). 3 A generalization of the Skorokhod problem In this section, we consider the open set Ogiven in section 2 but we assume that the coefficient bis independent of ω. We suppose given x0∈¯ Oand a function g∈C([0, T]; Rd) such that g0= 0.We want to solve the deterministic problem xt=x0+Zt 0 b(s, xs)ds +gt−kt,(3.1) kt=−Zt 0 ∇φ(xs)d|k|s,|k|t=Zt 0 1{xs∈∂O} d|k|s, t ∈[0, T].(3.2) 3 Definition 2. A solution of the problem (3.1)-(3.2) is a pair (x, k)of continuous functions defined on [0, T ]with values in Rd, such that xt∈¯ Ofor all t∈[0, T],kis of bounded variation on [0, T], and the equations (3.1)-(3.2) are satisfied for all t∈[0, T ]. From Theorem 2.1 and Remark 2.1 in Lions and Sznitman(4) we can assert the following result: Theorem 2. Let suppose b≡0and x0∈¯ Ogiven. Then, for any function g∈C([0, T]; Rd)such that g0= 0 there exists a unique pair (x, k)solution of the problem (3.1)-(3.2). Moreover, the mapping g7→ xis H¨older continuous of order 1/2on compact sets of C([0, T]; Rd). Remark 2. Observe that, as a direct consequence of the H¨older continuity of order 1/2 on compact sets of C([0, T]; Rd) of the the mapping g7→ x, we can assert that under the conditions of Theorem 2, if {gn} ⊂ C([0, T]; Rd) is a sequence of functions such that gn 0= 0 and gn→gin C([0, T]; Rd), then, if we denote by (xn, kn) (resp. (x, k)) the pair solution of (3.1)-(3.2) corresponding to b≡0 and gn(resp. g), we have that xn→xin C([0, T ]; ¯ O). We will see now that we can extend Theorem 2 to the case in which b6≡ 0. First at all, we have the following result: Theorem 3. Let be Osatisfying the conditions in section 2. Consider given a measurable and bounded function b: [0, T]ׯ O → Rd,such that for all t∈[0, T]the function b(t, ·)is continuous on ¯ O. Then, for each x0∈¯ Oand g∈C([0, T]; Rd)such that g0= 0 given, there exists at least one solution (x, k)of the problem (3.1)-(3.2). Proof. We will proceed in two steps. Step 1 Let also suppose that g∈C1([0, T]; Rd) and bis Lipschitz in xon ¯ O, i.e. there exists L > 0 such that |b(t, x)−b(t, x0)| ≤ L|x−x0| for all t∈[0, T] and all x, x0∈¯ O. In this case, the existence and uniqueness of solution to (3.1)-(3.2) can be deduced from the stochastic results in Lions and Sznitman(4). However, for more clarity, we give a completely deterministic proof. Denote by fthe derivative of g. For each y∈C([0, T ]; Rd) given, consider the problem xt=x0+Zt 0 (b(s, ys) + fs)ds −kt,(3.3) 4 kt=−Zt 0 ∇φ(xs)d|k|s,|k|t=Zt 0 1{xs∈∂O} d|k|s, t ∈[0, T].(3.4) Obviously, the function ˜gt=Zt 0 (b(s, ys) + fs)ds is continuous on [0, T], with ˜g0= 0, thus by Theorem 2, there exists a unique solution (x, k) of (3.3)-(3.4). It is enough to prove that there exists a unique fixed point for the mapping F:y∈C([0, T]; ¯ O)7→ x∈C([0, T]; ¯ O) defined by (3.3)-(3.4). Let x=Fy and x0=Fy0. Using (2.1), it is easy to see that exp {−C0(φ(xt) + φ(x0 t))}|xt−x0 t|2 ≤ −C0Zt 0 exp {−C0(φ(xs) + φ(x0 s))}(∇φ(xs), b(s, ys) + fs)|xs−x0 s|2ds −C0Zt 0 exp {−C0(φ(xs) + φ(x0 s))}(∇φ(x0 s), b(s, y0 s) + fs)|xs−x0 s|2ds +2 Zt 0 exp {−C0(φ(xs) + φ(x0 s))}(xs−x0 s, b(s, ys)−b(s, y0 s)) ds. (3.5) As xand x0take values in ¯ O, exp{−2C0max ¯ Oφ} ≤ exp{−C0(φ(xt) + φ(x0 t))} ≤ 1 for all t∈[0, T]. Moreover, ∇φ,band fare uniformly bounded, and so, using that bis Lipschitz, it is easy to obtain from (3.5) the existence of a constant C > 0, independent of y,y0and t, such that |xt−x0 t|2≤CZt 0|xs−x0 s|2+|ys−y0 s|2ds, for all t∈[0, T]. Therefore, sup r∈[0,t] |xr−x0 r|2≤CZt 0 sup r∈[0,s] |xr−x0 r|2+ sup r∈[0,s] |yr−y0 r|2!ds, 5 for all t∈[0, T],and from Gronwall’s lemma we have sup r∈[0,t] |xr−x0 r|2≤Cexp(CT)Zt 0 sup r∈[0,s] |yr−y0 r|2ds, (3.6) for all t∈[0, T]. It is known that (3.6) implies that a power of Fis a contraction in C([0, T]; Rd), and so there exists a unique fixed point for F. Step 2 Suppose now that we are in the conditions of the theorem. In this case, we can approach gby a sequence of functions gn∈C1([0, T]; Rd) such that gn 0= 0, converging to gin C([0, T ]; Rd). Furthermore, if we fix a regularizing sequence {ρn} ⊂ D(Rd), i.e. ρn:Rd→R, ρn≥0,supp(ρn)⊂BRd(0,1/n),ZRd ρndx = 1 and define bn(t, x) = ZRd ρn(y)˜ b(t, x −y)dy, (t, x)∈[0, T]×Rd,(3.7) with ˜ ba measurable and uniformly bounded extension of bto [0, T]×Rd, such that ˜ b(t, ·) is continuous in Rd, we obtain a sequence of measurable functions bn: [0, T]×Rd→Rdsuch that, in particular, for all t∈[0, T] the function bn(t, ·) is continuous on Rd, |bn(t, x)| ≤ sup y∈Rd |˜ b(t, y)|,∀x∈Rd,(3.8) |bn(t, x)−bn(t, x0)| ≤ Ln|x−x0|,∀x, x0∈Rd,(3.9) bn(t, ·)→b(t, ·) uniformly in ¯ O.(3.10) By Step 1, for each nwe have a unique solution (xn, kn) of the problem: xn t=x0+Zt 0 bn(s, xn s)ds +gn t−kn t, kn t=−Zt 0 ∇φ(xn s)d|kn|s,|kn|t=Zt 0 1{xn s∈∂O} d|kn|s, t ∈[0, T]. It is obvious that {bn(·, xn ·)}is bounded in L2(0, T;Rd), and thus the sequence {R· 0bn(s, xn s)ds}is bounded in C([0, T]; Rd) and equicontinuous. Therefore, it is easy to see that there exist a subsequence {xµ}⊂{xn}and an element B ∈ L2(0, T;Rd), such that bµ(·, xµ ·)*Bin L2(0, T;Rd) and 6 Z· 0 bµ(s, xµ s)ds →Z· 0 Bsds in C([0, T]; Rd). Then, according to Theorem 2, xµ→xin C([0, T ]; Rd), with (x, k) the solution of xt=x0+Zt 0 Bsds +gt−kt, kt=−Zt 0 ∇φ(xs)d|k|s,|k|t=Zt 0 1{xs∈∂O} d|k|s, t ∈[0, T]. But, as xµ→xin C([0, T]; Rd), it is easy to obtain from (3.8), (3.10), and the continuity of b(t, ·),that bµ(·, xµ ·)→b(·, x·) in L2(0, T;Rd). Thus, B=b(·, x·), and (x, k) is a solution of (3.1)-(3.2). In the proof of Theorem 3 we have seen that, under the conditions of the theorem, if bis also Lipschitz in x, then the solution of (3.1)-(3.2) is unique. In fact, we have the following result Theorem 4. Under the conditions of Theorem 3, suppose that there exists Lbx∈Rsuch that for all t∈[0, T]and all x, x0∈¯ O, (x−x0, b(t, x)−b(t, x0)) ≤Lbx|x−x0|2. Then, for each x0∈¯ Oand g∈C([0, T]; Rd)given such that g0= 0, there exists a unique solution (x, k)of the problem (3.1)-(3.2). Proof. Because of Theorem 3, we only have to check uniqueness. Let (x, k) and (x0, k0) two solutions of (3.1)-(3.2) corresponding to the same x0and g. Then, for all t∈[0, T] xt−x0 t=Zt 0 (b(s, xs)−b(s, x0 s)) ds −kt+k0 t, and consequently, exp {−C0(|k|t+|k0|t)}|xt−x0 t|2 =−C0Zt 0 exp {−C0(|k|s+|k0|s)}|xs−x0 s|2(d|k|s+d|k0|s) +2 Zt 0 exp {−C0(|k|s+|k0|s)}(xs−x0 s, b(s, xs)−b(s, x0 s)) ds +2 Zt 0 exp {−C0(|k|s+|k0|s)}(xs−x0 s,∇φ(xs)) d|k|s −2Zt 0 exp {−C0(|k|s+|k0|s)}(xs−x0 s,∇φ(x0 s)) d|k0|s.(3.11) 7 It is easy to see that, by (2.1), (3.2), and the hypotheses on b, we obtain from (3.11) exp {−C0(|k|t+|k0|t)}|xt−x0 t|2≤2|Lbx|Zt 0 exp {−C0(|k|s+|k0|s)}|xs−x0 s|2ds for all t∈[0, T], and thus, from Gronwall’s lemma, we obtain the claimed result. Remark 3. Consider the hypotheses of Theorem 4, and the sequence bn given by (3.7). Denote by (xn, kn) the unique solution of the problem xn t=x0+Zt 0 bn(s, xn s)ds +gt−kn t, kn t=−Zt 0 ∇φ(xn s)d|kn|s,|kn|t=Zt 0 1{xn s∈∂O} d|kn|s, t ∈[0, T]. Then, reasoning as in Step 2 of the proof of Theorem 3, and by the uniqueness of the solution (x, k) of (3.1)-(3.2), we can assert that all the sequence xnconverges to xin C([0, T]; Rd). 4 Proof of Theorem 1 For the proof, we will proceed in two steps. Step 1 Let σbe independent of x, i.e. σ(ω, t, x) = σ(ω, t), a.s. for all (t, x)∈[0, T]ׯ O. In this case, denote Mt=Zt 0 σ(s)dWs, and observe that a pair (X, k) of {Ft}-progressively measurable processes with values in Rdis a solution of (2.2)-(2.3) if and only if, a.s. ω∈Ω, (X(ω), k(ω)) is a solution of the problem Xt(ω) = x0+Zt 0 b(ω, s, Xs(ω)) ds +Mt(ω)−kt(ω),(4.1) kt(ω)=−Zt 0 ∇φ(Xs(ω))d|k(ω)|s,|k(ω)|t=Zt 0 1{Xs(ω)∈∂O}d|k(ω)|s,(4.2) for all t∈[0, T]. But, according to Theorem 4, for each ω∈Ω there exists a unique solution (X(ω), k(ω)) of (4.1)-(4.2). Thus, in order to prove that the random 8 pair (X, k) defined by (4.1)-(4.2) is the unique strong solution of (2.2)-(2.3), we must only see that X(and so k) is {Ft}-progressively measurable. To this end, observe that, by Theorem 3.1 and Remark 3.3 in Lions and Sznitman(4), the existence of strong solution to (2.2)-(2.3) is guaranteed if bis also Lipschitz in x. Consequently, if we fix a regularizing sequence {ρn} ⊂ D(Rd) and define for (t, x)∈[0, T]×Rd, bn(ω, t, x) = ZRd ρn(y)˜ b(ω, t, x −y)dy, a.s., with ˜ bthe extension of bwhose existence is observed in Remark 1, we obtain a sequence (Xn, kn) of {Ft}-progressively measurable processes such that a.s. they are solutions of Xn t(ω) = x0+Zt 0 bn(ω, s, Xn s(ω)) ds +Mt(ω)−kn t(ω), kn t(ω) = −Zt 0 ∇φ(Xn s(ω)) d|kn(ω)|s,|kn(ω)|t=Zt 0 1{Xn s(ω)∈∂O} d|kn(ω)|s, for all t∈[0, T]. Moreover, by Remark 3, a.s. Xn(ω) converges to X(ω) in C([0, T ]; Rd). Thus, in particular, X(and therefore, k) is {Ft}-progressively measurable. Step 2 In the conditions of Theorem 1. We proceed in a similar way to the proof of Lemma 3.1 in Lions and Sznitman(4). Denote by L4 Ft(Ω; C([0, T]; Rd)) the space of the elements of L4(Ω; C([0, T]; Rd)) that are {Ft}-progressively measurable. Then, the space L4 Ft(Ω; C([0, T]; Rd)) is a Banach subspace of L4(Ω; C([0, T]; Rd)). Consider the mapping ˆ F:L4 Ft(Ω; C([0, T]; Rd)) →L4 Ft(Ω; C([0, T]; Rd)) that to each Y∈L4 Ft(Ω; C([0, T]; Rd)) associates ˆ F(Y) = X, with (X, k) the strong solution of Xt=x0+Zt 0 b(s, Xs)ds +Zt 0 σ(s, Ys)dWs−kt,(4.3) kt=−Zt 0 ∇φ(Xs)d|k|s,|k|t=Zt 0 1{Xs∈∂O} d|k|s, t ∈[0, T],(4.4) whose existence and uniqueness is guaranteed by Step 1. Observe that, as Xt∈¯ O, and Ois bounded, we have that, of course, X∈L4 Ft(Ω; C([0, T]; Rd)). 9