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Potential Analysis (2005) 22: 375–393

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Numerical Approximation for a White Noise Driven SPDE with Locally Bounded Drift ROGER PETTERSSON1, and MIKAEL SIGNAHL2,

1 Matematiska och Systemtekniska Institutionen, Vejdes Plats 7, Växjö Universitet, 351 95 Växjö,

Sweden 2 Matematisk Statistik, Matematikcentrum, Box 118, Lunds Universitet, 221 00 Lund, Sweden (e-mail: [email protected]) (Received: 29 July 2003; accepted: 18 May 2004) Abstract. A numerical scheme for a stochastic partial differential equation of heat equation type is considered where the drift is locally bounded and the dispersion may be state dependent. Uniform convergence in probability is obtained. Mathematics Subject Classifications (2000): Key words: stochastic partial differential equations, Malliavin calculus.

1. Introduction We consider the initial-boundary-value problem for a nonhomogeneous, nonlinear heat equation driven by a Brownian sheet:  ∂ 2u ∂ 2W ∂u   = 2 + f (t, x, u(t, x)) + σ (t, x, u(t, x)) ,  ∂t ∂x ∂t∂x (1.1) u0 ∈ C[0, 1], u0 (0) = u0 (1) = 0, u(0, x) = u0 (x),    u(t, 1) = u(t, 0) = 0, 0 t T, where W is a Brownian sheet on [0, T ]×[0, 1]. A definition of a solution of (1.1) is given in Section 2. Gyöngy [8] considered a natural numerical scheme to (1.1), see (3.1) below. Uniform convergence in probability was obtained in the case where the drift f may not be Lipschitz continuous but the dispersion σ is constant (if both the drift and the dispersion are Lipschitz continuous, much stronger convergence, and a convergence rate, was obtained). In this note we weaken the assumption on the dispersion: the dispersion may depend on the state. Previous work also include [5] showing that Gyöngy’s convergence rate is the best possible. In [1] the finite element and difference methods are compared for the case σ ≡ 1 and a special form for f . Computational experiments indicate similar accuracy but the finite

Partially supported by the EU grant ref. ERBF MRX CT96 0057A. Corresponding author.

376

ROGER PETTERSSON AND MIKAEL SIGNAHL

element method is computationally more efficient. Shardlow [19] investigates the case σ ≡ 1 and f dependent on u and proves strong convergence rates for a finite difference scheme. We use ideas from [10] where they considered a time discrete numerical scheme for (1.1). The numerical scheme here is discretized in both time and space. The paper is organized as follows. In Section 2 we state the problem and conditions under which it is considered. We also recall some basic facts concerning the Malliavin calculus. Section 3 describes the numerical scheme and the proof that the approximate solutions converge uniformly in probability. Some proofs are left to an appendix. 2. Definitions and Assumptions Let W = {W (t, x): (t, x) ∈ [0, T ] × [0, 1]} be a Brownian sheet defined on a complete probability space (, F , P ). That is, W is a zero-mean Gaussian random field with covariance E(W (s, x), W (t, y)) = (s ∧ t)(x ∧ y),

s, t ∈ [0, T ], x, y ∈ [0, 1].

(2.1)

For each t ∈ [0, T ], Ft is the σ -field generated by the family of random variables {W (s, x): s ∈ [0, t], x ∈ [0, 1]} and the P -null sets. We assume F = FT and say that a random field u = {u(t, x): (t, x) ∈ [0, T ] × [0, 1]} is adapted if u(t, x) is Ft -measurable for each (t, x). The properties of W are described in [21]. We say that an Ft -adapted continuous random field u is a solution of (1.1) if for any test function ψ ∈ C 1,2 ([0, T ] × [0, 1]) such that ψ(t, 0) = ψ(t, 1) = 0 we have for all t ∈ [0, T ], 1 1 u(t, x)ψ(t, x) dx = u0 (x)ψ(0, x) dx 0 0 t 1 ∂ ∂ ψ(s, x) dx ds u(s, x) + + ∂x 2 ∂s 0 0 t 1 f (s, x, u(t, x))ψ(s, x) dx ds + 0 0 t 1 σ (s, x, u(s, x))ψ(s, x)W (dx, ds). (2.2) + 0

0

We consider (1.1) under the following assumptions. (E) There is a solution of (1.1); (PU) The pathwise uniqueness holds for (1.1); whenever u and v are random ) equipped , F, Ft , P fields carried by some filtered probability space ( such that u and v are solutions, with W in place with a Brownian sheet W of W on a stochastic interval [0, τ ), then u(t, ·) = v(t, ·) a.s. for all t ∈ [0, τ (ω)).

NUMERICAL APPROXIMATION FOR A WHITE NOISE DRIVEN SPDE

377

(LB) The functions f and σ are measurable and locally bounded functions, i.e., bounded on every compact set; (ND) The equation (1.1) is nondegenerate; for every R > 0 there is a λR > 0 such that σ 2 (t, x, r) λR , for all t, x and |r| R. (SL) ∂σ/∂r is a bounded function and Lipschitz continuous in r ∈ R. REMARK 2.1. Conditions on f and σ that guarantees existence of solutions have been considered in [6, 3, 7, 11, 12]. Articles devoted to showing pathwise uniqueness under different conditions on f and σ can be found in [7, 11]. The point of the non-degeneracy condition is to ensure the approximate solutions have densities which are uniformly integrable (in a sense made precise in Lemma 3.6) and is needed in our proof. Let us briefly recall some facts concerning Malliavin calculus for the Brownian sheet W . Detailed accounts can be found in [17, 14, 13, 20, 2, 15]. For h ∈ H ≡ L2 ([0, T ] × [0, 1]) we can write T 1 W (h) = h(t, x)W (dt, dx). 0

0

This is well-defined because of (2.1) noting that W (t, x) = W (1[0,T ]×[0,x] ) and W (h)L2 () = hH . For f ∈ Cb∞ (Rn ), the space of all infinitely differentiable functions with bounded derivatives, we can define Malliavin differentiation at (t, x) of the random variable F = f (W (h1 ), . . . , W (hn ))

(2.3)

as n ∂f (W (h1 ), . . . , W (hn ))hi (t, x). D(t,x) F = ∂x i i=1

In particular, we have D(t,x) W (s, y) = 1[0,s]×[0,y] (t, x). This differentiation procedure can be iterated obtaining a derivative operator D (k) of any order k = 1, 2, 3, . . . . Let S denote the set of random variables F for which there is a natural number n and functions h1 , . . . , hn ∈ H such that the representation (2.3) is valid. Since D (k) is a closable (unbounded) operator D (k) : Lp () → Lp (; H ⊗k ) it can be extended to a closed operator by closing S with respect to the norm 1/p

k p p (j ) F k,p = E|F | + ED F H ⊗j . j =1

The resulting domain is denoted by Dk,p .

378


3. Numerical Solution We consider a numerical solution of (1.1). Let {ti = iT /m}m i=0 be a partition of [0, T ] and {xj = j/n}nj=0 a partition of [0, 1]. We consider the iterative and implicit projection scheme un,m = {un,m (ti , xj )} which works as follows. We initialize by un,m (0, xj ) = u0 (xj ) for j = 0, . . . , n. Then, once un,m (ti , xj ) is known for j = 0, . . . , n, un,m (ti+1 , xj ), for j = 0, . . . , n, is the unique solution of un,m (ti+1 , xj ) = un,m (ti , xj ) T + n un,m (ti+1 , xj ) m T + f (ti , xj , un,m (ti , xj )) m T + σ (ti , xj , un,m (ti , xj ))n,m W (ti , xj ), m n,m n,m u (ti+1 , 0) = u (ti+1 , 1) = 0, for i = 0, . . . , m − 1,

(3.1)

where, for j = 1, . . . , n − 1, n un,m (·, xj ) is the usual approximation of the second derivative in space un,m (·, xj +1 ) − 2un,m (·, xj ) + un,m (·, xj −1 ) (1/n)2 and n,m W (ti , xj ) is similarly the approximate time-space derivative W (ti+1 , xj +1 ) − W (ti+1 , xj ) − W (ti , xj +1 ) + W (ti , xi ) . 1/n · T /m Note that the operator I − (T /m) n is invertible (see [8]) for every m 1, n 1 so there exists a random field un,m defined on the space time grid that solves (3.1). For (t, x) ∈ (ti , ti+1 ) × (xi , xi+1 ) we interpolate un,m linearly to get a continuous random field:

m un,m (t, xj ) = un,m (ti , xj ) + (t − ti ) un,m (ti+1 , xj ) − un,m (ti , xj ) , T (3.2)

un,m (t, x) = un,m (t, xj ) + n(x − xj ) un,m (t, xj +1 ) − un,m (t, xj ) . For a lattice point ti and any x ∈ [0, 1] we can write (see [8]) 1 n,m Gn,m u (ti , x) = ti (x, y)u0 (y) dy 0 ti 1

n,m Gn,m (s, y) dy ds + ti+1 −s (x, y)f s, y, u 0 0 ti 1

n,m + Gn,m (s, y) W (dy ds), ti+1 −s (x, y)σ s, y, u 0

0

(3.3)

379


where

m T t= t , T m

[nx] T 1 x= , t¯ = t + , x¯ = x + , n m n −[(m/T )t] n−1 T ϕjn (x)ϕj (y) Gn,m 1 − λnj t (x, y) = m j =1 jπ 2 n = −j 2 π 2 cjn λnj = −4 sin2 2n sin2 (j π/2n) 4 n c = 1 j 2 π2 √ (j π/2n) ϕj (x) = 2 sin j π x ϕjn (x) = ϕj (xk ) + n(x − xk )(ϕj (xk+1 ) − ϕj (xk )),

(3.4)

x ∈ [xk , xk+1 ],

for 1 j n − 1. The ϕjn ’s satisfy 1 ϕjn (y)ϕ n (y) dy = δj , j, = 1, . . . , n − 1,

(3.5)

0

for every integer n 1. We need the following assumption on n and m: (D) n2 T /m q, for some constant q < 1/2. The main result in this note is the following. THEOREM 3.1. Assume (E), (PU), (LB), (ND), (SL) and (D). Then n,m lim P sup sup |u (t, x) − u(t, x)| = 0, n→∞ m→∞

t∈[0,T ] x∈[0,1]

for every > 0. The proof consists of a series of lemmas. Recall that a function f satisfies a linear growth condition if there is a constant C such that |f (x)| C(1 + |x|) for all x. In that case f is clearly locally bounded. LEMMA 3.2 (Gyöngy [8, Proposition 3.5(i)]). Assume (D) and linear growth of f and σ . Then, for every p 1,

sup sup sup E |un,m (t, x)|2p < ∞. n,m t∈[0,T ] x∈[0,1]

Let, for ti = iT /m and t ∈ [ti , ti+1 ], 1 n,m Gn,m v (ti , x) = ti (x, y)u0 (y) dy, 0

m (t − ti ) v n,m (ti+1 , x) − v n,m (ti , x) , T wn,m (t, x) = un,m (t, x) − v n,m (t, x). v

n,m

(t, x) = v

n,m

(ti , x) +

380


LEMMA 3.3 (Gyöngy [8, Proposition 3.7]). Assume (D) and linear growth of f and σ . For every p 1 there is a constant Kp such that

E |wn,m (s, x) − wn,m (t, y)|2p Kp |t − s|1/4 + |x − y|1/2 , for all s, t ∈ [0, T ], x, y ∈ [0, 1].

LEMMA 3.4 (Gyöngy [8, Proposition 3.8]). Assume (D). For every α < 14 , β < and t > 0 there exists a constant Kα,β such that 1 1 n,m + β , sup |v (t, x) − uh (t, x)| Kα,β mα n x∈[0,1]

1 2

where

∞ 1

uh = 0

e−j

2 πt

ϕj (x)ϕj (y)u0 (y) dy.

j =1

For the next lemma we let C0 ((0, T ) × (0, 1)) denote the set of all real valued continuous functions on (0, T ) × (0, 1) having compact support. LEMMA 3.5. For every k 0 let zk = {zk (t, x): t ∈ [0, T ], x ∈ [0, 1]} be a continuous Ft k -adapted random field and let W k be a Brownian sheet carried by some filtered probability space (, F , (Ft k )t0 , P ). Assume that for every > 0 and T > 0, lim P sup sup |zk − z0 |(t, x) + |W k − W 0 |(t, x) = 0. k→∞

t∈[0,T ] x∈[0,1]

Let f = f (t, x, r) be a bounded Borel function on [0, T ] × [0, 1] × R and τ k : [0, T ] → [0, T ], ρ k : [0, 1] → [0, 1] and τ k − id∞ + ρ k − id∞ → 0 where id(x) = x. Denote f (t, x, z(t, x)) by f (z)(t, x). (a) Assume f is continuous. Then, as k → ∞, T

1

0

k

k

k

k

0

and T

T

1

1

f (z0 )ψ dx dt 0

k

k

k

k

0

k

(3.6)

0

P

T

1

f (z )(τ , ρ )ψ(τ , ρ )W (dx, dt) → k

0

P

f (z )(τ , ρ )ψ(τ , ρ ) dx dt → k

f (z0 )ψW 0 (dx, dt) 0

0

(3.7) for every measurable and bounded ψ.


381

(b) Assume for every (t, x) ∈ ((0, T )×(0, 1)) the law Qkt,x of zk (t, x) is absolutely k = continuous with respect to the Lebesgue measure λ on R and the density pt,x k dQt,x /dλ satisfies, for some ρ > 1, T 1 k pt,x (r)ρ ψ(t, x) dr dx dt < ∞ (3.8) sup k

0

0

R

when ψ ∈ C0 ((0, T ) × (0, 1)). Then (3.6) and (3.7) holds. Lemma 3.5 is a modification of Lemma 4.3 in [10]. LEMMA 3.6. Assume (SL), (D), f and σ are bounded and σ 2 λ > 0 globally. n,m be the density of un,m (t, x). Then for any ψ ∈ C0∞ ((0, T ) × (0, 1)), and Let pt,x any ρ > 1, T 1 n,m ρ sup pt,x (r) ψ(t, x) dr dx dt < ∞. n,m

0

0

R

LEMMA 3.7 (Gyöngy and Nualart [10, Lemma 5.1]). Let F ∈ D2,α be a random −2β variable with finite moments of all orders and such that E(DF H ) < ∞ for some α, β > 1 with 1/α + 1/β = 1/2. Let pF be the density of F . Then for ρ > 1, α ρ/α −2β ρ/β EDF H pFρ (r) dr K 1 + E D (2) F H ⊗H R

where the constant K depends on ρ and the moments of F . We will apply Lemma 3.7 to F = un,m (t, x) for α = β = 4. LEMMA 3.8. Assume (SL), (D), p 2, f ≡ 0 and σ bounded. Then un,m (t, x) ∈ D2,p for all n, m, t, x and p (3.9) sup sup E D (k) un,m (t, x)H < ∞, k = 1, 2. n,m t,x

. For convenience let γ (x) = x ∧ (1 − x). LEMMA 3.9. Assume (SL), (D), f ≡ 0 and 0 < λ σ 2 with σ bounded. Then

Kγ −28 (x)t −4 . sup E Dun,m (t, x)−8 H n,m

LEMMA 3.10. Assume (D). If 0 t < ti and n 4, then for some positive constant KT , only dependent on T , 1 n,m Gt −t (x, y)2 dy KT (ti − t)−1/2 . (3.10) i+1 0

382


Moreover, there exist positive constants KT1 and KT2 , only dependent on T , such that, when ε ∈ [0, ti ], ti 1 n,m √ 1 2 n,m Gt −s (x, t)2 dy ds K 2 ε, KT γ (x)εC (ε) (3.11) T i+1 ti −ε 0

where

    n − 1,

T , m C n,m (ε) = T 1    (n − 1) ∧ √ , ε . m ε √ Note that as (n, m) → ∞, C n,m (ε) → 1/ ε. We only use the weaker statement that, for ε ∈ [0, ti ], ti 1 √ 2 |Gn,m K1 ε ti+1 −s (x, y)| dy ds K2 ε. ε
0,

−4 E Dun,m (ti , x)2H ∞ 1 3 n,m 2 =4 dy y P Du (ti , x)H < y 0 ∞ 1 4 3 n,m 2 dy. (3.12) y P Du (ti , x)H < a +4 y a We can write n,m Ds,y un,m (ti , x) = Ss,y (ti , x)σ (s, y, un,m (s, y)),

0 s ti ,

n,m (ti , x) is the solution (unique by a Picard argument) to where Ss,y ti 1 n,m n,m Ss,y (ti , x) = Gti+1 −s (x, y) + Gn,m ti+1 −θ (x, η) s¯ ∧ti 0

∂σ n,m (θ , η, un,m (θ, η)) · Ss,y (θ , η)W (dη, dθ ). × ∂r

(3.13)

383


By σ 2 λ and (3.13), Du

n,m

(ti , x)2H

=

ti 0

λ

1

|Ds,y un,m (ti , x)|2 dy ds

0 ti 1

0

n,m |Ss,y (ti , x)|2 dy ds.

(3.14)

0

For any ε ti , ti

1

n,m |Ss,y (ti , x)|2 dy ds 0 0 1 n,m 1 ti Gt −s (x, y)2 dy ds i+1 4 ti −ε 0 2 ti 1 ti 1 ∂σ n,m n,m Gti+1 −θ (x, η) Ss,y (θ , η)W (dη, dθ ) dy ds − ∂r ti −ε 0

s¯ ∧ti 0

Kεγ 2 (x) − A,

(3.15)

where we used Lemma 3.10. By Chebysjev’s inequality, Burkholder’s inequality for Hilbert-valued stochastic integrals, (SL), Hölder’s inequality and Minkowski’s inequality for integrals P(A α) α −q E(Aq ) α

−q

Kq ∂r σ 2q ∞E

ti

1

ti

ti −ε 0

s¯

q

1 0

2 |Gn,m ti+1 −θ (x, η)|

n,m × |Ss,y (θ , η)|2 dη dθ dy ds ti 1 ti 1 −q q−1 2 α Kq ε E |Gn,m ti+1 −θ (x, η)| ti −ε 0 s¯ 0 q n,m 2 × |Ss,y (θ , η)| dη dθ dy ds ti 1 ti 1 2 |Gn,m α −q Kq εq−1 ti+1 −θ (x, η)| ti −ε 0 s¯ 0 q

n,m 2q 1/q × E|Ss,y (θ, η)| dη dθ dy ds q ti 1 ti 1 n,m −q q−1 2 |Gti+1 −θ (x, η)| dη dθ dy ds α Kq ε

α −q Kq εq−1

ti −ε 0 ti 1

ti −ε 0

s¯

0

(ti − s¯ )q/2 dy ds α −q Kq ε3q/2 ,

(3.16)

384


where we have used Lemmas 3.8 and 3.10. Let us estimate the probability in (3.12) using (3.14), (3.15) and (3.16) as

1 n,m 2 P Du (ti , x)H < P A > Kεγ 2 (x) − (λy)−1 y

−q Kq ε3q/2 Kεγ 2 (x) − (λy)−1 . Plugging this into (3.12) results in, for any a > 0, ∞

−q −8 n,m 4 y 3 ε3q/2 Kεγ 2 (x) − (λy)−1 dy. E Du (ti , x)H a + Kq a

For fixed y, choose 2 ε= 2 Kγ (x)λy

and

a=

2 Kγ 2 (x)λti

.

With these definitions y a ⇒ ε ti and the integral is bounded by ∞ −3q −q/2 y 3 y −3q/2 y q dy, Kq γ (x)λ a

where the integral is convergent if, say, q = 10. It is easy to see that, then, the righthand side is bounded by Kλ−4 γ −28 (x)ti−4 . Writing D(s,y) u(t, x) = θ D(s,y) u(ti , x) +(1 − θ )D(s,y) u(ti+1 , x) we arrive at

n,m D(s,y) u(t, x) = θ 1st Gn,m t¯−s (x, y) + (1 − θ )1st¯Gt¯¯−s (x, y) σ t 1 ∂σ Gt¯−τ (x, y) D(s,y) u(τ , x)W (dy, dτ ) + ∂r s¯ ∧t 0 t¯ 1 ∂σ Gt¯¯−τ (x, y) D(s,y) u(τ , x)W (dy, dτ ). + ∂r s¯ ∧t¯ 0 Proceeding now as in the proof of the grid point case the result follows. 2 Proof of Lemma 3.6. First assume f ≡ 0. By Lemmas 3.7 and 3.8,

ρ/4 ρ pt,x (r) dr K1 1 + (ED (2) un,m 4H )ρ/4 EDun,m −8 H R

ρ/4 K2 γ (x)−12 (1 + t −1 ) Kγ (x)−3ρ t −ρ/4

which yields the result. The general case with a bounded Borel function f follows by a Girsanov argument. Let T 1 f n,m (s, x, un,m (s, x))W (ds, dx) = exp − L σ 0 0 1 T 1 f 2 (s, x, un,m (s, x)) ds dx . − 2 0 0 σ

385


Introduce the probability measure Qn,m by dQn,m /dP = Ln,m . Then W n,m defined by f n,m (s, y, un,m (s, y)) dx ds + W (A), W (A) = A σ for all Borel sets A ∈ [0, T ] × [0, 1] is a Brownian sheet under Qn,m . Furthermore, n,m un,m (t, x) solves, under Qn,m , (3.3) with f ≡ 0 and W replaced by W n,m . Let pt,x n,m n,m be the density of ut,x under the original measure P and qt,x the density under Qn,m . Then, for N > 0, 1/α + 1/α = 1 and 1/β + 1/β = 1, n,m n,m pt,x (r)ρ 1{pt,x (r)